Operations With Integers

Hello students, welcome to today's mathematics lesson! I'm so happy to see you all here, ready to learn something new and exciting. Today, we are going to study Chapter 2, which is all about Operations with Integers. Now, you might have already learned about integers in your previous classes, but in this chapter, we are going to go much deeper and understand how to add, subtract, multiply, and even divide integers. Are you ready? Let's begin!

So students, let's start with a quick recap of what integers are. You remember that integers include all positive numbers, all negative numbers, and zero, right? So integers are like ... -5, -4, -3, -2, -1, 0, 1, 2, 3, 4, 5, and so on. The positive integers are greater than zero, the negative integers are less than zero, and zero is neither positive nor negative. Now, let's warm up with a fun puzzle that will help us understand integers better.

So students, let's look at Rakesh's Puzzle. Rakesh gives you a challenge. He says, "I have thought of two numbers. Their sum is 25, and their difference is 11." Can you tell me the two numbers? Now, you don't need to use any formulas or complicated algebra. Just try different pairs of numbers and check if they satisfy both conditions. Let's make a table and try some guesses.

If we try first number as 10 and second number as 15, then the sum is 10 + 15 = 25, which is correct. But the difference is 10 - 15 = -5, which is not 11. So that's not right. Let's try 20 and 5. Sum is 20 + 5 = 25, that's correct. Difference is 20 - 5 = 15, but we need 11. Not quite. Let's try 19 and 6. Sum is 19 + 6 = 25, correct. Difference is 19 - 6 = 13, still not 11. Let's try 18 and 7. Sum is 18 + 7 = 25, correct. Difference is 18 - 7 = 11. That's perfect! So the two numbers are 18 and 7. Great job if you found it!

Now, here's the interesting part. Rakesh gives you a second challenge. He says, "Think of two numbers whose sum is 25, but their difference is -11." Now, what does this mean? The difference being -11 means that if we subtract the second number from the first number, we get -11. Let's use the same method and try different pairs. You will notice that if you swap the numbers from the first puzzle, you get the answer to the second puzzle. That is, the first number is 7 and the second is 18! Because 7 + 18 = 25, and 7 - 18 = -11. See how the negative sign in the difference just tells us which number is bigger? When the difference is positive, the first number is bigger. When the difference is negative, the second number is bigger. This is a very important understanding, students. Always remember that when we say "difference", we mean first number minus second number.

Now, let's try a few more examples together to practice. For each pair, we need to find two numbers whose sum and difference are given.

For part (a), sum is 27 and difference is 9. Let the first number be x and second number be y. We have x + y = 27 and x - y = 9. Adding these two equations, we get 2x = 36, so x = 18. Then y = 27 - 18 = 9. So the numbers are 18 and 9. Check: 18 + 9 = 27, and 18 - 9 = 9. Perfect!

For part (b), sum is 4 and difference is 12. Now, if the sum is 4 but the difference is 12, that means the first number must be much bigger than the second. Using the same method, x + y = 4 and x - y = 12. Adding, 2x = 16, so x = 8. Then y = 4 - 8 = -4. So the numbers are 8 and -4. Check: 8 + (-4) = 4, and 8 - (-4) = 8 + 4 = 12. Yes! So one of our numbers is negative. This is perfectly fine because we're working with integers, which include negative numbers.

For part (c), sum is 0 and difference is 10. So x + y = 0 and x - y = 10. Adding, 2x = 10, so x = 5. Then y = 0 - 5 = -5. So the numbers are 5 and -5. Check: 5 + (-5) = 0, and 5 - (-5) = 5 + 5 = 10. Beautiful! Notice that when the sum is 0, the two numbers are opposites of each other.

For part (d), sum is 0 and difference is -10. This is just the opposite of the previous one. So x + y = 0 and x - y = -10. Adding, 2x = -10, so x = -5. Then y = 0 - (-5) = 5. So the numbers are -5 and 5. See how swapping the order changes the sign of the difference?

For part (e), sum is -7 and difference is -1. So x + y = -7 and x - y = -1. Adding, 2x = -8, so x = -4. Then y = -7 - (-4) = -7 + 4 = -3. So the numbers are -4 and -3. Check: -4 + (-3) = -7, and -4 - (-3) = -4 + 3 = -1. Correct!

For part (f), sum is -7 and difference is -13. So x + y = -7 and x - y = -13. Adding, 2x = -20, so x = -10. Then y = -7 - (-10) = -7 + 10 = 3. So the numbers are -10 and 3. Check: -10 + 3 = -7, and -10 - 3 = -13. Perfect!

Now students, I want you to practice this with your partner. One person can think of two integers and give their sum and difference, and the other person must figure out the integers. After some practice, you can even perform this as a magic trick for your family members and surprise them! They'll be amazed that you can guess the numbers just from their sum and difference.

Now, let's move on to a very interesting example that will help us understand integers better. This is the Carrom Coin example. Imagine a carrom coin placed on a board. Let's say we strike the coin to move it to the right. Each strike moves the coin a certain number of units of distance rightwards based on the force of the strike.

To begin with, the coin is at point 0. If the coin is struck twice, with the first strike moving it by 4 units and the second strike moving it by 3 units, what will be the final position of the coin? It's clear that the coin will be 4 + 3 = 7 units from 0. Simple, right?

Now, if the coin is struck twice, and if the two movements are known, can you give a formula for the final position of the coin? If the first strike moves the coin 'a' units to the right and if the second strike moves the coin 'b' units to the right, then the final position is P = a + b, where P is the distance of the coin from the starting point 0. This is exactly like adding positive integers.

But now, here's the twist. Suppose the coin can be struck to move it in either direction — left or right. How do we handle this? One way to do this is to consider different cases of the directions of the strikes — both are rightward, both are leftward, the first one is rightward and the second one is leftward, and the first one is leftward and the second one is rightward. But this can get complicated.

An efficient way to model this situation is to use positive and negative integers. First, let us model the line on which the coin moves as a number line. You remember the number line, don't you? It has zero in the middle, positive numbers to the right, and negative numbers to the left.

Let us consider the rightward movement positive and the leftward movement negative. So, if the coin moves right by 5 units, we represent it as +5. If it moves left by 7 units, we represent it as -7.

Suppose the first strike moves the coin rightward by 5 units from 0, and the second strike leftward by 7 units. Then we take the first movement as 5 units and the second movement as -7 units. What is the final position of the coin? This can be found by simply adding the two movements: 5 + (-7) = -2. So the coin is at -2, or it is 2 units to the left of 0. Beautiful, isn't it?

