Real Numbers: questions with answers

A factor-tree node is the product of its two children. The missing node has children 3 and 85, giving 255. Dividing its parent 510 by its sibling 2 confirms the same value.

The root has children 2 and 385, so it equals their product 770. Adding them gives 387, which is not the factor-tree rule.

A prime has exactly two distinct positive factors, and a composite has more than two. The only positive factor of 1 is 1, so it is neither.

Subtract each remainder from its own number, obtaining 18 and 24. The required divisor is their HCF, 6, which exceeds both remainders. Taking the HCF of the original numbers instead gives 1.

The size must divide 24 exactly and 56 exactly, or something would be left over, so it is a COMMON FACTOR of the two. The largest common factor is the HCF, which is 8 cm. The LCM 168 is larger than either measurement, so it cannot be a piece of them at all.

For positive $n$, $8^n = 2^{3n}$ has only the prime factor 2. Ending in 0 requires both 2 and 5. No positive power works, and $8^0 = 1$ also does not end in 0.

Any non-zero number raised to the power 0 equals 1, so $8^0 = 1$. Since 1 is odd, the claim '$8^n$ is even' fails at $n = 0$. Natural numbers start at 1 and never hit this case, but whole numbers start at 0, so the claim does not extend to whole numbers.

The HCF equals the smaller number exactly when the smaller number divides the larger one. In 16 and 80 it does, since the larger is 5 times the smaller, so every factor of the smaller is a common factor and the HCF is the smaller number itself. In the other two pairs the smaller number does not divide the larger, so their HCF is a proper factor of it and is strictly smaller.

Both numbers use the primes 2 and 3. Their HCF takes the smaller powers and equals 18. The greatest powers give 1296, so division by 72 is required.

For 8, 9 and 25 the HCF is 1 and the LCM is 1800, and 8 x 9 x 25 = 1800 as well, so the two agree. That happens because the three numbers are pairwise coprime. For 6, 72 and 120 the HCF is 6 and the LCM is 360, giving 2160, while the product is 51840 -- the chapter's own counterexample. For 2, 4 and 8 the HCF is 2 and the LCM is 8, giving 16, while the product is 64.

The product of 45 and 63 is 2835 and their LCM is 315. Since HCF times LCM equals the product, the quotient is their HCF, 9. The LCM is not the product, so the answer is not 1.

A finite display cannot rule out a repeating block beginning later. Some square roots, such as $\sqrt{9} = 3$, are rational. The contradiction proof, starting from $\sqrt{2} = \frac{a}{b}$ in lowest terms, rules out every integer fraction for $\sqrt{2}$.

AB and BC have whole-number lengths, so both are rational. Pythagoras gives AC as $\sqrt{10}$ cm. Since 10 is not a perfect square, AC is irrational despite having a decimal expansion.

Pythagoras gives AC as $\sqrt{13}$ cm. Since 13 is not a perfect square, AC is irrational. Adding the rational length 5 keeps the perimeter irrational. Finite calculator digits cannot establish non-repetition forever, and some square roots are rational.

Subtract $10x$ from both sides: $990x = 2303$, so $x = \frac{2303}{990}$. Moving $+10x$ left without changing its sign incorrectly gives $1010x$. Repeating decimals can be written as fractions of integers.