In this chapter we will give some rules, but the best rule is *strategy*. In other words, try to think what is the best way in each case and proceed accordingly, instead of blindly follow rules.
When we multiply an integer a by fraction b/c, two things can be done. The first is to multiply a·b and place the result in the numerator. For example, 2·(1/3) = (2·1)/3 = 2/3. But if we are given 4·(3/8) it is absurd to write 12/8. It is much better to cancel the 4 with the denominator, as in 4·3/(4·2) = 3/2 = 1[1/2].
Another example is 4·(5/12). Again, it is absurd to multiply 4·5. Better to deal with 4/12=1/3 and then we have 5/3=1[2/3].
When dealing with letters, just cancel the ones you can and multiply the numerator by the rest.
In the case of a·(b/c) we just write (a·b)/c. But if we have a·b·c·(d)/(b·e·a), then we cancel a and b and we get (c·d)/e.
The most trivial case is that of a·(1/a), which must give 1. And a·(b/a) = b.
If we want to divide a fraction by an integer, it is best to consider the inverse of such integer and see if we can cancel something from the numerator of the fraction.
For example, dividing 12/25 by 2 we do the inverse of 2 which is 1/2 and we consider 12·(1/2) = 12/2 = 6. Then the result is 6/25.
Another choice is just to write the integer as a factor in the fraction's denominator. For example, 12/25 divided by 4 is
12 12 3·4 3 ---- / 4 = ------ = ----- = ----- . 25 25·4 25·4 25
These are quite trivial operations.
When dividing a fraction by an integer, many times we don't have such an easy case as before. For example, we may want to divide 3/4 by 2.
Euler proposes to multiply rewrite 3/4 as (3·2)/(4·2) = 6/8, so that now the numerator 6 can be divided by 2, giving us the result 3/8.
But we can directly place the 2 as a factor of the denominator
3 3 3 --- / 2 = ----- = --- . 4 4·2 8
Don't try to look for "the one method". Consider all the angles and proceed each time differently according to the situation.
With letters, a/b divided by c is simple a/(b·c).
Finally Euler says here that to divide a fraction by an integer you can just put the integer in the fraction's denominator as a factor.
Example: divide 9/16 by 3. Don't be too fast writing 9/48. This would be following a rule in a mindless way. Instead, consider factorisation:
9 9 3·3 3 ---- / 3 = ------ = ------ = ----- . 16 16·3 16·3 16
This is a good example in which Euler's previous method (considering only the fraction's numerator and the integer) would directly lead to 9/3 = 3, and then we could directly arrive to 3/16.
Let's multiply a fraction a/b by another fraction c/d. It is just the concatenation of the two approaches done before. We just need to multiply a/b by c and divide it by d.
But let's not make something so easy to appear as a complicated thing. Just multiply the numerators together and multiply the denominators together.
a c a·c --- · --- = ----- . b d b·d
Example, (2/3)·(4/5) = (2·4)/(3·5) = 8/15.
In order to divide two fractions, if they happen to have the same denominators, then we can get rid of them and consider only the division of the numerators.
Example: 8/12 divided by 9/12 is just 8/9.
But in general, denominators are different. Euler proposes to transform the fractions as to have both with a common denominator.
For example, with letters we can have (a/b) divided by (c/d). Then we can transform them as (a·d)/(b·d) and (c·b)/(d·b), respectively. Once they have the same denominator, just divide the numerators: (a·d)/(c·b) = ad/bc.
This is equivalent to apply what I always imagine as a sandwich, so I call it the sandwich rule. Let's draw a conceptual sandwich:
+--------------+
| top bread | bread
+----------------------+ ------------
| tomato | tomato bread·bread
+----------------------+ = ------------------- = ------------- .
| tofu | tofu tomato·tofu
+----------------------+ ------------
| bottom bread | bread
+--------------+
Don't even try a mnemonic here, because they are usually stupid when there is a clear logic behind. Moreover, your are likely to associate bread with bottom and tomato/tofu with top, which can be fatal!
I confess I always think in terms of "bread up" and "the rest down". Which does not qualify even as a slice of toast, since it is upside down!.
Another way I sometimes think is in terms of a reflection of a big boat on a very calmed sea:
_ _ _
| | | | | |
+--+ +--+ +--+-+--+
| top of boat | air
+----------------------+
| bottom of boat /
-----------------------·-------sea's surface --- rotate around this axis
| reflected bottom \
+----------------------+
| reflected top | water
+--+ +--+ +--+-+--+
|_| |_| |_|
Then, if you use your mind to rotate the image around the indicated axis, you just obtain
(real top)
--------------
(real bottom) (real top)·(reflected top) top·top
--------------------- = ------------------------------- = ------------- .
(reflected bottom) (real bottom)·(reflected bottom) bottom·bottom
-----------------
(reflected top)
Apply this to (5/8) divided by (2/3), which could be written as (5/8) / (2/3).
We simply get (5·3) / (8·2) = 15/16.
Divided (3/4) / (1/2) and get (3·2)/(4·1) = 6/4 = 3/2 = 1[1/2].
Since the rule of multiplication is so clear, you can just see that dividing by a fraction is the same as multiplying by its inverse, so (bread)/(tomato) divided by (tofu/bread) is just bread/tomato multiplied by bread/tofu, which clearly gives the result above.
Example: (3/4) / (1/2) is just (3/4)·(2/1) = (3·2)/(4·1) = 3/2 = 1[1/2].
