The World of Numbers: questions with answers

Brahmagupta's rule is $4 \times 0 = 0$, and once a product has reached 0 nothing later can bring it back: $0 \times 6 \times 8 = 0$. So the whole product is $0$. Leaving the 0 out because it 'does nothing' gives $192$ — that is the ADDITION rule, $a + 0 = a$, applied where it does not belong. Multiplying is not adding, which is why the two rules had to be stated separately.

The two zero rules meet in one expression here and pull opposite ways. $5 \times 0 = 0$ destroys its product, and then $0 + 7 = 7$ leaves the rest alone, so the answer is $7$. Treating the 0 as though it changed nothing gives $5 + 7 = 12$, and treating it as though it wiped the whole expression gives 0 — but a 0 only destroys what it MULTIPLIES.

In Brahmagupta's words, this is a debt ($rina$) added to a debt, and the result is a larger debt: 7 owed and 11 more owed is 18 owed. Nothing was repaid, so the trader cannot end up with a fortune, and no debt was cancelled against another, so the difference 4 never arises.

$(-5) \times (-8) = 40$ is the correct one: the product of two debts is a fortune. $(-8) \times (-14) = -112$ breaks that same rule, which is why the two products carry opposite signs and only one of them can be right. $(-5) + (-14) = 19$ treats two debts added as a fortune, and $19 - (-14) = 5$ takes away a debt of 14 and ends up 14 poorer instead of 14 richer.

It is, and the reason is simply that it STOPS. A decimal with twelve places is that many ten-thousand-millionths — $0.856141622914$ is a whole number over $10^{12}$ — so it is a ratio of integers however unruly its digits look. Length is not the test and neither is the absence of a pattern; the chapter's own $23.560185612239874790120$ is rational for exactly this reason. Only a decimal that runs on FOR EVER without repeating is irrational, and this one does not run on at all.

It is. The chapter's rule is that a rational number's decimal either STOPS or REPEATS, and this one repeats: the block 74 comes round again for ever. That is enough to turn it back into a fraction — $x = 0.\overline{74}$ gives $100x - x = 74$, so $x = \frac{ 74 }{99}$. Never ending is not the test; $0.\overline{3}$ never ends either and is $\frac{1}{3}$.

The definition asks whether the number CAN be written as $\frac{p}{q}$ with $p$ and $q$ integers and $q \neq 0$ — not whether it is written that way already. Putting it over 1 does it: $9 = \frac{9}{1}$. That is why the summary says the rational numbers include the natural numbers and the integers. Writing it over 0 would not do, because $q$ may not be 0, and a number does not have to have a decimal part to be rational.

Both numbers on the page are integers, so this passes every part of the definition except the one that matters: $q \neq 0$. Dividing by 0 gives no number at all — nothing multiplied by 0 gives 18 — so $\frac{18}{0}$ is not a rational number with an unusual value, it is not a number. The clause is part of the definition for exactly this reason.

The greatest common factor of 60 and 75 is 15, and dividing both parts by it gives $\frac{4}{5}$, which cannot be reduced further because 4 and 5 share no factor. Dividing by 3 alone stops halfway at $\frac{20}{25}$ — still reducible. Taking 15 away from both parts gives $\frac{45}{60}$, a different number altogether. And dividing only the top gives $\frac{4}{75}$, which is not even close: whatever is done to one part must be done to the other.

6 divides 18 exactly 3 times, so dividing both parts by 6 gives $\frac{1}{3}$. Dividing only the top gives $\frac{1}{18}$ — which LOOKS reduced, because its numerator is 1, and is not, because the denominator was left as it was. 2 is a common factor here as well, but taking 2 away from both parts gives $\frac{4}{16}$, and subtracting is not dividing: it keeps the gap between the parts and destroys the ratio. Dividing only the bottom gives $\frac{6}{3}$.

The distance between two numbers is $|a - b|$: here $|12 - (-4)| = |12 + 4| = 16$. Walking it gives the same answer — 4 steps from $-4$ up to 0, then 12 more up to 12. Adding the two SIGNED numbers instead gives 8, which is smaller, because the negative one cancels part of the other; and a distance can never be $-16$, since a distance has no direction.

Both numbers are on the same side of 0, so the distance is what is left when the smaller is taken from the larger: $|7 - 11| = 4$. Adding them gives 18, which is further than 11 is from 0 at all — adding is what the previous kind of question needed, not this one. And doing the subtraction the other way round gives $-4$, a negative distance, which is what the bars exist to prevent.

This is the chapter's own step. Add and halve: $3 + \frac{9}{2}$ is $\frac{15}{2}$, and half of that is $\frac{15}{4}$ — the same averaging the chapter uses. Writing the whole number as $\frac{3}{1}$ and then adding the numerators to each other and the denominators to each other gives $\frac{12}{3}$, which lies between the two but is not the middle of them. The step can always be taken, and taken again on the result.

Add the two and halve: $\frac{5}{4}$ is $\frac{10}{8}$, so the sum is $\frac{13}{8}$, and half of that is $\frac{13}{16}$. Adding the numerators to each other and the denominators to each other gives $\frac{8}{12}$ instead — that fraction does lie between the two, which is why it looks right, but it is not the middle. And there is always such a number: between any two rational numbers lies another, and so infinitely many.

It does. The test is run on the denominator in LOWEST TERMS, and $\frac{63}{14}$ reduces to $\frac{9}{2}$ once both parts are divided by 7. 2 is built only from 2s and 5s, so the denominator can be turned into a power of 10 and the decimal stops — just as $\frac{3}{20} = \frac{3}{2^2 \times 5} = \frac{15}{100} = 0.15$. Reading the primes off 14 instead finds a factor of 7 and calls a terminating decimal repeating; that 7 is not there once the fraction is reduced.