# Why is gcd important in number theory?

#### honestrosewater

Gold Member
I saw someone say this in another thread. Why is it so important? My best guess is that it has something to do with prime factorization, but that's a pretty wild guess.

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#### matt grime

Homework Helper
it was me, right?

Z is a PID The ideal generated by a and b is the ideal generated by thier gcd. sort of a fundamental result. hcf(a,b) is the least, strictly positive element of the set of all things of the form as+bt, s,t in Z, we get euclid's algorithm from thinking about gcds, we start thinking about modulo arithmetic units, coprime objects, groups, rings, ideals, inverses, this leads us on to is factorization always unique, what are primes really. it is just the starting block of number theory, along with the observation that there are a distinguished set of objects called primes.

#### neurocomp2003

yup part of the foundations of numbertheory...coding it is a beauty.

#### honestrosewater

Gold Member
matt grime said:
it was me, right?
Right.
Z is a PID The ideal generated by a and b is the ideal generated by thier gcd. sort of a fundamental result. hcf(a,b) is the least, strictly positive element of the set of all things of the form as+bt, s,t in Z, we get euclid's algorithm from thinking about gcds, we start thinking about modulo arithmetic units, coprime objects, groups, rings, ideals, inverses, this leads us on to is factorization always unique, what are primes really. it is just the starting block of number theory, along with the observation that there are a distinguished set of objects called primes.
Okay, that sounds important.
I just read a little about PIDs. An ideal in Z generated by a is just {a*z : z in Z}? I can't figure out what you mean by: The ideal generated by a and b is the ideal generated by thier gcd. What is the ideal generated by a and b? The union of the ideal generated by each?

#### matt grime

Homework Helper
it is the set of all things of the form as+bt for s,t in Z.

#### honestrosewater

Gold Member
matt grime said:
it is the set of all things of the form as+bt for s,t in Z.
Okay, that makes more sense. Do you know where I can find a proof of "hcf(a,b) is the least, strictly positive element of the set of all things of the form as+bt, s,t in Z" - does it have a common name I could look up? It sounds interesting.

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#### matt grime

Homework Helper
it is easy to prove - try it. if you know eulcid's algorithm then it is straightforward.

#### honestrosewater

Gold Member
Okay, I'll look up Eulcid's algorithm and give it a try.

#### honestrosewater

Gold Member
I don't see how to use it. I haven't really done anything in number theory- i.e. I have like no basic theorems. I've got:
For all a in Z, there exists some b and c in Z such that bc = a. (b = a and c = 1 gives me this quickly.) So letting cx = a and cy = b, cxs + cyt = c(xs + yt), so any common factor c of a and b divides as + bt. And I know some c exists because it can just be 1. And I know hcf(a, b) > 0, because 1 is always a common factor. But that's really all I have. Is it all correct?
When hcf(a, b) = as + bt = c(xs + yt), I know (xs + yt) is in Z, so c and (xs + yt) are both common factors of hcf(a, b) - and neither are 0. But I seriously can't see how to prove that any common factor of a and b is also a factor of hcf(a, b). How can I prove it?
Am I going in the wrong direction?

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#### matt grime

Homework Helper
For all a in Z, there exists some b and c in Z such that bc = a. (b = a and c = 1 gives me this quickly.)
what are a and b? not as i gave them

So letting cx = a and cy = b, cxs + cyt = c(xs + yt), so any common factor c of a and b divides as + bt. And I know some c exists because it can just be 1.
what is c and what are you doing with it?

And I know hcf(a, b) > 0, because 1 is always a common factor. But that's really all I have. Is it all correct?
When hcf(a, b) = as + bt = c(xs + yt), I know (xs + yt) is in Z, so c and (xs + yt) are both common factors of hcf(a, b) - and neither are 0. But I seriously can't see how to prove that any common factor of a and b is also a factor of hcf(a, b). How can I prove it?

what are you trying to prove?

the theorem i stated was if d is hcf(a,b) then d is the least positive element of the form as+bt for s and t integers.

by euclid's algorithm we know that d=as+bt for some choice of s and t, so all that remians is to show is that d is minimal. well, if c=as+bt for some s and t obviously d divides the RHS and hence c, thus c, if is positive must be greater than d. end of proof.

#### Muzza

Let $$S = \{ax + by; x, y \in \mathbb{Z}\}$$. Obviously S contains some positive (non-zero) elements, so there is a smallest positive element $$d \in S$$. Let $$d = ax_0 + by_0$$.

Define $$P = \{kd; k \in \mathbb{Z}\}$$.

Obviously $$P \subseteq S$$.

To show $$S \subseteq P$$ you need the Euclidean algorithm. Suppose that $$ax_1 + by_1 = u \in S$$. Write $$u = qd + r$$, with $$q, r$$ integers and $$0 \leq r < d$$. Notice that $$r = qd - u \in S$$. But we can't have $$0 < r$$, since then r would be a positive element in S, smaller than the smallest positive element in S! Thus r = 0 and $$u = qd \in P$$, so that $$S \subseteq P$$.

Now all you need to establish is that d is a divisor of both a and b (this follows from the equality of P and S), and that every other common divisor of a and b is smaller than d (you can prove that by showing that if w divides a and b, then w divides d).

