I'm trying to do this proof by contradiction. I know I have to use a lemma to establish that if x is divisible by 3, then x2 is divisible by 3. The lemma is the easy part. Any thoughts? How should I extend the proof for this to the square root of 6?

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I&#039;m trying to do this proof by contradiction. I know I have to use a lemma to establish that if x is divisible by 3, then x2 is divisible by 3. The lemma is the easy part. Any thoughts? How should I extend the proof for this to the square root of 6?

Maricela Alarcon · Accepted Answer

Say 3 is rational. Then 3 can be represented as ab, where a and b have no common factors. So 3=a2b2 and 3b2=a2. Now a2 must be divisible by 3, but then so must a (fundamental theorem of arithmetic). So we have 3b2=(3k)2 and 3b2=9k2 or even b2=3k2 and now we have a contradiction. What is the contradiction?

Elois Puryear · Accepted Answer

suppose 3 is rational then 3=ab for some (a,b) suppose we have ab in simplest form. 3=ab a2=3b2 if b is even, then a is also even in which case a/b is not in simplest form. if b is odd then a is also odd. Therefore: a=2n+1 b=2m+1 (2n+1)2=3(2m+1)2 4n2+4n+1=12m2+12+3 4n2+4n=12m2+12m+2 2n2+2n=6m2+6m+1 2(n2+n)=2(3m2+3m)+1 Since n2+n) is an integer, the left hand side is even. Since (3m2+3m) is an integer, the right hand side is odd and we have found a contradiction, therefore our hypothesis is false.

I'm trying to do this proof by contradiction. I know I have to use a l

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