In general, if the first strike moves the coin 'a' units (which is positive if the strike is to the right and negative if the strike is to the left), and the second strike 'b' units (which is positive if the strike is to the right and negative if the strike is to the left), then the final position 'P', after the two strikes, is again P = a + b.

So students, this is a very important concept. When we model movements on a number line using positive and negative integers, we can simply add the integers to find the final position. The positive integers represent movement to the right, and negative integers represent movement to the left.

Now, let's answer some questions based on this model.

Question 1: If the first movement is -4 and the final position is 5, what is the second movement? We have first movement + second movement = final position. So, -4 + second movement = 5. Therefore, second movement = 5 - (-4) = 5 + 4 = 9. So the second movement is 9 units to the right.

Question 2: If there are multiple strikes causing movements in the order 1, -2, 3, -4, ..., -10, what is the final position of the coin? We need to add all these movements: 1 + (-2) + 3 + (-4) + 5 + (-6) + 7 + (-8) + 9 + (-10). Let's calculate this step by step. 1 + (-2) = -1. Then -1 + 3 = 2. Then 2 + (-4) = -2. Then -2 + 5 = 3. Then 3 + (-6) = -3. Then -3 + 7 = 4. Then 4 + (-8) = -4. Then -4 + 9 = 5. Then 5 + (-10) = -5. So the final position is -5, which means the coin is 5 units to the left of the starting point.

So students, by modeling the movements as numbers, both positive and negative, we are able to capture two pieces of information — the distance or magnitude, and the direction, rightwards or leftwards. For example, when we say the movement is -4, the magnitude is 4 and the direction is leftward.

Now, let's recall how we used the token model to understand integers in Grade 6. We used green tokens to represent positive 1 and red tokens to represent negative 1, that is -1. Together, they make zero, as they cancel each other out. This is called a zero pair. A green token and a red token together make zero because +1 + (-1) = 0.

Now, let's use this token model to understand subtraction of integers. Let's find (+7) – (+18). To subtract 18 from 7, that is, (+7) – (+18), we need to remove 18 positives from 7 positives. But there are not enough tokens to remove 18 positives! We only have 7 positive tokens. So what do we do?

We put in enough zero pairs so that we can remove 18 positives. How many? We have 7 positives and we need 11 more to make 18. So we need to put in 11 zero pairs. Each zero pair has one green and one red token. After putting in 11 zero pairs, we have 7 + 11 = 18 positive tokens, and 11 negative tokens. Now we can remove 18 positives. What is left? There are 11 negatives, meaning -11. So, 7 – 18 = -11.

This is exactly what we get when we think of subtraction as adding the additive inverse. We had also seen that subtracting a number is the same as adding its additive inverse. The additive inverse of an integer a is represented as -a. So the additive inverse of 18 is represented as -(18) = -18, and the additive inverse of -18 is represented as -(-18) = 18.

So students, let's verify these statements using tokens.

Statement (a): 7 – 18 = 7 + (-18). We just found that 7 – 18 = -11. And 7 + (-18) is also 7 - 18 = -11. So this is true. The additive inverse of 18 is -18, so subtracting 18 is the same as adding -18.

Statement (b): 4 – (-12) = 4 + 12. Now, 4 – (-12) means we subtract -12 from 4, which is the same as removing 12 negatives from 4 positives. But we have only 4 positives, so we need to add zero pairs. Let's think about this differently. 4 – (-12) means we want to find what number we get when we subtract -12 from 4. Since -12 is the same as 12 negatives, subtracting -12 means removing 12 negatives. But we don't have any negatives to remove! So we add 12 zero pairs first. After adding 12 zero pairs, we have 4 positives and 12 negatives. Now we remove 12 negatives. What is left? There are 4 positives left, which is +4. So 4 – (-12) = 4. And 4 + 12 is also 4 + 12 = 16. Wait, that's not matching. Let me think again.

Actually, students, let's be more careful. 4 – (-12) means we start with 4 and we subtract negative 12. Subtracting negative 12 means we remove 12 negatives. But we don't have any negatives in our starting collection. So we need to add 12 zero pairs first. After adding 12 zero pairs, we have 4 positives and 12 negatives. Now we remove 12 negatives. What is left? We have 4 positives and 0 negatives, which is simply 4. So 4 – (-12) = 4. But 4 + 12 = 16. So there's something wrong here.

Wait, I think I made a mistake. Let me reconsider. When we say 4 – (-12), we can think of it as 4 + (additive inverse of -12). The additive inverse of -12 is 12. So 4 + 12 = 16. But our token method gave us 4. Where is the error?

The error is in how we interpreted the token removal. When we subtract a negative number, we are actually removing negatives, which is equivalent to adding positives. Let me redo this carefully.

We want to compute 4 – (-12). This means we start with 4 positives. We need to remove 12 negatives. But we don't have any negatives. So we add 12 zero pairs. Now we have 4 positives and 12 negatives. Now we remove 12 negatives. After removing, we have 4 positives and 0 negatives. So we have 4 positives, which is +4. But this doesn't match 4 + 12 = 16.

Hmm, there's still something wrong. Let me think about this from a different angle. Actually, students, the token model has some limitations when it comes to subtracting negative numbers. Let's think about it algebraically instead.

We know that subtraction is the inverse of addition. So if a + b = c, then c - a = b. Similarly, for 4 – (-12), we want to find a number x such that 4 = x + (-12). If x + (-12) = 4, then x = 4 + 12 = 16. So 4 – (-12) = 16. And indeed, 4 + 12 = 16. So the statement 4 – (-12) = 4 + 12 is correct.

But why did the token method give us 4? The issue is that when we remove negatives, we're actually removing the red tokens, which should increase our value. But in my token method, I think I made an error in the interpretation. Let me think about it once more.

Actually, when we have 4 positives and we add 12 zero pairs, we have 4 positives and 12 negatives. When we remove 12 negatives, we're left with just 4 positives. But wait, that's not right. When we add 12 zero pairs, we're adding 12 positives and 12 negatives. So we have 4 + 12 = 16 positives and 12 negatives. Then when we remove 12 negatives, we're left with 16 positives. So the answer is 16! That makes sense now. I apologize for the confusion earlier.

So students, the key insight is that when we add zero pairs, we add both positives and negatives. So after adding 12 zero pairs to our original 4 positives, we have 16 positives and 12 negatives. Then when we remove 12 negatives, we're left with 16 positives. So 4 – (-12) = 16, which is indeed equal to 4 + 12.

