Adding and subtracting fractions is easy and direct when they have the same denominator: we just add or subtracting the numerators while keeping that common denominator.
2/7 + 3/7 is just (2+3)/7 = 5/7.
4/7 - 2/7 is just (4-2)/7 = 2/7.
2/3 - 5/3 is just (2-5)/3 = -3/3 = -1.
2/4 - 3/4 + 1/4 = (2-3+1)/4 = 0/4 = 0.
When fractions don't have equal denominators, we then recall that magnificent flexibility of fractions in which we can rewrite them (while keeping their value) so that they end up having the same denominator.
1/2 + 3/4 cannot be directly added, so we take the first fraction and transform it to 2/4 (by multiplying down and up by 2). Then, 2/4 + 3/4 = (2+3)/4 = 5/4.
That was an easy case because by multiplying one of them we could get the same denominator as the other. The denominators were 2 and 4, and we know that 2·2 = 4.
For adding 1/2 + 1/3 we don't have such a direct way. The denominators are 2 and 3, and we cannot get 3 by multiplying 2 by a whole number. What do we do, then?
We then multiply the denominators, both of them, by increasing numbers 2,3,4,..., until we hit the same denominator.
+--------------+---------------+---------------+ | multiply by: | denominator 1 | denominator 2 | +--------------+---------------+---------------+ | 1 | 2 | 3 | +--------------+---------------+---------------+ | 2 | 4 | *6* | +--------------+---------------+---------------+ | 3 | *6* | 9 | +--------------+---------------+---------------+ | 4 | 8 | 12 | +--------------+---------------+---------------+
There was not need to reach 4 as multiplier! Already by multiplying denominator 1 by 3 and multiplying denominator 2 by 2 we obtain a common denominator, which is 6.
1 1 1·3 1·2 3 2 5 --- + --- = ----- + ----- = --- + --- = --- . 2 3 2·3 3·2 6 6 6
Another example: add 3/4 + 5/8. In this case, notice again how we can directly transform one denominator into the other, so it is an easy case: just (3·2)/(4·2) + 5/8 = 6/8 + 5/8 = 11/8, or 1[3/8] if you like.
Another example: 3/6 + 7/9. You could say that the first fraction is not in its least terms, and that it should be 1/2, but here it does not matter much, or does it? If we do now a "race" to reach a common denominator, now in a horizontal form,
1 2 3 6: 6 12 *18* race ends 9: 9 *18* race ends
Bingo! We have 18 as common denominator, so we can do
3 7 3·3 7·2 9 14 23 --- + --- = ----- + ----- = ---- + ---- = ---- . 6 9 6·3 9·2 18 18 18
A very important thing that can be seen here is that we could have extended our "race" beyond what was necessary:
1 2 3 4 5 6 7 8 9 6: 6 12 *18* 24 30 *36* 42 48 *54* etc 9: 9 *18* 27 *36* 45 *54* etc
We see how there are other choices: we could have chosen 36 or 54 as common denominators. And many more (infinite) are waiting up there. What we have done here is to choose the minimum common denominator.
A *multiple* of a number is the result of multiplying the number itself by an integer. For example, multiples of 3 are 3·2=6, 3·3=9, 3·4=12, and so on. Then, the "race" that we have written up there is a multiple race: we just wrote the multiples of both denominators until we hit the *least common multiple*, also called lcm. If you remember elementary school, you must recall that this concept was of paramount importance.
There is also a dirty trick that can be done to add two fractions without caring about whether the common denominator is the least or not: just multiply denominator 1 by denominator 2 and denominator 2 by denominator 1 (don't forget to also multiply their corresponding numerators as to keep the value of the fraction invariant). Then, for this same example of 3/6 + 7/9, we multiply the first fraction by 9 and the second fraction by 6:
3 7 3·9 7·6 27 42 69 23·3 23 --- + --- = ----- + ----- = ---- + ---- = ---- = ------ = ---- . 6 9 6·9 9·6 54 54 54 18·3 28
Many students prefer "not to think" and apply this method. It has the benefit of avoiding the search for the least common multiple, but numbers usually grow bigger and then we have to reduce the fraction to its least terms. So I am not sure if it is worth. However, this is an excellent approach to follow when adding fractions made of letters, as we will see later.
