Why are we able to use a particular value and assume that it applies to all other cases in algebra?

Question

I was working on a question and a though came across my mind.

In the question, we had to prove that if n is odd, the gcd(3n, 3n + 2) = 1. So what I did was I said let a = gcd(3n, 3n + 2). Then we know a | 3n and a | 3n + 2 so a must divide any linear combination of 3n and 3n + 2, i.e., a | y(3n) + z(3n+2). Now, suppose y =-1 and z = 1 then a | 2 and so a must only equal 1 as n is odd and so a ≠ 2.

My question is, why are we allowed to take a particular value of y = -1 and z =1 and say that a divides a particular linear combination and then say that the rule that comes from that particular case, applies to all other cases? I.e., why can we say that because the a | 2 and so a must equal 1, then the gcd = 1. What if the linear combination was divisible by some other value that also divides 3n and 3n + 2?

Because I know that it does work through trial and error of various values, but I want to know why. Is there some sort of mathematical principal I am unaware of?

If someone could shed some light on this, it'd be greatly appreciated.

Thank you in advance.

P.s.My professors working out has been attached below.

Prove that if n is odd, then gcd(3n,3n+2)=1. What if n is even?

Solution. Let d = gcd(3n,3n + 2). Since d | 3n and d | 3n + 2, it follows that d | (3n+2)−3n = 2. Since the only divisors of 2 are 1 and 2, it follows that d = 1 or 2. But if n is odd, then 3n is also odd, so we cannot have d=2(since d|3n). It follows that d = 1 if n is odd. (If n is even, then 2 divides both 3n and 3n+2, so d = 2 in this case.)

Welcome to MSE. For some basic information about writing mathematics at this site see, e.g., basic help on mathjax notation, mathjax tutorial and quick reference, main meta site math tutorial and equation editing how-to. — José Carlos Santos, Apr 26 '22 at 06:25
"must divide any linear combination", i.e. all possible combinations, and, in particular, $y=-1, z=1$. — dEmigOd, Apr 26 '22 at 06:34
Yes, but how is it that we can say a = 1 and not any other higher value for other linear combinations? — Oofy2000, Apr 26 '22 at 06:36
@Oofy2000 "assume that it applies to all other cases" $;-;$ There is no such assumption in the proof you quote. It's just a series of direct implications. If you don't follow one of the steps, then ask about that specific step and explain what the problem is that you see with it. — dxiv, Apr 26 '22 at 07:22
Via Euclidean algorithm reduction step $,\gcd(m,m+2) = \gcd(m,2) = 2,$ if $m$ is even, else $1$ for odd $m$ (as in OP when $m = 3n,,$ for odd $n).,$ Your professor is essentially repeating the proof of the linked Euclidean algorithm reduction step (in the special case of the OP). — Bill Dubuque, Apr 26 '22 at 08:19

score 1 · Accepted Answer · answered Apr 26 '22 at 07:28

This is something that might in some spaces be called the "axiom schema of replacement", and it boils down to:

If a statement $P(x)$ is true for all values of $x$, then if you choose a value of $x$ then $P$ must be true when you substitute that value in.

It relates to proof by contradiction - if there is a value of $x$ you can choose that makes $P(x)$ false, then clearly $P(x)$ is not true for every value of $x$.

In this case, it means that since $a$ divides every linear combination of $3n$ and $3n+2$, then if we choose the most restrictive linear combination possible it will still be true that $a$ divides it. And since we happen to be able to make the value $(3n)(-1) + (3n+2)(1) = 2$, we know that $a$ still has to divide this linear combination, otherwise it doesn't divide all of them.

It doesn't matter if $a$ divides every other linear combination, because the theorem isn't "The GCD divides every linear combination except maybe one or two" - either it divides anything you can imagine, or it wasn't actually the GCD to begin with.

I cannot thank you enough for that explanation. Absolutely brilliant. Thank you — Oofy2000, Apr 26 '22 at 07:41

Why are we able to use a particular value and assume that it applies to all other cases in algebra?

1 Answers1