Number Theory Worked ExamplesIGCSE Maths 0580

Prime: exactly two distinct positive factors. $9 = 3 \times 3$ and $27 = 3^3$ are not prime; $11$ has only $1$ and $11$.

Cube: an integer cubed. $27 = 3^3$, and $64 = 4^3$.

Irrational: cannot be written as a fraction of integers. $45 = 9 \times 5$ is not a perfect square, so $\sqrt{45} = 3\sqrt{5}$ is irrational.

Both square and cube: $64 = 8^2$ and $64 = 4^3$, so $64$ satisfies both.

A number is rational if it can be written as $\frac{a}{b}$ with $a, b$ integers and $b \neq 0$.

(a) $\sqrt{36} = 6 = \frac{6}{1}$ — rational, because $36$ is a perfect square.

(b) $\frac{22}{7}$ is already a fraction of integers — rational. It approximates $\pi$ but is not $\pi$.

(c) $0.\dot{3} = \frac{1}{3}$ — rational. Every recurring decimal is rational.

(d) $\pi - \pi = 0 = \frac{0}{1}$ — rational, even though $\pi$ itself is irrational.

Let $x$ be the recurring decimal.

Two digits recur, so multiply by $10^2 = 100$ to shift by one full cycle.

Subtract the first equation from the second. The recurring tails cancel exactly.

Solve, then cancel by the HCF of $27$ and $99$, which is $9$.

Only the $6$ recurs — the $41$ does not. Handle this by creating two equations whose recurring tails line up exactly.

Multiply by $100$ to move the point past the non-recurring digits.

Multiply by $1000$ to move it one further place — one full cycle of the recurring digit.

Subtract. The identical recurring tails cancel.

Solve and cancel by the HCF of $375$ and $900$, which is $75$.

A number is a perfect square exactly when every prime index is even.

Check each index: $2^2$ is already even. $3^3$ is odd, so one more $3$ is needed. $5^1$ is odd, so one more $5$ is needed.

Multiply the missing factors together.

Check the result really is a square.

A number is a perfect cube exactly when every prime index is a multiple of $3$.

$2^2$ needs one more $2$ to reach $2^3$. $3^3$ is already a multiple of $3$. $5^1$ needs two more $5$s to reach $5^3$.

Multiply the missing factors together.

Check.

(a) Turn the three rational numbers into exact values first. The reciprocal of $0.4$ is $1 \div 0.4$, the recurring decimal is a mixed number, and $64 = 4^3$.

Place $\sqrt{30}$ between the square roots of the square numbers either side of $30$. Since $5.5^2 = 30.25$, it lies just below $5.5$.

$\pi = 3.14\ldots$, so it sits just past $3$. Now match each value to a point: $C$ is two-thirds of the way from $1$ to $2$, $E$ is at $2.5$, $A$ is just past $3$, $D$ is at $4$ and $B$ is between $5$ and $6$.

(b) $\sqrt{30}$ is irrational because $30$ is not a square number, and $\pi$ is irrational. The other three are exact fractions or integers, so they are rational.

(a) Enclose $P$ in a $3$ by $3$ square. The four corner pieces left over are right-angled triangles with shorter sides $1\text{ cm}$ and $2\text{ cm}$. Subtract them.

(b) The side of a square is the square root of its area. $5$ is not a square number ($4 < 5 < 9$), so $\sqrt{5}$ cannot be written as a fraction of integers: it is irrational.

(c) Enclose $Q$ in a $7$ by $7$ square. This time the corner triangles have shorter sides $3\text{ cm}$ and $4\text{ cm}$.

$25$ is a square number, so the side of $Q$ is rational even though the square is tilted.

(d) Write the ratio and rationalise the denominator by multiplying top and bottom by $\sqrt{5}$.

Finish with the conclusion: the ratio equals $\sqrt{5}$, which is irrational (part (b)), so the ratio is irrational.

(a) A flat square layer with $k$ cubes along each side uses $k^2$ cubes, so the number of cubes must be a square number. $27$ lies between two consecutive square numbers, so it is not square.

(b) The larger cube has $n^3$ small cubes, and this must be a square number. Since $n^2$ is already square, $n^3 = n^2 \times n$ is square exactly when $n$ itself is square.

The smallest square number greater than $1$ is $4$. Check that it works and find the side of the layer.

(c) Ria's number is both a cube and a square. In prime factors every index must be a multiple of $3$ and even, so a multiple of $6$. The number is a sixth power.

Only $729$ lies between $500$ and $1000$. Check both arrangements.

(a) Natural numbers are the counting numbers $1, 2, 3, \dots$ — whole and positive. $-13$ and $-1.5$ are negative, and $\tfrac{2}{7}$ and $0.\dot{4}$ are not whole. Simplify the roots first: $\sqrt{64} = 8$, which is natural.

(b) A prime has exactly two factors, $1$ and itself, and must be positive. $21 = 3 \times 7$ is not prime, and $-13$ is negative, so it is not prime even though $13$ is.

(c) $\sqrt{40}$ is irrational because $40$ is not a square number ($36 < 40 < 49$). $\sqrt{64}$ is not irrational — it equals $8$.

(d) Rational numbers can be written as a fraction of two integers. $\tfrac{2}{7}$, $0.\dot{4} = \tfrac{4}{9}$ and $-1.5 = -\tfrac{3}{2}$ are all rational and none is an integer.

(a) One million is $1\,000\,000$. Multiply $3.06$ by one million: the decimal point moves $6$ places to the right.

(b) Split the number at the thousands: $40$ thousands and $17$ ones. The hundreds digit is $0$, so no hundreds are said.

(c) Build it in parts: two hundred and five thousand is $205\,000$; add thirty.