#### honestrosewater

Gold Member
matt grime said:
what are a and b? not as i gave them
what is c and what are you doing with it?
What are you trying to prove?
For all a in Z, there exists some b and c in Z such that bc = a. (b = a and c = 1 gives me this quickly.)
I was using them as brand new variables. I need that result for the next step.
So letting cx = a and cy = b, cxs + cyt = c(xs + yt), so any common factor c of a and b divides as + bt.
I want to prove that any common factor of a and b is also a factor of as + bt. So, using the previous step, I let a = cx and b = cy, where c, x, and y are integers. c is a common factor of a and b. By substituting cx for a and cy for b, as + bt = cxs + cyt = c(xt + yt). So any common factor of a and b is also a factor of as + bt. Doesn't that work?
by euclid's algorithm we know that d=as+bt for some choice of s and t, so all that remians is to show is that d is minimal. well, if c=as+bt for some s and t obviously d divides the RHS and hence c, thus c, if is positive must be greater than d. end of proof.
Okay, but how do you know that d = hcf(a, b)?

#### matt grime

Homework Helper
because i decl;ared d to be? or do you mean, how do i know that d can be expressed as an integral combination of a and b? welll, that is what euclid's algorithm proves for us automatically. and you used b for two different things. the first one ought to have been an x (which it becomes halfway through the prrof)

#### honestrosewater

Gold Member
Muzza said:
(you can prove that by showing that if w divides a and b, then w divides d).
I'll have to read the rest again. I seriously can't see how to prove that any common factor of a and b is also a factor of hcf(a, b).

#### matt grime

Homework Helper
Euclid's sodding algorithm does it for you!

there are integers s and t such that hcf(a,b)=as+bt

and divisor of a and b divides the RHS (you proved this yourself!) and hence divdes hcf(a,b)

#### honestrosewater

Gold Member
matt grime said:
because i decl;ared d to be? or do you mean, how do i know that d can be expressed as an integral combination of a and b? welll, that is what euclid's algorithm proves for us automatically. and you used b for two different things. the first one ought to have been an x (which it becomes halfway through the prrof)
I mean
by euclid's algorithm we know that d=as+bt for some choice of s and t,
Okay, but is this not true for any other common factor of a and b? That is, if c is any common factor of a and b, does c = as + bt for some s and t?

#### honestrosewater

Gold Member
matt grime said:
Euclid's sodding algorithm does it for you!

there are integers s and t such that hcf(a,b)=as+bt

and divisor of a and b divides the RHS (you proved this yourself!) and hence divdes hcf(a,b)
Oh, right.

#### matt grime

Homework Helper
honestrosewater said:
I mean
Okay, but is this not true for any other common factor of a and b? That is, if c is any common factor of a and b, does c = as + bt for some s and t?

no, no other factor is expressible in this form. as i proved. the hcf of a and b always divides as+bt.

#### honestrosewater

Gold Member
matt grime said:
no, no other factor is expressible in this form. as i proved. the hcf of a and b always divides as+bt.
Okay, yes, sorry, I get it, just took a little while. I'll be more careful about using variables consistently too. Thanks.

#### AKG

Homework Helper
I think you can look at this proof in 4 stages.

The first stage is trivial if you know the Euclidean Algorithm, it is just that hcf(a,b) = sa + tb for some integers s, t.

Next, you notice that S = {sa + tb: s, t in Z} is generated entirely by its least positive element. Suppose d is the least positive element, and c is in S and is positive, but d does not divide c. Then the division algorithm tells us that there is x such that 0 < x < d and some postive integer f such that c = df + x. But then x = c - df, which must be in S (you can easily check this), and x < d, contradicting our choice of d, so d really does divide all c in S.

Next is to notice that such d divides a and b. This follows directly from the previous "stage" since a and b are the combinations of the form 1a + 0b and 0a + 1b, so they're in S, and so d divides them. So d is a common factor of a and b.

It remains to show that d is the greatest common factor of a and b. The greatest common factor of two numbers has the property that it is divisible by all common factors of a and b. So we show that if x is a common factor of a and b, then it divides d. If x divides a and x divides b, then xc = a, xf = b. But then d = sa + tb = s(xc) + t(xf) = x(sc + tf), so x divides d, as required.

#### honestrosewater

Gold Member
Okay, I understand that one too. Thanks.

#### AKG

Homework Helper
I think you can prove the Euclidean Algorithm easily too. It can be proven as other recursive algorithms are proven: show that it works for a "base case", show that the general case leads to the base case, and that if it works at one stage of the process, then it works at the previous stage. If we want to find gcd(a,b), then the base case is when either a|b or b|a. In this case, we choose either a or b, and either choice is a "linear" combination of a and b. If neither a divides b nor b divides a, then without loss of generality, say a < b. Define c as b (mod a). So 0 < c < a, and b = ka + c for some natural k. Is gcd(c,a) equal to gcd(a,b)? Define x = gcd(c,a) and y = gcd(a,b). x divides every positive "linear" combination of c and a (that is, any number of the form sc + ta > 0 for integers s, t) since if xc' = c and xa' = a, then:

sc + ta = s(xc') + t(xa') = x(sc' + ta')

But a is obviously a linear combination of a and c, and since b = ka + c, b is also a linear combination of the two. So x divides a and b, hence x|y. y divides a and b, and since c = b - ka is a positive linear combination of b and a, y|c, so y divides both a and c, hence y|x. x|y and y|x give us x = y.

It remains to prove that the Euclidean Algorithm will eventually take us to a base case. Well we can see that if at one stage we are taking a and b as arguments with a < b, then at the next stage we're taking a and c = b (mod a) < a, so now we have two numbers less than or equal to a. You can see that this process keeps replacing the larger number of the pair with a new number that is smaller than the smaller of the original pair. This process can't go on forever; in the "worst" case, we will come down to something like a and 1, and of course this satisfies the base case conditions because 1|a, since 1 divides everything.

Unless I made a mistake in the above, you can prove this entire theorem from scratch.

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