So we have verified that subtracting a number is the same as adding its additive inverse. This is a very important property of integers.

Now students, let's recap what we've learned so far. We learned about finding two numbers when their sum and difference are given. We learned about modeling movements on a number line using positive and negative integers. And we learned about the token model for addition and subtraction of integers, and how subtraction is the same as adding the additive inverse. Make sure you understand these concepts before we move on to multiplication.

Now, let's move on to the next section, which is Multiplication of Integers. This is where things get really interesting!

We used the token model to represent addition and subtraction of integers. We now explore how to model multiplication of integers using tokens.

Suppose we put some positive tokens into an empty bag as shown in the figure. How many positives are in the bag now? There are 8 positives in the bag. We can see this as adding 2 positives to the bag 4 times. Thus, the operation is, 4 × 2 = 8. We have seen this kind of multiplication of positive integers before. It's just like repeated addition. 4 times 2 means add 2 four times, which gives 8.

Now, can we use tokens to give meaning to multiplications like 4 × (-2)? This is a new kind of multiplication where one of the numbers is negative. Let's see how.

For every new operation, we start with an empty bag. 4 × (-2) can be interpreted as placing 2 negatives into an empty bag 4 times. We use red tokens for negatives, so we place 2 negatives into an empty bag 4 times. There are now 8 red tokens or 8 negatives in the bag, meaning -8. So, 4 × (-2) = -8.

Similarly, we can find the values of 4 × (-6) and 9 × (-7). For 4 × (-6), we place 6 negatives into the bag 4 times, which gives 24 negatives, or -24. For 9 × (-7), we place 7 negatives into the bag 9 times, which gives 63 negatives, or -63.

Now, how do we interpret (-4) × 2? When the multiplier is positive, we place tokens into the bag. When the multiplier is negative, we remove tokens from the bag. So, for (-4) × 2, we need to remove two positives or two green tokens from the bag 4 times.

But there are no tokens in the bag, because we start with an empty bag. Just as in the case of subtraction, to remove 2 positives from an empty bag, we need to first place 2 zero pairs inside and then remove the 2 positives. We need to do this 4 times.

So, we place 2 zero pairs, which gives us 2 positives and 2 negatives. Then we remove 2 positives. After removing, we have 0 positives and 2 negatives. We do this 4 times in total. After doing this 4 times, we have 0 positives and 8 negatives, which is -8. So, (-4) × 2 = -8.

Now, students, you might be wondering why we are removing green tokens and not red tokens. The reason is that when the multiplier is negative, we are removing tokens. And since the multiplicand is 2, which is positive, we are removing positive tokens, that is, green tokens.

Now, what happens when both the integers in the multiplication are negative? How do we model (-4) × (-2) with tokens?

For (-4) × (-2), we need to remove 2 negatives from the bag 4 times. Since there are no red tokens in the bag, we need to place 2 zero pairs and remove 2 negatives, and we need to do this 4 times.

So, we place 2 zero pairs, which gives us 2 positives and 2 negatives. Then we remove 2 negatives. After removing, we have 2 positives and 0 negatives. We do this 4 times in total. After doing this 4 times, we have 8 positives and 0 negatives, which is +8. So, (-4) × (-2) = 8.

So students, let's summarize what we've learned so far about multiplication of integers using the token model:

4 × 2 = 8 (positive times positive is positive) 4 × (-2) = -8 (positive times negative is negative) (-4) × 2 = -8 (negative times positive is negative) (-4) × (-2) = 8 (negative times negative is positive)

This is very interesting! We can see that when we multiply two integers, the sign of the product depends on the signs of the two numbers being multiplied.

Now, let's verify these results using another method. Let's look at the patterns in integer multiplication.

Consider the multiplication table for 3. We have: 4 × 3 = 12 3 × 3 = 9 2 × 3 = 6 1 × 3 = 3 0 × 3 = 0

What do you notice in this pattern? Can you describe it? We can see that, when the multiplicand is positive, for every unit decrease in the multiplier, the product decreases by the multiplicand. For example, from 4 × 3 = 12 to 3 × 3 = 9, the product decreased by 3. From 3 × 3 = 9 to 2 × 3 = 6, the product decreased by 3 again. And so on.

Now, will this pattern continue when the multiplier goes below zero and becomes a negative number? Let's see:

4 × 3 = 12 3 × 3 = 9 2 × 3 = 6 1 × 3 = 3 0 × 3 = 0 (-1) × 3 = -3 (-2) × 3 = -6 (-3) × 3 = -9 (-4) × 3 = -12

Yes indeed! The same pattern continues when the multiplier becomes a negative number. For every unit decrease in the multiplier, the product decreases by the multiplicand. From 0 × 3 = 0 to (-1) × 3 = -3, the product decreased by 3. From (-1) × 3 = -3 to (-2) × 3 = -6, the product decreased by 3 again. And so on. This pattern holds for all integers.

Now, what is the pattern when the multiplicand is a negative integer? Let's look at the multiplication table for -3:

4 × (-3) = -12 3 × (-3) = -9 2 × (-3) = -6 1 × (-3) = -3 0 × (-3) = 0 (-1) × (-3) = 3 (-2) × (-3) = 6 (-3) × (-3) = 9 (-4) × (-3) = 12

This is the inverse of the previous pattern. When the multiplicand is negative, for every unit decrease of the multiplier, the product increases by the multiplicand. For example, from 4 × (-3) = -12 to 3 × (-3) = -9, the product increased by 3 (which is the absolute value of -3). From 3 × (-3) = -9 to 2 × (-3) = -6, the product increased by 3 again. And so on.

Will this pattern continue when the multiplier goes below zero and becomes a negative integer? Yes! Even when the multiplier is negative, the same pattern is observed. When the multiplicand is negative, for every unit decrease in the multiplier, the product increases by the multiplicand.

We can see from these patterns that what is true for multiplication when the integers are positive is also true when the integers are negative. This is a very powerful observation!

Now, let's look at the complete times 3 table when the multipliers and multiplicands are positive, and when they are negative. We can see that:

- When both the multiplier and the multiplicand are positive, the product is positive. - When both the multiplier and the multiplicand are negative, the product is positive. - When one of the multiplier or the multiplicand is positive and the other is negative, their product is negative.

So students, here's the rule for multiplying two integers:

If both integers have the same sign (both positive or both negative), their product is positive. If the two integers have different signs (one positive and one negative), their product is negative.

This is exactly what we observed in our token model as well!

Now, let's look at some examples to practice this rule.

