MIT 18.02 Multivariable Calculus, Fall 2007 16 Online
OpenStudy (anonymous):

Hi, i'm trying to learn how to do proper mathematical proofs for my assignments, but i can't seem to grasp the correct logic. Could anyone give me any hints or recommend any good material to help (clear books, software, videos, etc...). @Mathematics

OpenStudy (anonymous):

Do you have a specific problem you would like to work out?

OpenStudy (anonymous):

yes. It says to - Prove that if a|c and b|c and gcd(a,b) = 1, then ab|c.

OpenStudy (anonymous):

Alright. So we are going to do whats called a Direct Proof. This is where we take whats given, and manipulate it to get what we want to prove. First lets actually say what we are given.

OpenStudy (anonymous):

We are given:$a\mid c, b\mid c, (a,b)=1$but what does this actually mean? $a\mid c\iff c=ak_1,k_1\in \mathbb Z$this is telling us that a divides c means that c is equal to a times some integer. Likewise:$b\mid c\iff c=bk_2,k_2\in \mathbb Z$

OpenStudy (anonymous):

Im going to use Bezout's Identity for the last part. Bezout's Identity states if the gcd(a,b)=1, then there exist integers x and y such that:$ax+by=1$ So now that ive said what each of the 3 things given to us means, now lets try to think of how i can combine them to prove that:$ab\mid c$

OpenStudy (anonymous):

Taking Bezout's Identity and multiplying by c gives:$ax+by=1\iff acx+bcy=c$ But I know that:$c=ak_1=bk_2$Substituting those values in I obtain:$acx+bcy=c\Longrightarrow abk_2+bak_1=c$

OpenStudy (anonymous):

The left side of the equation is divisible by ab because:$abk_2+bak_1=ab(k_2+k_1)$Therefore, because the left side is divibsle by ab, the right side is as well, and we get:$ab\mid c$

OpenStudy (anonymous):

And that's the final proof?

OpenStudy (anonymous):

yes. Every step is either justified, or is just an algebraic step. thats what makes is a solid proof.

OpenStudy (anonymous):

Ok. Thanks!

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