# Math Help - [SOLVED] Proving the irrationality of √3.

1. ## [SOLVED] Proving the irrationality of √3.

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.

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

2. Hello,
Originally Posted by ilikedmath
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 3q² = 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.
Perfect

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.
Yes, this is a way of proving it. But you can prove it for any prime number p :

p divides a² iff then p divides a.

Suppose the prime factorisation of a :

$a=\prod_{i=1}^n p_i^{\alpha_i}$ (where $p_i$ are all prime numbers dividing a and $p_i \neq p_j \quad \forall i \neq j$)
We have :
$a^2=\prod_{i=1}^n p_i^{2 \alpha_i}$ (using basic exponents rules)

Any prime number dividing a² has an even power. So if p divides a², at least p² also divides a².
So p divides a.

Huh...my proof is experimental, so if you have any remark on this...

3. ## thank you

thanks for looking at this and for the proof of primes there.