Instead of two fractions, we may need to add or subtract many more at once. So we need to find a common denominator for all of them. Ideally, we make a race for their least common multiple. If we want to add 1/2 + 2/3 + 3/4 + 4/5 + 5/6, the race can look daunting:
2 3 4 5 6 7 8 9 10 11 12 13
2: 4 6 8 10
3: 6 9 12 15
4: 8 12 16 20 wait a minute!!!!
5: 10 15 20 25
6: 12 18 24 30
As we have seen, we can spend a lot of time building this "race". We better a find a more powerful method.
The denominators are 2, 3, 4, 5, and 6. Or better: 2, 3, 2·2, 5 and 2·3. Look how, if all need to be equal, 2 is missing another 2, a 3 and a 5. Also, 3 is missing two 2's and a 5. Also, 2·2 is missing a 3 and a 5 and finally, 2·3 is missing a 2 and a 3. With this in mind, the least common multiple is just 2·2·3·5 = 60.
1 2 3 4 5 1·(2·3·5) 2·(2·2·5) 3·(3·5) 4·(2·2·3) --- + --- + --- + --- + --- = ---------- + ---------- + -------- + ---------- + 2 3 4 5 6 2·(2·3·5) 3·(2·2·5) 4·(3·5) 5·(2·2·3) 5·(2·5) 30+40+45+48+50 213 + --------- = --------------- = ----- . 6·(2·5) 60 60
To practise the notation with the supplement (Euler calls the supplement "fractional remainder" but I prefer to use Newton's term here), we can rewrite 213/60 = 3[33/60]=3[11/20].
As we said before, when we work with letters, that "dirty" trick becomes the cleanest trick you can make! For example, let's say we want to add a/b + c/d. Then, multiply each fraction by its partner's denominator:
a c a·d c·b ad + bc --- + --- = ----- + ----- = ---------- . b d b·d d·b bd
We have not used a subtraction example yet, so let's do p/q - r/s:
p r p·s r·q ps - qr --- - --- = ----- - ----- = ---------- . q s q·s s·q sq
Let's deal with a fascinating question: given two fractions, which is greater and which is smaller? In other words, inequalities again!
Imagine we have 2/3 and 5/7 and we want to establish their inequality relation with < or >, or perhaps =, who knows!
For now, let's connect them with a generic (and invented) symbol like #. For example, we write 2 # 3 before knowing whether # is a =, a < or a >. In the future, we will have unknowns for numbers, and the most universal letter for such a (single) unknown is x. But what about comparison symbols? They also deserve a comparison unknown symbol! So let's use # and try to identify which value it has.
In such a trivial case, it is clear that # = < (according to our sloppy notation for symbols).
But what about the example that Euler is proposing? 2/3 # 5/7.
Euler tells us to convert both fractions to a common denominator (no matter if it is the minimum or not). In this case, the denominators are 3 and 7, both already primes, so for a common denominator (actually for the minimum one), we see that 3 lacks a 7 and 7 lacks a 3. This leads us to (2·7)/(3·7) # (5·3)/(7·3) => 14/21 # 15/21. Once the denominators are equal, we can get rid of them and write 14 # 15, so it is clear that # = <.
What about a more direct trick? Just cross multiply the two fractions!
2 5 --- # --- => 2·7 # 5·3 => 14 # 15 => # = < . 3 7
By "cross" multiplication we mean to multiply the left numerator by the right denominator and *keep the result on the left* and to multiply the right numerator by the left denominator and *keep the result on the write*.
Why have I remarked to keep everything in its proper place? Because if, by chance, you inadvertently flip right and left sides, the symbol in the middle becomes flipped again! And remember that -< = > and that -> = <. So if you mess up and end up writing 5·3 # 2·7 you will get the wrong #. Instead, you could have written 5·3 (-#) 2·7, and then yes, 15 (-#) 14 implies that (-#) = > => # = <.