Example (a): 4 × (-3). The signs are different (positive times negative), so the product is negative. The magnitude is 4 × 3 = 12. So the product is -12.

Example (b): (-6) × (-3). The signs are the same (both negative), so the product is positive. The magnitude is 6 × 3 = 18. So the product is 18.

Example (c): (-5) × (-1). Both negative, so product is positive. Magnitude is 5 × 1 = 5. So the product is 5.

Example (d): (-8) × 4. Different signs, so product is negative. Magnitude is 8 × 4 = 32. So the product is -32.

Example (e): (-9) × 10. Different signs, so product is negative. Magnitude is 9 × 10 = 90. So the product is -90.

Example (f): 10 × (-17). Different signs, so product is negative. Magnitude is 10 × 17 = 170. So the product is -170.

Now, let's consider the expression 1 × a. We know that the value of this expression is 'a' for all positive integers. Is this true for all negative integers too?

Using the token model, we put 'a' negatives into the bag just once. After this, the bag contains 'a' negatives. For example, if 'a' is -5 (5 negatives), then the bag contains 5 negatives, i.e., -5. So, 1 × a = a (for all integers a, both positive and negative).

What is the value of the expression -1 × a? When 'a' is positive, then from our observations on integer multiplication, the product has the same magnitude as 'a' but is negative. When 'a' is negative, then the product has the same magnitude as 'a' but is positive. In each case, we notice that the product is the additive inverse of the multiplicand 'a'. Thus, -1 × a = -a (for all integers a).

Now, students, let's verify the commutative property of multiplication for integers. In the case of integers, is the product the same when we swap the multiplier and the multiplicand? Let's try this for some numbers.

Observe the following pairs of multiplications: 3 × -4 = -12 and -4 × 3 = -12. The product is the same! -30 × 12 = -360 and 12 × -30 = -360. The product is the same! -15 × -8 = 120 and -8 × -15 = 120. The product is the same! 14 × -5 = -70 and -5 × 14 = -70. The product is the same!

What do you notice in these pairs of multiplication statements? The product is the same when we 'swap' the multiplier and multiplicand. Earlier, we have seen a similar property with addition. Will this always happen?

The magnitude of the product does not change when the multiplier and the multiplicand are swapped. This is because the magnitude of the product depends only on the magnitudes of the multiplier and the multiplicand, and we know that the product of two positive integers does not change when the numbers being multiplied are swapped.

Does the sign of the product change if we swap the multiplier and multiplicand? If both are positive or both are negative, the product is positive before and after swapping. So the sign does not change in this case. If one is positive and the other negative, the product is negative before and after swapping. So the sign does not change in this case either.

Hence, the product does not change when the multiplier and multiplicand are swapped, whatever their signs may be. Thus, multiplication is commutative for integers. In general, for any two integers, a and b, we can say that a × b = b × a.

Now, students, let me tell you about an interesting historical fact. Just like for addition and subtraction of integers, an ancient Indian mathematician named Brahmagupta also articulated explicit rules for integer multiplication and division. He did this in his book called Brāhmasphuṭasiddhānta, written in 628 CE. He used the notions of fortune or dhana for positive values and debt or riṇa for negative values. This was the first time that rules for multiplication and division of positive and negative numbers were articulated, and was an important step in the development of arithmetic and algebra!

Brahmagupta's rules were: "The product or quotient of two fortunes is a fortune." This means positive times positive is positive, and positive divided by positive is positive. "The product or quotient of two debts is a fortune." This means negative times negative is positive, and negative divided by negative is positive. "The product or quotient of a debt and a fortune is a debt." This means negative times positive is negative, and negative divided by positive is negative. "The product or quotient of a fortune and a debt is a debt." This means positive times negative is negative, and positive divided by negative is negative.

Aren't these the same rules we just learned? It's amazing that these rules were discovered over 1400 years ago!

Now, let's look at some real-life examples to understand how these operations are used.

Example 1: An exam has 50 multiple choice questions. 5 marks are given for every correct answer and 2 negative marks for every wrong answer. What are Mala's total marks if she had 30 correct answers and 20 wrong answers?

We use positive and negative integers. The mark for each correct answer is a positive integer 5 and for each wrong answer is a negative integer -2.

Marks for 30 correct answers = 30 × 5 = 150. Marks for 20 wrong answers = 20 × (-2) = -40.

Thus the arithmetic expression for 30 correct answers and 20 wrong answers is: 30 × 5 + 20 × (-2) = 150 + (-40) = 110.

Mala got 110 marks in the exam.

Now, what are the maximum possible marks in the exam? There are 50 questions, and each correct answer gives 5 marks, so maximum marks = 50 × 5 = 250. What are the minimum possible marks? If all answers are wrong, then marks = 50 × (-2) = -100. So the minimum possible marks are -100.

Example 2: There is an elevator in a mining shaft that moves above and below the ground. The elevator's positions above the ground are represented as positive integers and positions below the ground are represented as negative integers.

(a) The elevator moves 3 metres per minute. If it descends into the shaft from the ground level (0), what will be its position after one hour?

We can model this using subtraction. The elevator moves at 3 metres per minute. So in one hour it moves 180 metres (60 × 3). If it started at ground level (0 metres) and descended, we should subtract 180 from 0. 0 - 180 = -180. So, it will reach the -180 metre position, which is 180 metres below the ground.

Alternatively, let us say that the speed and direction of the elevator are represented by an integer (metres per minute). It is +3 when it is moving up and it is (-3) when moving down. Since the elevator is moving down, the speed is (-3) metres per minute. It moves for 60 minutes. So it goes 60 × (-3) = (-180). The position of the elevator after 60 minutes is 180 metres below the ground level.

(b) If it begins to descend from 15 m above the ground, what will be its position after 45 minutes?

Starting Position = 15. Distance Travelled = The elevator moves down at the speed of 3 metres per minute for 45 minutes, that is, 45 × (-3). So, Ending Position = 15 + (45 × (-3)) = 15 + (-(45×3)) = 15 + (-135) = (-120).

The elevator will be 120 metres below the ground.

Now students, let's move on to the Division of Integers. We have earlier seen how division can be converted into multiplication. For example, (-100) ÷ 25 can be reframed as, 'what should be multiplied to 25 to get (-100)?'. That is, 25 × ? = (-100). We know that 25 × (-4) = (-100). Therefore, (-100) ÷ 25 = (-4).

Similarly, (-100) ÷ (-4) can be reframed as, 'What should be multiplied to (-4) to get (-100)?' (-4) × ? = (-100). We know that (-4) × 25 = (-100). Therefore, (-100) ÷ (-4) = 25.

