QuanticaPrealgebraOpen in the courseSubtraction turned out to be addition, since $a-b$ is just $a+(-b)$, and the opposite of $b$ was the partner that made it work. That leaves the last operation, division. Here is a guess worth holding onto. If subtraction was another form of addition, division will be another form of multiplication, and it will need a partner of its own.
Problem
A blueprint app has one button that scales every length by 5, blowing a drawing up to five times its size. You tap it by accident. You need a second scaling that puts the drawing back exactly as it was, neither bigger nor smaller. Multiplying by $1$ is the scaling that changes nothing, so what number must you scale by to undo the \(\times 5\)?
Show a hint
- What single number multiplies with 5 to give 1?
Show the full solution
Scaling up by 5 and back down acts like multiplying by 1, so you need the number that turns 5 into 1.$$5 \times \frac{1}{5} = 1$$The undo scales by \(\boxed{\frac{1}{5}}\), the reciprocal of 5.
Problem
Here is a pattern worth predicting before you compute. Last chapter, the opposite of the opposite of a number brought you right back to the number, since two flips across zero return home. Now guess the reciprocal of a reciprocal, then test the guess. What is the reciprocal of \(\frac{1}{6}\)?
Show a hint
- The reciprocal of \(\frac{1}{6}\) is the number that multiplies with \(\frac{1}{6}\) to give 1.
- Flip \(\frac{1}{6}\) over.
Show the full solution
Flip \(\frac{1}{6}\) and it becomes 6.$$\frac{1}{6} \times 6 = 1$$So the reciprocal of \(\frac{1}{6}\) is \(\boxed{6}\). Flipping twice returns the original, so the reciprocal of a reciprocal is the number itself.
Problem
One number has no reciprocal at all. Try to find the reciprocal of $0$, the partner that would multiply with 0 to give 1. Hunt for it, and say what goes wrong.
Show the full solution
The reciprocal of 0 would have to turn 0 into 1, but \(0 \times (\text{any number}) = 0\), never 1. So 0 has no reciprocal, the only number without one.
Problem
Last chapter, exactly one number was its own opposite, the number 0, sitting right on its own mirror image. Now ask the matching question for reciprocals. How many numbers are their own reciprocal, equal to the number that multiplies with them to give 1? Test small cases and see what you find.
Show a hint
- A number \(x\) is its own reciprocal when \(x \times x = 1\).
- Try 1. Then remember the sign rule and try \(-1\).
Show the full solution
A number is its own reciprocal when it times itself gives 1. \(1 \times 1 = 1\), and \((-1) \times (-1) = 1\), so 1 and \(-1\) both qualify. Nothing else does, since any number above 1 has a reciprocal below 1. That leaves exactly \(\boxed{2}\) such numbers.
With the reciprocal in hand, we can build division out of multiplication. First we check the new rule against a division you already know the answer to, then use it on a division that is awkward to do the old way.
Problem
A relay team of $4$ runners splits a $12$ km course into equal legs. Plain division says each leg is \(12 \div 4 = 3\) km, no surprise. Now compute it the new way instead. Take \(\frac{1}{4}\) of 12, that is \(12 \times \frac{1}{4}\). How long is each leg this way?
Show the full solution
Each runner covers one quarter of 12 km, found by multiplying.$$12 \times \frac{1}{4} = \frac{12}{4} = \boxed{3}$$Dividing 12 by 4 and multiplying 12 by \(\frac{1}{4}\) are two names for one number.
Problem
Undo a $\times 6$: compute $$24 \div 6.$$
Show a hint
- Dividing by 6 undoes multiplying by 6.
Show the full solution
\(24 \div 6 = \boxed{4}\), since \(4 \times 6 = 24\).
Problem
A jar holds 3 cups of flour. Your scoop holds \(\frac{1}{4}\) cup. How many full scoops fill the jar? In symbols, find $$3 \div \frac{1}{4}.$$
Show a hint
- Dividing by \(\frac{1}{4}\) means multiplying by the reciprocal of \(\frac{1}{4}\).
- The reciprocal of \(\frac{1}{4}\) is 4, so you are really computing \(3 \times 4\).
Show the full solution
Turn the division into multiplication by the reciprocal of \(\frac{1}{4}\), which is 4.$$3 \div \frac{1}{4} = 3 \times 4 = \boxed{12}$$The answer grew past 3 because we divided by a number below 1, whose reciprocal is bigger than 1.
Problem
How many halves fill 6? Find $$6 \div \tfrac{1}{2}.$$
Show a hint
- Dividing by $\tfrac12$ is multiplying by $2$.
Show the full solution
\(6 \div \tfrac{1}{2} = 6 \times 2 = \boxed{12}\), so twelve halves make six.
Problem
Over $6$ equal stages of a dive, a submarine’s depth changed by a total of $-240$ m (a downward move counts as negative, as it did two lessons ago). The stages were identical, so what was the depth change per stage? Compute $$-240 \div 6.$$
Show a hint
- Division is multiplication by the reciprocal, so this is \(-240 \times \frac{1}{6}\).