I usually prefer to think the previous rule than inverting one fraction and then multiplying. It is the same, in fact, but sometimes you prefer a point of view over another. But the best is to keep as many points of view as possible.
For example, dividing (3/7) / (1/2) I automatically invert 1/2, since the inverse of the inverse of 2 is just 2.
Another difficulty many students have is locating the bread or boat rule when they are given an integer divided by a fraction or a fraction divided by the integer. They become puzzled trying to find what is bread or what is tomato.
Solution: just divide by 1 when in doubt!
Examples: 3 / (2/5) ---> transform it to (3/1) / (2/5) and now it is clear. Also, if having (2/5) / 3, transform it to (2/5) / (3/1) and that's all you need to apply those rules.
Again, Euler deals with the issue of dividing by 0. If dividing 1 by (1/10) is just 10, then 1 / (1/10000) = 10000. Then, if we divide by a very small quantity like 1/10000000000, we will get an enormous result as 10000000000. In the limit of dividing by a vanishing quantity, we obtain indefinitely large results.
A fraction divided by itself must give one, since (a/b) / (a/b) = (a/b) · (b/a) = ab / ab = 1.
What is half of 3/4? This is another obstacle for many students. How to express this in fractions? The answer is very easy: just substitute the word "of" by a multiplying symbol. Then "half of" becomes "(1/2)·".
The half of 2/3 = (1/2) · (2/3).
The fourth of 3/4 = (1/4) · (3/4).
The double of 7/3 = 2 · (7/3).
We can also advance the percentage notation, since it is exactly the same.
The 50% of 2/3 = (50/100) · (2/3) = (1/2) · (2/3).
The 25% of 3/4 = (25/100) · (3/4) = (1/4) · (3/4).
The 200% of 7/3 = (200/100) · (7/3) = 2 · (7/3).
The sign rules for multiplication and division of fractions are the same as for integers. It is better to separate the absolute values and the operation with signs.
But there are always cases that can look confusing, so let's explore some important scenarios. Make sure you fully understand all cases with all steps:
-3 3 3 ---- = ---- = - --- . 4 -4 4
-1 + 2 1 - 2 -(1 - 2) 1 - 2 --------- = - ------- = --------- = ------- = etc . 3 + 4 3 + 4 3 + 4 -3 - 4
-2 2 2 2 ---- --- - --- ----- 3 -3 3 3 -5 -------- = ------ = ----- = -------- = ---- . 4 4 -4 -4 6 --- --- --- - ---- 5 5 -5 -5
((-2) / 3 ) / (4/5) = (2/(-3)) / (4/5) = -(2/3) / ((-4)/(-5)) = etc.
Multiply x/6 by 2x/9. We get (x·2x) / (6·9) = x² / (3·9) = x² / 27.
Multiply x/2 by 4x/5 and by 10x/21.
We get (x·4x·10x) / (2·5·21) = (4·10·x³) / (2·5·21) = (4x³) / 21.
See how we need to calm our will to multiply before it is time. It is always wise to consider retaining factorised forms.
Multiply x/a by (x+a)/(a+c).
We get (x·(x+a)) / (a·(a+c)) = (x²+ax) / (a²+ac).
Multiply 3x/2 by 3a/b.
We get (3x·3a) / (2b) = (9ax) / (2b).
Multiply 2x/5 by 3x²/(2a).
We get (2x·3x²) / (5·2a) = (3x³) / (5a).
Take the product of 2x/a, 3ab/c and 3ac/(2b).
We get (2x·3ab·3ac) / (a·c·2b) = 9ax.
Product of b + bx/a and a/x.
We get (ba+bx)/a multiplied by a/x, so (ab+bx) / x.
Product of (x²-b²)/(bc) by (x²+b²)/(b+c).
We get (x⁴-b⁴) / (b²c + bc²).
Product of x, (x+1)/a and (x-1)/(a+b).
We get (x·(x+1)(x-1)) / (1·a·(a+b)) = (x³-x) / (a²+ab).
Divide x/3 by 2x/9.
We get (x·9) / (3·2x) = 3/2 = 1[1/2].
Divided 2a/b by 4c/d.
We get (2a·d) / (b·4c) = ad / (2bc).
Divide (x+a)/(2x-2b) by (x+b)/(5x+a).
We get ( (x+a)·(5x+a) ) / ( 2·(x-b)·(x+b) ) = (5x² + 6xa + a²) / (2x² - 2b²).
Divide (2x²)/(a³+x³) by x/(x+a).
We get (2x²(x+a)) / (x·(a³+x³)) = (2x²+2ax) / (x³+a³).
Divide 7x/5 by 12/13.
We get (7x·13) / (5·12) = (91x) / 60.
Divide 4x²/7 by 5x.
We get (4x²·1) / (7·5x) = 4x / 35.
Remember to write 5x as 5x/1.
Divide (x+1)/6 by 2x/3.
We get ((x+1)·3) / (6·2x) = (x+1) / (4x).
Divide (x-b)/(8cd) by (3cx)/(4d).
We get ((x-b)4d) / (8cd·3cx) = (x-b) / (6c²x).
Divide (x⁴-b⁴)/(x²-2bx+b²) by (x²+bx)/(x-b).
We get
(x⁴-b⁴)(x-b) (x²+b²)(x²-b²)(x-b) (x²+b²)(x+b) (x²+b²) -------------------- = --------------------- = ------------- = -------- . (x²-2bx+b²)(x²+bx) (x-b)(x-b)(x²+bx) x(x+b) x