What Is the Flaw in This Proof of the Greatest Common Divisor?

[kathrynag]

Homework Statement

Suppsoe a, b$$\in$$natural numbers, and d=GCD(a,b). Then d^2=GCD(a^2,b^2). I need to find where the proof goes wrong.

Homework Equations

The Attempt at a Solution

By hypothesis, we have that d divides a and d divides b, so there are integres s and t with a=ds and b=dt. Then a^2=d^2s^2 and so d^2 divides a^2. Similarily d^2 divides b^2. Thus d^2 is a common divisor of a^2 and b^2, as desired.

 

[HallsofIvy]

No, not "as desired". What you "desire" is the greatest common divisor. Saying that d^2 is a common divisor does not mean it is the GREATEST common divisor.

 

[kathrynag]

No, not "as desired". What you "desire" is the greatest common divisor. Saying that d^2 is a common divisor does not mean it is the GREATEST common divisor.

Ok, that made a lot more sense now that I see that!