Similarly, we know that (-25) × (-2) = 50. Therefore, 50 ÷ (-25) = (-2).

Can you summarise the rules for integer division looking at the above pattern? In general, for any two positive integers a and b, where b ≠ 0, we can say that: a ÷ (-b) = -(a ÷ b) (-a) ÷ b = -(a ÷ b) (-a) ÷ (-b) = a ÷ b

So students, the rule for division is exactly the same as the rule for multiplication: - If both the dividend and divisor have the same sign (both positive or both negative), the quotient is positive. - If the dividend and divisor have different signs (one positive and one negative), the quotient is negative.

Now, let's practice some division problems.

Example (a): 14 × (-15). This is multiplication, not division. Wait, let me check the question. Actually, this is from the "Figure it Out" section. Let me find the values:

(a) 14 × (-15). Different signs, so product is negative. Magnitude is 14 × 15 = 210. So the answer is -210.

(b) -16 × (-5). Same signs (both negative), so product is positive. Magnitude is 16 × 5 = 80. So the answer is 80.

(c) 36 ÷ (-18). Different signs, so quotient is negative. Magnitude is 36 ÷ 18 = 2. So the answer is -2.

(d) (-46) ÷ (-23). Same signs (both negative), so quotient is positive. Magnitude is 46 ÷ 23 = 2. So the answer is 2.

Now, let's look at some more real-life problems.

Problem 2: A freezing process requires that the room temperature be lowered from 32°C at the rate of 5°C every hour. What will be the room temperature 10 hours after the process begins?

The temperature is decreasing at 5°C per hour. So after 10 hours, the temperature will decrease by 10 × 5 = 50°C. Starting from 32°C, the final temperature is 32 - 50 = -18°C. So the room temperature will be -18°C after 10 hours.

Problem 3: A cement company earns a profit of ₹8 per bag of white cement sold and a loss of ₹5 per bag of grey cement sold. Represent the profit/loss as integers.

(a) The company sells 3,000 bags of white cement and 5,000 bags of grey cement in a month. What is its profit or loss?

Profit from white cement = 3,000 × 8 = ₹24,000. Loss from grey cement = 5,000 × (-5) = -₹25,000. Total profit/loss = 24,000 + (-25,000) = -₹1,000. So the company has a loss of ₹1,000.

(b) If the number of bags of grey cement sold is 6,400 bags, what is the number of bags of white cement the company must sell to have neither profit nor loss?

Let the number of bags of white cement be x. Profit from white cement = x × 8 = 8x. Loss from grey cement = 6,400 × (-5) = -32,000. For neither profit nor loss, we need 8x + (-32,000) = 0. So 8x = 32,000, which gives x = 4,000. So the company must sell 4,000 bags of white cement to have neither profit nor loss.

Now, let's look at some expressions using integers. What is the value of the expression 5 × -3 × 4? Does it matter whether we multiply 5 × -3 and then multiply the product with 4, or if we multiply -3 × 4 first and then multiply the product with 5?

Let's check: (5 × -3) × 4 = -15 × 4 = -60. 5 × (-3 × 4) = 5 × -12 = -60.

The product is the same! Let's take a few more examples of multiplication of 3 integers and check this property. What do we observe?

We can see that the product is the same when we 'group' the multiplications in these two ways. So, integer multiplication is associative, just like integer addition. In general, for any three integers a, b, and c, a × (b × c) = (a × b) × c.

In the expression 5 × -3 × 4, try to multiply 5 and 4 first and then multiply the product with -3: (5 × 4) × -3. 5 × 4 = 20, and 20 × -3 = -60. This also gives the same product.

Are there orders in which 5 × -3 × 4 can be evaluated? Will the product be the same in all these cases? Yes! Multiply the expression 25 × -6 × 12 in all the different orders and check if the product is the same in all cases. The product remains the same when 3 or more numbers are multiplied in any order. This is the associative property of multiplication for integers.

Now, let's look at the distributive property. Consider the expression 5 × (4 + (-2)). As in the case of positive integers, is this expression equal to 5 × 4 + 5 × (-2)?

Let's check: 5 × (4 + (-2)) = 5 × 2 = 10. 5 × 4 + 5 × (-2) = 20 + (-10) = 10. Yes, it is equal! We call this the distributive property.

Check if the distributive property holds for (-2) × (4 + (-3)) (that is, if this expression equals (-2) × 4 + (-2) × (-3)), and for a few other such expressions of your choice. What do we observe? We see that the distributive property seems to hold for integers, as well. Will this always happen? Yes, it will!

In the case of positive integers, we used a rectangular arrangement of objects to visually understand why the distributive property holds. We can use the same setup even in the case of integers by using green tokens for positive numbers and red tokens for negative numbers. For example, consider the following rectangular arrangement of tokens. We see that the overall arrangement represents 4 × (2 + (-3)), and it is clear that this also equals the sum of 4 × 2 and 4 × (-3).

Can you visually show the distributive property for an expression like -4 × (2 + (-3))? Using the fact that multiplying a number by -4 is adding the inverse of the number 4 times, we can show that this also works.

Thus, for any integers a, b, and c, we have a × (b + c) = (a × b) + (a × c).

Now students, let's look at some pattern machines. Two pattern machines are given below. Each machine takes 3 numbers, does some operations and gives out the result.

Find the operations being done by Machine 1. The operation done by Machine 1 is (first number) + (second number) – (third number). Written as an expression, this will be a + b – c, where a is the first number, b is the second number, and c is the third number.

For example, 5 + 8 – 3 = 10, and (-4) + (-1) – (-6) = -4 - 1 + 6 = 1.

So, the result of the last group will be, (-10) + (-12) – (-9) = -10 - 12 + 9 = -22 + 9 = -13.

Find the operations being done by Machine 2 and fill in the blank. Make your own machine and challenge your peers in finding its operations.

Now, let's practice some more problems from the "Figure it Out" section.

1. Find the values of the following expressions: (a) (-5) × (18 + (-3)) = (-5) × (15) = -75. (b) (-7) × 4 × (-1). First, (-7) × 4 = -28. Then -28 × (-1) = 28. So the answer is 28. (c) (-2) × (-1) × (-5) × (-3). Let's multiply step by step. (-2) × (-1) = 2. Then 2 × (-5) = -10. Then -10 × (-3) = 30. So the answer is 30.

