What is the proof for a = gcd(a, b) when a|b?

hoopsmax25
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Homework Statement



Suppose a, b ∈ N and a|b. Prove that a = gcd(a, b).

Homework Equations


Seems easy intuitively but actually proving it is giving me problems.


The Attempt at a Solution


I have been trying to use the fact that gcd(a,b)=na + mb here m and n are integeres but got stuck.
 
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Well we clearly have a|a and a|b, and the largest divisor of a number is itself. This should follow immediately.
 
Yeah and i understand that but we are asked to prove it, not explain why.
 
Well you can probably prove by contradiction that the largest divisor of a number is itself. I mean a|b => a=gcd(a,b) is so trivial that there's very little to say, how formal does your professor expect your proof to be? You can take it all the way down to pure logic if you really wanted to.
 
The difference between a proof and an explanation actually becomes increasingly fuzzy the farther you get in math.
 
A proof of this is certainly possible.

But first, we would have to know how exactly you defined a|b and gcd(a,b).
 
Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...
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