Since we are with tricks, let's ask another question, one that Euler does not ask but one I consider very interesting: given the fractions a/b and c/d, how does the fraction (a+c)/(b+d) relate to the other two?
Or even simpler, to warm up: given a fraction a/b, how is it related to (a+1)/(b+1)? Is the latter greater or smaller than the former?
Example: given 2/3, what about (2+1)/(3+1) = 3/4? Which one is greater? If we cross multiply 2/3 # 3/4 we get 8 # 9, so this means # = <. But this was just a particular case! We want to have generic conclusions, so let's work with letters.
Let's compare, in general, a/b with (a+n)/(b+n).
a (a+n) --- # ----- => a(b+n) # b(a+n) => ab + an # ba + bn => an # bn => a # b. b (b+n)
What an interesting result! It says that if a < b, then a/b < = (a+n) / (b+n) no matter the value of n. Just switch < by > to see what happens for a > b. Or even see the trivial case of # = =.
If a < b we know that a/b < 1, so let's start adding increasing values of n in their correct order:
a a+1 a+2 a+3 a+n --- < ----- < ----- < ----- < ... < ----- < 1. b b+1 b+2 b+3 b+n
Be sure to fully understand why all these terms are in this order and why they are all smaller than 1.
Be aware here as well that we are dealing only with a,b being both positive numbers! And we are considering n to be positive as well!
Continuing with positive a,b,n let now a be > b, so
a a+1 a+2 a+3 a+n --- > ----- > ----- > ----- > ... > ----- > 1. b b+1 b+2 b+3 b+n
Again, by adding +n up and down, we get closer and closer to 1.
What is the problem with having negatives values of a or b? First, let's discuss cases with negative a and positive b.
In this case, it is always the case that a < b, so that we obtain the first succession, approaching to 1 from bottom side. Nothing changes here.
The real "problem" comes from allowing b to be negative. When this happens, there is a value of n (which is defined positive for now) for which the denominator becomes 0. Then we encounter an "explosive" behaviour.
Depending on the sign of the numerator, the denominator going to 0 will explode towards +∞ or -∞. But don't worry! If we keep adding greater and greater values of n, the value of the fraction *comes back* from ∞. In fact, if it explodes to +∞, it comes back from -∞, and vice versa.
Once accepting this weird round trip excursion, we can generalize the signs for a, b and n and let them positive or negative.
Let's have arbitrary integers a, b and n. Then, as the denominator is (b+n), let's call the problematic value of n as n* which is n*=-b. Once we know this "problematic" value, we can distinguish the right and the left side to it.
To the right side, by adding positive n's, we will always tend towards +1. And to the left side, we will always tend towards +1 as well. Only that if in one wing we approach top-down, in the other we approach bottom-up.
Let's work some examples. If we being by 3/7, we see that n*=-7. Then, if we start adding positive n, like 4/8, 5/9, 6/10, etc, we know we are on the right wing, and we approach towards 1 bottom-up, since 3/7 is below 1. If, on the other hand, we subtract values, like 2/6, 1/5, 0/4, -1/3, -2/2, -3/1, -4/0 and boom! We have evolved towards -∞. From there we enter the left wing, in which next values of -5/(-1), -6/(-2), -7/(-3), -8/(-4) etc, tend towards +1, this time top-down, as any of these values suggest (for example, -8/(-4) = 2 > 1).
If we begin with 4/3, we know that the problematic value is n*=-3. So for positive n we are always on the right wing, and we approach to 1 as 4/3, 5/4, 6/5, 7/6, etc, so it is a top-down approach. If we add negative n, we go 4/3, 3/2, 2/1 and boom, 1/0. We have hit +∞, and will come back from -∞ to have 0/(-1), -1/(-2), (-2)/(-3) and so on, which tends towards +1 again, this time in a bottom-up fashion.
The final step is to allow adding different values at denominator and numerator. So we depart from a/b and we perform (a+m/(b+n) where m and can be different integers. What happens then?
Again there is a problematic zone of n*=-b, and we distinguish its right from its left wing.
The behaviour is all very similar: we hit infinite and come again from it with sign changed. What changes now is that we don't tend to +1. Instead, we know tend (for both wings) towards m/n. And this is valid for every sign for a, b, m and n.
