Question bank — Addition, subtraction, multiplication, division of fractions

False — only true when both are positive and less than 1. For example $\frac{3}{2}\times \frac{4}{3}=2$, bigger than either; and two negatives multiply to a positive, e.g. $-\frac{1}{2}\times -\frac{1}{2}=\frac{1}{4}$.

False — a negative times a negative is positive (two flips of the number line undo each other): $-\frac{2}{3}\times -\frac{3}{4}=\frac{6}{12}=\frac{1}{2}$.

True for positive fractions — you're combining pieces, so the total exceeds either part (e.g. $\frac{1}{4}+\frac{1}{4}=\frac{1}{2}$, bigger than each $\frac{1}{4}$). This fails once a negative is involved.

False — the sign of the sum follows the larger piece. $\frac{1}{4}+\left(-\frac{1}{2}\right)=-\frac{1}{4}$, because you subtract the halves' worth and the negative wins.

False — it grows only when you divide by a positive fraction less than 1. Dividing by a fraction greater than 1 shrinks it: $\frac{1}{2}÷\frac{3}{2}=\frac{1}{3}$.

False — this adds unequal-sized pieces as if equal. Test $\frac{1}{2}+\frac{1}{2}$: it gives $\frac{2}{4}=\frac{1}{2}$, but two halves make a whole, $1$.

False — multiplication takes a fraction of a fraction (an area), which needs no matching piece sizes. You multiply straight across.

True — cancelling common factors first just applies the golden rule ($\frac{ak}{bk}=\frac{a}{b}$) early; the final value is identical either way.

False — the reciprocal of $\frac{0}{5}=0$ would require dividing by 0, which is undefined. Every fraction except zero has a reciprocal.

True — flipping the divisor multiplies by its reciprocal, which undoes the multiply; it fails only if $c=0$ (dividing by zero).

True — the reciprocal of a negative fraction is still negative, so a positive times a negative is negative: $\frac{5}{6}÷\left(-\frac{2}{3}\right)=\frac{5}{6}\times \left(-\frac{3}{2}\right)=-\frac{5}{4}$.

False — they all mark the same point on the number line. Convention is to keep the sign out front or on the numerator, never on the denominator.

False — both give the same value; the LCM (smallest common denominator) just keeps numerators smaller. $\frac{5}{6}-\frac{3}{4}$ is $\frac{1}{12}$ via LCM and $\frac{2}{24}=\frac{1}{12}$ via $bd$.

Wrong — you can't subtract numerators and denominators separately. Rewrite over a common bottom: $\frac{8}{12}-\frac{3}{12}=\frac{5}{12}$.

Wrong — the answer is simply negative: $\frac{1}{4}-\frac{2}{4}=-\frac{1}{4}$. Subtracting more than you have gives a debt, a perfectly valid rational number.

Wrong — the $-$ sign must ride along with the numerator: $\frac{8}{12}+\frac{-3}{12}=\frac{8-3}{12}=\frac{5}{12}$. Opposite signs mean the numerators subtract.

Wrong — the reciprocal of $-\frac{2}{3}$ keeps its sign: it's $-\frac{3}{2}$, not $\frac{3}{2}$. Correct: $\frac{5}{6}\times \left(-\frac{3}{2}\right)=-\frac{5}{4}$.

Wrong — $\frac{3}{-4}$ is really $-\frac{3}{4}$, so it's $-\frac{3}{4}+\frac{1}{4}=-\frac{2}{4}=-\frac{1}{2}$. Normalise the negative denominator to the numerator before adding.

Wrong — you flip the divisor (the one you divide by), not the first. Correct: $\frac{3}{4}\times \frac{5}{2}=\frac{15}{8}$.

Wrong conclusion — multiplying by $\frac{2}{3}<1$ takes part of $\frac{4}{5}$, so the result shrinks. Indeed $\frac{8}{15}<\frac{4}{5}$ since $\frac{4}{5}=\frac{12}{15}$.

Wrong — you changed the denominator but forgot to scale the numerators. $\frac{5}{6}=\frac{20}{24}$, $\frac{3}{4}=\frac{18}{24}$, giving $\frac{38}{24}=\frac{19}{12}$.

Wrong — you may only cancel a numerator with a denominator, not two numerators or two denominators. Cancel the 4 with 8 and the 3 with 9: $\frac{1}{2}\times \frac{1}{3}=\frac{1}{6}$.

Wrong — this is fine: $0\times \frac{3}{2}=0$. It's $\frac{2}{3}÷0$ that is undefined, since no number times 0 gives $\frac{2}{3}$.

Addition counts pieces, so the pieces must be the same size (matched denominators); multiplication takes a fraction of a fraction — an area — which needs no matching.

Multiplying by a negative flips the number line about $0$; flipping twice returns you to the original direction, so the two minus signs cancel and the result is positive (see the flip figure).

All of $-\frac{3}{4}$, $\frac{-3}{4}$, $\frac{3}{-4}$ are equal, but a sign hidden in the denominator is easy to lose while adding or flipping; pulling it onto the numerator (or out front) keeps sign-tracking honest.

A negative fraction points the opposite way, so combining them is a tug-of-war: you subtract the smaller pull from the larger, and the answer takes the sign of the larger.

Multiplying $\frac{c}{d}$ by its reciprocal $\frac{d}{c}$ gives 1, which cancels the divisor and leaves the multiplication that division undoes.

Cutting each piece into $k$ smaller pieces gives $k$ times as many pieces, each $k$ times smaller — the total amount of "pie" is unchanged.

You're asking "how many of these tiny chunks fit?" — and many small chunks fit into a fixed amount, so the count is large.

Cancelling early is the golden rule applied in reverse ($\frac{ak}{bk}=\frac{a}{b}$); the value never changes, but the numbers stay small.

Writing $1=\frac{k}{k}$ and multiplying keeps the value (times 1 changes nothing) while rescaling both parts — this is exactly the golden rule.

It's $-\frac{1}{4}$ — over the common bottom $\frac{1}{4}-\frac{2}{4}=-\frac{1}{4}$. Subtracting a larger fraction just gives a negative result, a legitimate rational number.

It's $-\frac{5}{4}$ — flip the divisor keeping its minus: $\frac{5}{6}\times \left(-\frac{3}{2}\right)$; positive ÷ negative is negative.

It's $-\frac{3}{4}$ — move the minus off the denominator onto the numerator (or out front); the value and its number-line position are unchanged.

It's $-\frac{1}{6}$ — rewrite as $\frac{4}{6}-\frac{5}{6}=-\frac{1}{6}$; the negative half wins because it's the larger piece.

It's $\frac{1}{2}$ — two negatives multiply to a positive, and $\frac{3\times 2}{4\times 3}=\frac{6}{12}=\frac{1}{2}$.

It's $\frac{5}{7}$ — since $1=\frac{1}{1}$, its reciprocal is itself, so dividing by 1 leaves any number unchanged.

Undefined — the divisor $\frac{0}{d}=0$ has no reciprocal, and no number times 0 can equal a nonzero $\frac{a}{b}$.

It's $0$ — zero pieces plus zero pieces is zero, regardless of denominators; you can write it as $\frac{0}{36}=0$.

It's $0$ — taking zero copies of $\frac{3}{5}$, or taking $\frac{3}{5}$ of nothing, leaves nothing.

Always $1$ — dividing any nonzero quantity by itself asks "how many fit into itself?", which is exactly one.

No — the denominator says how many equal parts make a whole, and you cannot split a whole into zero parts, so $\frac{5}{0}$ is undefined.

Both equal 1, so the difference is $0$ — a fraction with equal top and bottom is one whole, so this is $1-1$.