Swing the arc

The idea

Mark OA = 1 unit on the number line and AB = 1 unit perpendicular to it at A, so OB = √2; an arc with centre O and radius OB cuts the number line at P, which represents √2.

Where is √2?

We know a square of side 1 has a diagonal of length √2. Can we place √2 on the number line using only a ruler and a compass?

Between which two integers will √2 land?

• 0 and 1
• 1 and 2
• 2 and 3

Four moves

Press the buttons in order to build the length √2 and mark it on the line.

P is √2

Mark OA = 1 unit on the number line and draw AB = 1 unit perpendicular to it at A, so OB = √2. An arc with centre O and radius OB cuts the number line at P, which represents √2.

OB is the diagonal of a unit square: OB² = 1² + 1² = 2.

Notes

Mark OA = 1 and draw AB = 1 perpendicular to it, so OB = √2. The arc with centre O and radius OB meets the number line at P, which represents √2.

Check yourself

Which length is equal to √2?

In the right triangle OAB, OA = 1 and AB = 1. What is OB²?

Answer: 2

OB² = 1² + 1² = 2, so OB = √2.

The arc has centre O and radius OB. Where does it meet the number line?

Between which two integers does the point P lie?

• OA. OA = 1 unit.
• OB — correct. Yes. OB is the hypotenuse of a right triangle with both legs 1.
• AB. AB = 1 unit.
• At A, with OA = 1. The arc is √2 units from O, and A is only 1 unit from O.
• At the point 2. The radius is √2, which is less than 2.
• At P, with OP = OB = √2 — correct. Yes. Every point of the arc is √2 units from O.
• 1 and 2 — correct. Yes. √2 is more than 1 (since 1² = 1 < 2) and less than 2 (since 2² = 4 > 2).
• 0 and 1. P is further from O than A, and OA = 1.
• 2 and 3. OP is shorter than 2: the arc ends before the point 2.