Irrational numbers - Answer key

Fluency

Rational ( $7 = \dfrac{7}{1}$ , an integer).
Rational ( $0.5 = \dfrac{1}{2}$ , a terminating decimal).
Rational ( $\sqrt{9} = 3$ ).
Rational (already a fraction of integers).
Rational ( $\sqrt{25} = 5$ ).
Irrational ( $7$ is not a perfect square, so $\sqrt{7}$ is non-terminating and non-repeating).
Rational ( $0.\overline{6} = \dfrac{2}{3}$ ; repeating decimals are rational).
Irrational ( $\pi$ is non-terminating and non-repeating).
Rational ( $-3 = \dfrac{-3}{1}$ ).
Rational ( $\sqrt{0.25} = 0.5 = \dfrac{1}{2}$ ).
Irrational (the digits do not terminate and do not repeat; in fact this is $\sqrt{2}$ ).

Fluency

Reasoning

Kim is wrong. $0.\overline{9} = 1$ exactly, and $1$ is rational. Infinite digits do not make a number irrational - only non-repeating infinite digits do.
Not correct. $\dfrac{22}{7}$ is a rational approximation to $\pi$ . The true $\pi$ never terminates or repeats; $\dfrac{22}{7}$ does repeat ( $3.\overline{142857}$ ) and so it cannot equal $\pi$ .
Not always irrational. Example: $\sqrt{2} \times \sqrt{2} = 2$ , which is rational. So the product of two irrationals can be rational.
Wrong. $\sqrt{49} + \sqrt{2} = 7 + \sqrt{2}$ , and adding a rational to an irrational gives an irrational.

Problem solving

Exact: $\sqrt{2}$ m. Approximate: $1.41$ m.
Exact: $30\pi$ cm. Approximate: $94$ cm.
If $r$ is rational, $r^2$ is rational; area $= \pi r^2$ is rational times irrational ( $\pi$ ), which is irrational.
Exact: $13$ cm. Method: $5^2 + 12^2 = 25 + 144 = 169 = 13^2$ . Classification: rational (a Pythagorean triple - $13$ is a whole number).

Year 8 core - answers