We already know that 2/2, 3/3, 4/4, 5/5, etc are all integers. Specifically, equal to 1.
We also know that 2/1, 4/2, 6/3, 8/4, etc are all equal to 2.
Similarly, 3/1, 6/2, 9/3, etc are all equal to 3.
What we said before makes us think the following: that we can rewrite the same value expressed as many different fractions.
All fractions 1/2, 2/4, 3/6, 4/8 have the same value.
In general, a fraction a/b can be rewritten as
a 2a 3a --- = ---- = ---- = etc. b 2b 3b
To see this even more clearly, consider that the letter or box c contains the exact value of the division a/b. Then we can write
a a·b --- = c => ----- = c·b => a = c·b . b b
So if we multiply c by 2, we also obtain two times a: 2·a = (2c)·b, or 3·a = (3c)·b, etc. We can now "transfer" the 2 to the b factor to rewrite all as
a = c· b 2a = c·2b 3a = c·3b 4a = c·4b 5a = c·5b .
which is the same we wrote in the previous point (86).
Although it is true that many (infinite) different fractions can give the same value, it is often convenient to simply the fraction until we cannot further simplify it.
For example, consider 8/12, so we do
8 2·4 4 2·2 2 --- = ------ = --- = ----- = --- . 12 2·6 6 2·3 3
Every time the denominator and the numerator contain the same factor, we can cancel it out. The final result, 2/3, cannot be further reduced. We say that 2/3 is a fraction in *its least terms*. This concept of "least terms" of a fraction is of the highest importance.
We can proceed similarly with letters. Let's develop an example:
mab ab a ----- = ---- = --- . mbc bc c
In order to reduce a fraction to its least terms, we need to find a *common divisor*, i.e. a common factor in both the denominator and numerator. When we cannot find a common divisor, the fraction is already at its least terms.
Consider 48/120. It is clear that there is a common divisor 2, so we can divide both by 2, obtaining 24/60. Again, 2 is a common divisor, so divide all by two and get 12/30. Again, 2 is a common divisor, leading to 6/15. Now, 3 is a common divisor, leading to 2/3, which is in its least terms already.
The order in which you find the common factors is irrelevant, although I personally like to cancel all common divisors equal to 2 first.
Language is tricky and you can end up saying that, since 2/3 = 4/6, "the two fractions 2/3 and 4/6 are equal". Well, it would be better to say that the two fractions are different, but that they have the same value. This is not very important, but Euler certainly makes this distinction.
He says that the invariance of the value of a fraction by multiplying up and down by the same number is "the principal foundation of the doctrine of fractions".
This latter point is to remind us that whole numbers can also be considered as fractions. For example, 6 can be expressed as 6/1 or 12/2.
Euler propose to reduce some expressions to their lowest terms. It is weird that the author makes a great effort to introduce all concepts in order, being pedagogical to the extreme, while in these questions we are asked to know several algebra tricks which are a little advanced. It is as if someone has placed these questions there without Euler knowing it. Otherwise I cannot understand the level that they require at this point of the book. However, let's do them and learn some tricks. But if you feel discouraged, skip them now and return to them in a second reading.
cx + x² ----------- . ca² + a²x
Here, for example, we are asked to reduce this fraction when we have not talked about powers! But let's do it anyway and learn from it.
cx + x² x(c+x) x ----------- = -------- = ---- . ca² + a²x a²(c+x) a²
Here we have extracted a common factor, up x and down a². It gives (c+x) as a common divisor, which can be cancelled out.
x³ -b²x x(x²-b²) x(x-b)(x-b) x(x-b) x² - xb --------------- = --------- = ------------- = -------- = --------- . x² + 2bx + b² (x+b)² (x+b)(x+b) (x+b) x + b
Many tricks here. The first is to observe that the square of a sum is (a+b)² = (a² + 2ab + b²). The second is to know that the product of a sum multiplied by the subtraction has the following shortcut: (a+b)(a-b) = a² - b². These are very frequent tricks that you need to know by heart sooner or later, so why not now?
x⁴ - b⁴ (x² + b²)(x² - b²) x² + b² ----------- = -------------------- = --------- . x⁵ - b²x³ x³(x² - b²) x³
Here we needed to see that x⁴ - b⁴ is the subtraction of two squares, so that we can use the trick explained before.
x² - y² (x² - y²) 1 --------- = -------------------- = --------- . x⁴ - y⁴ (x² - y²)(x² + x²) x² + y²
a⁴ - x⁴ (a²+x²)(a²-x²) a² + x² --------------------- = ---------------- = --------- . a³ - a²x - ax² + x³ (a - x)(a²-x²) a - x
Here, in addition to the usual trick of subtracting squares, we needed to have the "ultra vision" of seeing that (a³ - a²x - ax² + x³) can be factored into (a-x) and (a²-x²). Don't get discouraged if you don't see this at first. It takes time and practise. But I will give you a hint for when to suspect this factorising.
Notice how in a³-a²x-ax²+x³ we have two entities, a and x. And while one (a) is always decreasing in power (from a³ to a² to a¹=a to finally a⁰=1), the other (x) is always increasing (from x⁰=1 to x¹=x to x² to finally x³). I can tell you that the author has written the terms in such an elegant way *on purpose*.
Imagine we multiply (a-2x)(3a-x)(4a-5x), which, after some painful multiplication, will give you 12a³ - 43a²x + 43ax² - 10x³. Don't pay attention to the numbers, but notice only how the exponents of a and x evolve. Exactly as we described before.
In our case we are presented with a³ - a²x - ax² + b³ so we know that we are just multiplying terms like (a+x), (a-x), (x-a) or (-x-a), since all coefficients are either 1 or -1. In fact, these four candidates reduce to two, since the other two differ only by a global sign. The solution is that the expression is equal to (a-x)(a-x)(a+x).
This is not the "ideal" time to explain this, but since Euler is suggesting these exercises, we may want to dig into some detail. Also, keep in mind that mathematics cannot be explained in a perfect linear progression.
5a⁵ + 10a⁴x + 5a³x² 5a³(a²+2ax+x²) 5a³(a+x)² 5a⁴ + 5a³x ------------------------- = -------------------------- = ------------------ = ---------------- . a³x + 2a²x² + 2ax³ + x⁴ x(a³x⁰+2a²x¹+2a¹x²+a⁰x⁴) x(a+x)(a²+ax+x²) a²x + ax² + x³
Here the most difficult part was to see that (a+x) factor in the denominator. This time we could not be sure that the factor had only 1 or -1 coefficients. Or could we? Yes, if we rewrite (a³x⁰+2a²x¹+2a¹x²+a⁰x⁴) as (a³+a²x+a²x+ax²+ax²+x⁴). Here we have 6 terms and 6=2·3, which means this can be the product of (two terms) by (three terms). We are suspecting this only because all terms are either +1 or -1. We can try to guess the two-term factor as (a+x) or (a-x).