2. Find the values of the following expressions: (a) (-27) ÷ 9 = -3. (b) 84 ÷ (-4) = -21. (c) (-56) ÷ (-2) = 28.

3. Find the integer whose product with (-1) is: (a) 27. We need a × (-1) = 27, so a = -27. (b) -31. We need a × (-1) = -31, so a = 31. (c) -1. We need a × (-1) = -1, so a = 1. (d) 1. We need a × (-1) = 1, so a = -1. (e) 0. We need a × (-1) = 0, so a = 0.

4. If 47 – 56 + 14 – 8 + 2 – 8 + 5 = -4, then find the value of -47 + 56 – 14 + 8 – 2 + 8 – 5 without calculating the full expression.

Notice that the second expression is just the negative of the first expression. If we take the negative of the first expression, we get -(47 - 56 + 14 - 8 + 2 - 8 + 5) = -47 + 56 - 14 + 8 - 2 + 8 - 5. So the answer is -(-4) = 4.

5. Do you remember the Collatz Conjecture from last year? Try a modified version with integers. The rule is — start with any number; if the number is even, take half of it; if the number is odd, multiply it by -3 and add 1; repeat. An example sequence is shown below.

-7 → 22 → 11 → 32 → -16 → -8 → -4 → -2 → -1 → 4 → 2 → 1

Try this with different starting numbers: (-21), (-6), and so on. Describe the patterns you observe.

6. In a test, (+4) marks are given for every correct answer and (-2) marks are given for every incorrect answer. (a) Anita answered all the questions in the test. She scored 40 marks even though 15 of her answers were correct. How many of her answers were incorrect? How many questions are in the test?

Let the number of incorrect answers be x. Total questions = 15 + x. Marks from correct answers = 15 × 4 = 60. Marks from incorrect answers = x × (-2) = -2x. Total marks = 60 + (-2x) = 60 - 2x. We are given that this equals 40. So 60 - 2x = 40, which gives 2x = 20, so x = 10. So Anita had 10 incorrect answers. Total questions = 15 + 10 = 25.

(b) Anil scored (-10) marks even though he had 5 correct answers. How many of his answers were incorrect? Did he leave any questions unanswered?

Let the number of incorrect answers be y. Marks from correct answers = 5 × 4 = 20. Marks from incorrect answers = y × (-2) = -2y. Total marks = 20 + (-2y) = 20 - 2y. We are given that this equals -10. So 20 - 2y = -10, which gives 2y = 30, so y = 15. So Anil had 15 incorrect answers. Total questions answered = 5 + 15 = 20. Since the problem doesn't specify the total number of questions, we cannot say if he left any questions unanswered. But if we assume he answered all questions, then there were 20 questions in the test.

8. Imagine you're in a place where the temperature drops by 5°C each hour. If the temperature is currently at 8°C, write an expression which denotes the temperature after 4 hours.

The temperature drops by 5°C per hour, so the change in temperature after 4 hours is 4 × (-5) = -20°C. The final temperature = 8 + (-20) = 8 - 20 = -12°C. So the expression is 8 + 4 × (-5) = -12.

9. Find 3 consecutive numbers with a product of (a) -6, (b) 120.

(a) Let the three consecutive integers be n, n+1, n+2. Their product is n(n+1)(n+2) = -6. We need to find three consecutive integers whose product is -6. Let's try some values. If n = -3, then (-3)(-2)(-1) = -6. Yes! So the numbers are -3, -2, -1. (b) n(n+1)(n+2) = 120. Let's try n = 4: 4 × 5 × 6 = 120. Yes! So the numbers are 4, 5, 6.

10. An alien society uses a peculiar currency called 'pibs' with just two denominations of coins — a +13 pibs coin and a –9 pibs coin. You have several of these coins. Is it possible to purchase an item that costs +85 pibs?

Yes, we can use 10 coins of +13 pibs and 5 coins of -9 pibs to make a total of +85. Because 10 × 13 + 5 × (-9) = 130 - 45 = 85.

Using the two denominations, try to get the following totals: (a) +20. We need 13a - 9b = 20 for some integers a and b. One solution is a = 2, b = 1: 2 × 13 - 1 × 9 = 26 - 9 = 17, not 20. Try a = 5, b = 3: 5 × 13 - 3 × 9 = 65 - 27 = 38. Try a = 8, b = 6: 8 × 13 - 6 × 9 = 104 - 54 = 50. This is getting complicated. Let's use a systematic approach. We want 13a - 9b = 20. Modulo 9, this gives 13a ≡ 20 (mod 9), which is 4a ≡ 2 (mod 9). Multiplying by the inverse of 4 mod 9, which is 7, we get a ≡ 14 (mod 9), so a ≡ 5 (mod 9). So a = 5, 14, 23, ... For a = 5, 13 × 5 = 65, so we need -9b = 20 - 65 = -45, so b = 5. So one solution is a = 5, b = 5, giving 5 × 13 + 5 × (-9) = 65 - 45 = 20. Yes!

(b) +40. 13a - 9b = 40. Using the same method, 4a ≡ 4 (mod 9), so a ≡ 1 (mod 9). So a = 1, 10, 19, ... For a = 1, 13 × 1 = 13, so we need -9b = 40 - 13 = 27, so b = -3, which is not valid (b should be non-negative). For a = 10, 13 × 10 = 130, so we need -9b = 40 - 130 = -90, so b = 10. So a = 10, b = 10 works: 10 × 13 + 10 × (-9) = 130 - 90 = 40.

(c) -50. 13a - 9b = -50. This gives 13a = 9b - 50. One solution is a = 1, b = 9: 13 - 81 = -68, not -50. a = 4, b = 10: 52 - 90 = -38. a = 7, b = 13: 91 - 117 = -26. a = 10, b = 16: 130 - 144 = -14. a = 13, b = 19: 169 - 171 = -2. a = 16, b = 22: 208 - 198 = 10. a = 19, b = 25: 247 - 225 = 22. a = 22, b = 28: 286 - 252 = 34. a = 25, b = 31: 325 - 279 = 46. a = 28, b = 34: 364 - 306 = 58. This is getting complicated. Let's try a different approach. We want 13a - 9b = -50, or 9b - 13a = 50. Modulo 13, this gives 9b ≡ 50 (mod 13), which is 9b ≡ 11 (mod 13). The inverse of 9 mod 13 is 3, so b ≡ 33 (mod 13), so b ≡ 7 (mod 13). So b = 7, 20, 33, ... For b = 7, 9 × 7 = 63, so 13a = 63 - 50 = 13, so a = 1. So a = 1, b = 7 works: 1 × 13 + 7 × (-9) = 13 - 63 = -50. Yes!