- One negative factor flips the sign of the product, just as it did for multiplication.
Show the full solution
Rewrite the division as multiplication by \(\frac{1}{6}\). One negative factor keeps the result negative.$$-240 \div 6 = -240 \times \frac{1}{6} = \boxed{-40}$$Division uses the same sign rules as multiplication, because it is multiplication by the reciprocal.
Problem
Structure beats grinding, as always. Compute $$(16 \times 21) \div (8 \times 3)$$ without doing the two multiplications first. Look for parts of the top that pair off neatly with parts of the bottom.
Show a hint
- Dividing by \(8 \times 3\) is multiplying by \(\frac{1}{8} \times \frac{1}{3}\), so pair the 16 with the 8 and the 21 with the 3.
- \(16 \div 8 = 2\) and \(21 \div 3 = 7\).
Show the full solution
Pair each top factor with the bottom factor it cancels.
$$(16 \times 21) \times \frac{1}{8} \times \frac{1}{3} = \left(16 \times \frac{1}{8}\right) \times \left(21 \times \frac{1}{3}\right) = 2 \times 7 = \boxed{14}$$
Grinding out \(336 \div 24\) gives 14 as well, but pairing first means you never build the big products.
Problem
Pair, do not grind: $$(25 \times 18) \div (5 \times 9).$$
Show a hint
- Pair 25 with 5, and 18 with 9.
Show the full solution
Pair the factors. \(25 \div 5 = 5\) and \(18 \div 9 = 2\), so the value is \(5 \times 2 = \boxed{10}\).
Problem
Sometimes the top and bottom share a factor so ugly you would dread computing it. Find $$(637 \times 50) \div (637 \times 10).$$ The 637 is bait. Do not take it.
Show a hint
- 637 sits on both the top and the bottom. What does \(637 \times \frac{1}{637}\) equal?
- Whatever 637 is, it cancels to 1, leaving only \(50 \div 10\).
Show the full solution
The 637 on top meets a \(\frac{1}{637}\) on the bottom, and that pair cancels to 1.
$$637 \times 50 \times \frac{1}{637} \times \frac{1}{10} = 1 \times (50 \div 10) = \boxed{5}$$
A factor sitting on both the top and the bottom always cancels to 1, however big it is, so you never multiply it out.
Problem
A signal runs through a pipeline that multiplies it by $14$, then by $27$, then by $10$. Three filters then divide it by $7$, by $9$, and by $5$. By what single factor does the signal change overall? Compute $$(14 \times 27 \times 10) \div (7 \times 9 \times 5)$$ by pairing, not by grinding.
Show a hint
- Turn every division into multiplication by a reciprocal, then hunt for pairs.
- 14 with 7, 27 with 9, 10 with 5. Each pair divides cleanly.
Show the full solution
Spread the three divisions out as reciprocals and pair each top factor with the bottom factor that cancels it.
$$(14 \div 7) \times (27 \div 9) \times (10 \div 5) = 2 \times 3 \times 2 = \boxed{12}$$
The signal comes out 12 times bigger, and none of the six original multiplications had to be carried out.
Problem
Addition and multiplication let you reorder and regroup at will. Division is stricter, and it is worth catching in the act. First, is \(12 \div 3\) the same as \(3 \div 12\)? Then settle a grouping question. Compute \(24 \div (4 \div 2)\) and compare it to \((24 \div 4) \div 2\). Enter the value of $$24 \div (4 \div 2).$$
Show a hint
- Work inside the parentheses first. \(4 \div 2 = 2\), so the left side is \(24 \div 2\).
- Compare with \((24 \div 4) \div 2 = 6 \div 2\).
Show the full solution
Inside the parentheses, \(4 \div 2 = 2\), so
$$24 \div (4 \div 2) = 24 \div 2 = \boxed{12}$$
Move the parentheses and you get \((24 \div 4) \div 2 = 6 \div 2 = 3\), a different answer from the same three numbers. Order fails too, since \(12 \div 3 = 4\) while \(3 \div 12 = \frac{1}{4}\). Division flips only the number you divide by, so shuffling the numbers changes which one gets flipped.
Problem
Mind the signs. $$(-56 \times 9) \div (7 \times (-3)).$$
Show a hint
- Negative over negative is positive, then pair off.
Show the full solution
Two negatives make the result positive, and then \((56 \div 7) \times (9 \div 3) = 8 \times 3 = \boxed{24}\).
Problem
Put every tool in one place. Compute $$(-45 \times 16) \div (-9 \times 4)$$ in your head. Mind the signs, convert the division, then pair off.
Show a hint
- Two negative numbers are involved, so the result is positive. Now just handle the sizes.
- Write it as \(45 \times 16 \times \frac{1}{9} \times \frac{1}{4}\) and pair 45 with 9, then 16 with 4.
