I am currently in Introduction to Real Analysis I. This is a homework problem I'm struggling with.

Here is how I went about my proof of "√3 is irrational":

Claim: √3 is irrational.

Proof: Assume √3 is rational. Thus √3 = p/q where p, q are integers and have no common factors. Since √3 = p/q then 3 = p²/q². Thus 3q² = p². Hence p^2 is divisible by 3 which implies p is divisible by 3 [see(*)proof]. Since p is divisible by 3, p = 3k for some integer k. Now 2q² = p² becomes

3q² = (3k)²

3q² = 9k²

q² = 3k². Therefore q² and hence q are divisible by 3. C! (contradiction) to p and q having no common factors. Therefore √3 is irrational. QED.

Here is the additional proof I am including(*):

Claim: a² is a multiple of 3 → a is a multiple of 3

Proof: Assume a is not a multiple of 3, for some integer a. So a = 3k is not true for any integer k. So a = 3k + 1 or a = 3k + 2 for some integer k.

Case I: Assume a = 3k + 1.

a² = 9k² + 6k + 1

a² = 3(some integer) + 1

a² does not equal 3k for any integer k

Therefore a² is not a multiple of 3

Case II: Assume a = 3k + 2.

a² = 9k² + 12k + 4

a² = 9k² + 12k + 3 + 1

a² = 3(some integer) + 1

a² = 9k² + 12k + 4

Therefore a² is not a multiple of 3.

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Any help, suggestions, corrections, and tips are appreciated. Thanks for your time and help!