(d) +8. 13a - 9b = 8. Modulo 9, 4a ≡ 8 (mod 9), so a ≡ 2 (mod 9). So a = 2, 11, 20, ... For a = 2, 13 × 2 = 26, so -9b = 8 - 26 = -18, so b = 2. So a = 2, b = 2 works: 2 × 13 + 2 × (-9) = 26 - 18 = 8.

(e) +10. 13a - 9b = 10. Modulo 9, 4a ≡ 10 (mod 9), which is 4a ≡ 1 (mod 9). So a ≡ 7 (mod 9). So a = 7, 16, 25, ... For a = 7, 13 × 7 = 91, so -9b = 10 - 91 = -81, so b = 9. So a = 7, b = 9 works: 7 × 13 + 9 × (-9) = 91 - 81 = 10.

(f) -2. 13a - 9b = -2. Modulo 9, 4a ≡ -2 (mod 9), which is 4a ≡ 7 (mod 9). So a ≡ 7 × 7 (mod 9) = 49 (mod 9) = 4 (mod 9). So a = 4, 13, 22, ... For a = 4, 13 × 4 = 52, so -9b = -2 - 52 = -54, so b = 6. So a = 4, b = 6 works: 4 × 13 + 6 × (-9) = 52 - 54 = -2.

(g) +1. 13a - 9b = 1. Modulo 9, 4a ≡ 1 (mod 9). The inverse of 4 mod 9 is 7, so a ≡ 7 (mod 9). So a = 7, 16, 25, ... For a = 7, 13 × 7 = 91, so -9b = 1 - 91 = -90, so b = 10. So a = 7, b = 10 works: 7 × 13 + 10 × (-9) = 91 - 90 = 1.

(h) Is it possible to purchase an item that costs 1568 pibs? We need 13a - 9b = 1568. This is a linear Diophantine equation. Since 13 and 9 are coprime (their greatest common divisor is 1), there will be infinitely many solutions. So yes, it is possible!

11. Find the values of: (a) (32 × (-18)) ÷ ((-36)). First, 32 × (-18) = -576. Then -576 ÷ (-36) = 16. So the answer is 16. (b) (32) ÷ ((-36) × (-18)). First, (-36) × (-18) = 648. Then 32 ÷ 648 = 32/648 = simplify by dividing by 4: 8/162 = simplify by dividing by 2: 4/81. So the answer is 4/81. (c) (25 × (-12)) ÷ ((45) × (-27)). First, 25 × (-12) = -300. Then 45 × (-27) = -1215. Then -300 ÷ (-1215) = 300/1215 = simplify by dividing by 15: 20/81. So the answer is 20/81. (d) (280 × (-7)) ÷ ((-8) × (-35)). First, 280 × (-7) = -1960. Then (-8) × (-35) = 280. Then -1960 ÷ 280 = -7. So the answer is -7.

12. Arrange the expressions given below in increasing order. (a) (-348) + (-1064) = -1412. (b) (-348) - (-1064) = -348 + 1064 = 716. (c) 348 - (-1064) = 348 + 1064 = 1412. (d) (-348) × (-1064) = positive (both negative), magnitude = 348 × 1064. Let's calculate: 348 × 1000 = 348,000, 348 × 64 = 22,272, so total = 370,272. (e) 348 × (-1064) = negative (positive times negative), magnitude = 348 × 1064 = 370,272, so the answer is -370,272. (f) 348 × 964 = 348 × 964. Let's calculate: 348 × 1000 = 348,000, 348 × 36 = 12,528, so total = 360,528.

Now, let's arrange these in increasing order (from smallest to largest): The smallest is (e) -370,272. Next is (a) -1412. Next is (b) 716. Next is (c) 1412. Next is (f) 360,528. The largest is (d) 370,272.

So the order is: (e), (a), (b), (c), (f), (d).

13. Given that (-548) × 972 = -532656, write the values of: (a) (-547) × 972. This is (-548 + 1) × 972 = (-548 × 972) + (1 × 972) = -532656 + 972 = -531684. (b) (-548) × 971. This is (-548) × (972 - 1) = (-548 × 972) - (-548 × 1) = -532656 + 548 = -532108. (c) (-547) × 971. This is (-547) × 971 = [(-548 + 1) × 971] = (-548 × 971) + (1 × 971) = -532108 + 971 = -531137. Alternatively, we can think of it as (-547) × 971 = [(-548) × 971] + [1 × 971] = -532108 + 971 = -531137.

14. Given that 207 × (-33 + 7) = -5382, write the value of -207 × (33 - 7) = ______.

First, -33 + 7 = -26. So 207 × (-26) = -5382. Now, 33 - 7 = 26. So -207 × 26 = -(207 × 26) = -5382. So the answer is -5382.

15. Use the numbers 3, -2, 5, -6 exactly once and the operations '+', '–', and '×' exactly once and brackets as necessary to write an expression such that— (a) the result is the maximum possible (b) the result is the minimum possible

We need to use each number 3, -2, 5, -6 exactly once, and each operation +, -, × exactly once. We can use brackets as needed.

For maximum possible result, we want to make the product as large as possible and add positive numbers, subtract negative numbers (which is like adding). So we want to multiply two large positive numbers and add the other two. The largest numbers are 5 and 3. So let's try (5 × 3) + (-2) - (-6) = 15 + (-2) + 6 = 19. Or (5 × 3) + (-6) - (-2) = 15 + (-6) + 2 = 11. Or (5 × (-2)) + 3 - (-6) = -10 + 3 + 6 = -1. Or (5 × (-6)) + 3 - (-2) = -30 + 3 + 2 = -25. Or (3 × (-2)) + 5 - (-6) = -6 + 5 + 6 = 5. Or (3 × (-6)) + 5 - (-2) = -18 + 5 + 2 = -11.

What about (5 + 3) × (-2) - (-6) = 8 × (-2) + 6 = -16 + 6 = -10. Or (5 + (-2)) × 3 - (-6) = 3 × 3 + 6 = 9 + 6 = 15. Or (5 + (-6)) × 3 - (-2) = (-1) × 3 + 2 = -3 + 2 = -1. Or (3 + (-2)) × 5 - (-6) = 1 × 5 + 6 = 11. Or (3 + (-6)) × 5 - (-2) = (-3) × 5 + 2 = -15 + 2 = -13. Or (3 + (-6)) × (-2) - 5 = (-3) × (-2) - 5 = 6 - 5 = 1.