Show the full solution
One negative on top and one on the bottom make the result positive, so drop both signs and pair the factors.
$$45 \times 16 \times \frac{1}{9} \times \frac{1}{4} = (45 \div 9) \times (16 \div 4) = 5 \times 4 = \boxed{20}$$
Signs first, then division turned into multiplication, then pairs that cancel. Three lessons in one short computation.
Practice these ideas
Practice
Compute $$(9 + 18 + 27 + 36) \div (36 + 27 + 18 + 9).$$
Show the solution
The numbers above the line and below it are identical, only listed in reverse, so the two sums are equal. Any nonzero amount divided by itself is \(\boxed{1}\), you never have to add a single thing.
Practice
Compute $$5{,}600{,}000 \div 70{,}000$$ by cancelling, not by long division.
Show the solution
Strike four matching zeros from the top and the bottom, and \(5{,}600{,}000 \div 70{,}000\) becomes \(560 \div 7 = \boxed{80}\). Cancelling equal factors of 10 from top and bottom never changes a quotient.
Practice
Compute $$(317 \times 317) \div 317$$ without ever multiplying 317 by itself.
Show the solution
Dividing a product by one of its own factors simply hands back the other factor, so the answer is \(\boxed{317}\). Multiplying by a number and then dividing by it is a round trip straight back to the start.
Practice
A ribbon is $6$ metres long. How many \(\frac{1}{5}\)-metre bookmarks can you cut from it? Compute $$6 \div \frac{1}{5}.$$
Show the solution
Five fifths fit in every metre, so six metres give \(6 \times 5 = \boxed{30}\) bookmarks. Dividing by \(\frac{1}{5}\) means multiplying by its reciprocal, 5, and the count grows past 6 because the piece you cut is tiny.
Practice
Compute $$20 \div \frac{2}{5}.$$
Show the solution
Flip the divisor and multiply. \(20 \div \frac{2}{5} = 20 \times \frac{5}{2} = \frac{100}{2} = \boxed{50}\).
Practice
What number is 4 less than the value of \(48 \div \frac{3}{4}\)?
Show the solution
\(48 \div \frac{3}{4} = 48 \times \frac{4}{3} = 64\), and \(64 - 4 = \boxed{60}\). Dividing by a number below 1 makes the result bigger, which catches almost everyone off guard the first time.
Practice
Compute $$(13 \times 40 \times 9) \div (9 \times 13)$$ by cancelling first.
Show the solution
The 9 and the 13 on top each meet a copy on the bottom and cancel to 1, leaving the lone \(\boxed{40}\). Neither big product was ever worth forming.
Practice
Compute $$(20 \times 50 \times 90) \div (2 \times 5 \times 9).$$
Show the solution
Pair and divide. \(20 \div 2 = 10\), \(50 \div 5 = 10\), and \(90 \div 9 = 10\), so the whole expression is \(10 \times 10 \times 10 = \boxed{1000}\). Scaling every factor up by ten multiplies the quotient by ten, once for each factor.
Practice
Compute $$(4{,}000{,}000 + 36) \div 4$$ by sharing the division across the sum.
Show the solution
Divide each piece on its own. \(4{,}000{,}000 \div 4 + 36 \div 4 = 1{,}000{,}000 + 9 = \boxed{1{,}000{,}009}\). Splitting the dividend into friendly chunks beats one monstrous long division.
Practice
Compute $$(72 + 56 + 40 + 24) \div (9 + 7 + 5 + 3)$$ by comparing the two sums.
Show the solution
Every top term is eight times its partner (\(72 = 8 \times 9\), \(56 = 8 \times 7\), \(40 = 8 \times 5\), \(24 = 8 \times 3\)), so the whole top is eight times the whole bottom. A quantity divided by one-eighth of itself is \(\boxed{8}\), and not a single sum had to be worked out.
Practice
Compute $$96 \div \frac{1}{2} \div \frac{1}{2} \div \frac{1}{3},$$ working left to right.
Show the solution
Working left to right, \(96 \div \frac{1}{2} = 192\), then \(192 \div \frac{1}{2} = 384\), then \(384 \div \frac{1}{3} = \boxed{1152}\). Two doublings and a tripling multiply 96 by \(2 \times 2 \times 3 = 12\).
Practice
Compute $$(45 \times 28) \div (35 \times 18)$$ without multiplying either product out.
Show the solution
Cancel in stages. 45 and 35 share 5, leaving 9 and 7, and 28 and 18 share 2, leaving 14 and 9. Now \((9 \times 14) \div (7 \times 9)\), where the 9 cancels and \(14 \div 7 = 2\), so the answer is \(\boxed{2}\). Clearing shared factors first turns four awkward numbers into a single digit.
Practice
Watch what division does as the divisor shrinks. $12\div1=12$, then $12\div\frac12=24$, then $12\div\frac14=48$, then $12\div\frac{1}{100}=1200$. The closer the divisor gets to $0$, the larger the quotient grows, with no ceiling anywhere. That is the real reason you cannot divide by $0$, since there is no single number the quotient would settle on, only values that grow without end.