As final conclusions, we can say that the terms
a a + m a + 2m a + 3m m --- , ----- , ------ , ------ , ..., (mind denominator = 0) ------> --- b b + n b + 2n b + 3n n
will end up in m/n. This tendency will be continuous if we don't cross the problematic zone, or we will explode to ∞ if we hit it.
Say you are given two fractions, a/b and m/n, and you want to produce a fraction that lies between the other two. You can just write (a+mq)/(b+nq) for any value of q, as long as there is not a 0-denominator between them. You can also write (qa+m)/(qb+n).
For example, you are given 3/5 and 5/8 and you want a fraction between them. Since we are dealing with all positive quantities, we have no dangerous zones in between, and (3+5)/(5+8) = 8/13 will be between 3/5 and 5/8. But also (3+2·5)/(5+2·8) = 13/21. In fact, we also see how this one lies between 8/13 and 5/8.
Do we want now another one, but this time between 3/5 and 8/13? No problem: for example (3·2+5)/(5·2+8) = 11/18. But we could have used the 8/13 to do (3+8)/(5+13) which gives the same result.
Fractions are really wild creatures! And now we have learned how to compare them, how they evolve, and how to insert new fractions in between.
And if we want to get a fraction that is outside the interval given by 3/5 and 5/8? First, see that 3/5 # 5/8 => 24 # 25, so 3/5 < 5/8. This means that if you want to go to the right of 5/8 you can just do (5+1)/(8+1) = 6/9, and if you want to go to the left of 3/5 you can do (3-1)/(5-1) = 2/4.
We return to Euler's teaching by proposing to subtract a fraction from an integer. For example, 3 - 2/3. We first transform the integer into 9/3, so that 9/3 - 2/3 = 7/3 = 2[1/3].
We don't need to convert the whole 3 into a fraction! We could do 3 - 2/3 = 2 + 1 - 2/3 = 2 + 3/3 - 2/3 = 2 + 1/3 = 2[1/3]. This is much simpler!
Euler reminds us to be aware of fractions for which the numerator is greater than the denominator (called improper fractions). For example, 17/12 is better written as 1[5/12].
What about adding numbers written is this compact form? For example, add 3[1/2] + 2[2/3]. Notice how 3[1/2] is just (3·2+1)/2=7/2 and how 2[2/3] is just (2·3+2)/3=8/3.
First we add the supplements, or as Euler calls them, the fractional residues (i.e. the residue divided by the divisor). So we add 1/2 + 2/3 = 3/6 + 4/6 = 7/6 = 1[1/6]. So we have now 3 integer contributions: 3, 2 and 1. The final result is, then, (3+2+1)[1/6] = 6[1/6]. The conventional sum would be 7/2 + 8/3 = (7·3)/6 + (8·2)/6 = (21+16)/6=37/6, which is equivalent to 6[1/6].
We are proposed to reduce each set of fractions to a common denominator.
(2x)/a and b/c. We clearly get (2cx) / (ac) and (ab) / (ac).
a/b and (a+b)/c. Again we perform the trick of multiplying each fraction by its partner's denominator to get (ac) / (bc) and (ab+b²) / (bc).
(3x)/(2a), (2b)/(3c) and d. The three denominators do not share any factor so let's add to each one what the others have.
We get (9cx) / (6ac), (4ab) / (6ac) and (6acd) / (6ac).
3/4, 2x/3 and a + 2x/a.
We convert the third term to (a²+2x)/a and then we see that the three denominators do not share any factor, so again we add to each one what the others have.
We get (9a) / (12a), (8ax) / (12a) and (12a²+24x) / (12a).
1/2, a²/3 and (x²+a²)/(x+a). Again the denominators do not share anything.
We get, (3x+3a) / (6x+6a), (2a²x+2a³) / (6x+6a) and (6x²+6a²) / (6x+6a).
b/(2a²), c/(2a) and d/a. In this case, the three denominators share an a. This means that 2a² is our common denominator.
We get (a²b) / (2a²), (ca) / (2a²) and (2da) / (2a²).