What about 5 × (3 + (-2)) - (-6) = 5 × 1 + 6 = 5 + 6 = 11. Or 5 × (3 + (-6)) - (-2) = 5 × (-3) + 2 = -15 + 2 = -13. Or 3 × (5 + (-2)) - (-6) = 3 × 3 + 6 = 9 + 6 = 15. Or 3 × (5 + (-6)) - (-2) = 3 × (-1) + 2 = -3 + 2 = -1. Or (-2) × (5 + 3) - (-6) = -2 × 8 + 6 = -16 + 6 = -10. Or (-6) × (5 + 3) - (-2) = -6 × 8 + 2 = -48 + 2 = -46.

What about 5 × 3 + (-2) - (-6) = 15 - 2 + 6 = 19. This seems to be the maximum so far. Can we do better? What about 5 × 3 - (-2) + (-6) = 15 + 2 - 6 = 11. Or 5 × 3 - (-6) + (-2) = 15 + 6 - 2 = 19. Or (5 + (-2)) × (3 + (-6)) = 3 × (-3) = -9. Or (5 + (-6)) × (3 + (-2)) = (-1) × 1 = -1.

What about (5 × (-2)) + (3 - (-6)) = -10 + 9 = -1. Or (5 × (-6)) + (3 - (-2)) = -30 + 5 = -25. Or (3 × (-2)) + (5 - (-6)) = -6 + 11 = 5. Or (3 × (-6)) + (5 - (-2)) = -18 + 7 = -11.

What about 5 × (3 - (-2)) + (-6) = 5 × 5 - 6 = 25 - 6 = 19. Or 5 × (3 - (-6)) + (-2) = 5 × 9 - 2 = 45 - 2 = 43. Wait, this is 43! But we need to check if we used each operation exactly once. We used × for 5 × (3 - (-6)), and we used + for adding (-2). But we haven't used the - operation. Let me check the expression: 5 × (3 - (-6)) + (-2). This uses ×, -, and +. And it uses each number 5, 3, -6, -2 exactly once. And the result is 5 × 9 + (-2) = 45 - 2 = 43. This is much larger than 19! So the maximum possible is at least 43.

Can we do even better? What about 5 × (3 - (-6)) - (-2) = 5 × 9 + 2 = 45 + 2 = 47. This uses ×, -, and -. But we need to use +, -, and × exactly once. So this doesn't work because we used - twice and didn't use +.

What about 5 × (3 - (-2)) - (-6) = 5 × 5 + 6 = 25 + 6 = 31. This is less than 43.

What about (5 + 3) × (-2) - (-6) = 8 × (-2) + 6 = -16 + 6 = -10. No.

What about (5 + (-2)) × 3 - (-6) = 3 × 3 + 6 = 9 + 6 = 15. No.

What about 5 × 3 + (-6) - (-2) = 15 - 6 + 2 = 11. No.

What about 5 × (-2) + 3 - (-6) = -10 + 3 + 6 = -1. No.

What about 3 × 5 + (-6) - (-2) = 15 - 6 + 2 = 11. No.

What about (5 - (-2)) × 3 + (-6) = 7 × 3 - 6 = 21 - 6 = 15. No.

What about (5 - (-6)) × 3 + (-2) = 11 × 3 - 2 = 33 - 2 = 31. No.

What about (3 - (-2)) × 5 + (-6) = 5 × 5 - 6 = 25 - 6 = 19. No.

What about (3 - (-6)) × 5 + (-2) = 9 × 5 - 2 = 45 - 2 = 43. This is the same as before!

So the maximum seems to be 47? But wait, we can't use - twice. Let me check the problem again: "use the operations '+', '–', and '×' exactly once". So we must use each of +, -, and × exactly once. So 5 × (3 - (-6)) + (-2) uses ×, -, and + exactly once. And it gives 43. Is there any expression that gives more than 43?

What about 5 × (3 + (-2)) - (-6) = 5 × 1 + 6 = 5 + 6 = 11. No. What about 5 × (3 + (-6)) - (-2) = 5 × (-3) + 2 = -15 + 2 = -13. No. What about (5 + 3) × (-2) - (-6) = 8 × (-2) + 6 = -16 + 6 = -10. No. What about (5 + (-2)) × 3 - (-6) = 3 × 3 + 6 = 9 + 6 = 15. No. What about (5 + (-6)) × 3 - (-2) = (-1) × 3 + 2 = -3 + 2 = -1. No. What about (3 + 5) × (-2) - (-6) = 8 × (-2) + 6 = -16 + 6 = -10. No. What about (3 + (-2)) × 5 - (-6) = 1 × 5 + 6 = 5 + 6 = 11. No. What about (3 + (-6)) × 5 - (-2) = (-3) × 5 + 2 = -15 + 2 = -13. No.

What about 5 × 3 + (-2) - (-6) = 15 - 2 + 6 = 19. No. What about 5 × 3 - (-2) + (-6) = 15 + 2 - 6 = 11. No. What about 5 × 3 - (-6) + (-2) = 15 + 6 - 2 = 19. No. What about 3 × 5 + (-2) - (-6) = 15 - 2 + 6 = 19. No. What about 3 × 5 - (-2) + (-6) = 15 + 2 - 6 = 11. No. What about 3 × 5 - (-6) + (-2) = 15 + 6 - 2 = 19. No.

What about (5 × 3) + (-2) - (-6) = 15 - 2 + 6 = 19. No. What about (5 × 3) - (-2) + (-6) = 15 + 2 - 6 = 11. No. What about (5 × 3) - (-6) + (-2) = 15 + 6 - 2 = 19. No.

What about 5 × (-2) + 3 - (-6) = -10 + 3 + 6 = -1. No. What about 5 × (-2) - 3 + (-6) = -10 - 3 - 6 = -19. No. What about 5 × (-6) + 3 - (-2) = -30 + 3 + 2 = -25. No. What about 5 × (-6) - 3 + (-2) = -30 - 3 - 2 = -35. No.

What about 3 × (-2) + 5 - (-6) = -6 + 5 + 6 = 5. No. What about 3 × (-2) - 5 + (-6) = -6 - 5 - 6 = -17. No. What about 3 × (-6) + 5 - (-2) = -18 + 5 + 2 = -11. No. What about 3 × (-6) - 5 + (-2) = -18 - 5 - 2 = -25. No.

So the maximum seems to be 43, from the expression 5 × (3 - (-6)) + (-2) or (3 - (-6)) × 5 + (-2).

Now for the minimum, we want the result to be as small as possible (most negative). We want to