1. ## proof by contradiction

Hi guys, having problems solving this problem. Question is prove by contradiction that $\sqrt{3}$ is irrational. Any help would be highly appreciated.

2. Ok here is my attempt.
Suppose that $\sqrt{3}$ is rational. Therefore $\sqrt{3}$ = $\frac{m}{n}$ for some integers m, n with n $\neq$ 0.
We may suppose without loss of generality (wlog) that m, n have no common factor (i.e. are coprime). Now $m^2$ = $3n^2$ and so $m$ is even. Put $m = 2k$. Then $4k^2$ = $2n^2$ and so $n^2$ = $2k^2$
which implies that n is even. This contradicts the assumption that m, n are coprime.
Hence $\sqrt{3}$ must be irrational.

Is this correct?

3. Hello, jvignacio!

You've mixed in 2's, 4's, and even numbers.

Prove by contradiction that $\sqrt{3}$ is irrational.
Assume that $\sqrt{3}$ is rational.

Then: . $\sqrt{3} \:=\:\frac{a}{b}$ for positive integers $a$ and $b$, where $a$ and $b$ are coprime. .[1]

Square the equation: . $\left(\sqrt{3}\right)^2 \:=\:\left(\frac{a}{b}\right)^2 \quad\Rightarrow\quad 3 \:=\:\frac{a^2}{b^2} \quad\Rightarrow\quad 3b^2 \:=\:a^2$

Since the left side is divisible by 3, the right side is also divisible by 3.
. . Hence: . $a \,=\,3c$ .[2]

The equation becomes: . $3b^2 \:=\:(3c)^2 \quad\Rightarrow\quad 3b^2 \:=\:9c^2 \quad\Rightarrow\quad b^2 \:=\:3c^2$

Since the right side is divisible by 3, the left side is also divisible by 3.
. . Hence: . $b \,=\,3d$ .[3]

But [2] and [3] say that $\text{GCD}\,(a,b) \,=\,3$ . . . which contradicts [1].

. . $Q.E.D.$

4. Originally Posted by Soroban

Since the left side is divisible by 3, the right side is also divisible by 3.
. . Hence: . $a \,=\,3c$ .[2]
dont understand how you got to the a = 3c part?

thanks

5. Square the equation: .

Since the left side is divisible by 3, the right side is also divisible by 3.
. . Hence: . .
dont understand how you got to the a = 3c part?
That's because, if $a^2$ is divisible by 3, then so is $a$. In order for 3 to divide $a\cdot a$, one could try breaking 3 into two factors: $3=pq$ so that $p\mid a$ and $q\mid a$, but $pq\not\;\mid a$ ( $x\mid y$ means " $x$ divides $y$"). However, 3 is prime, and this is impossible.

Since $3\mid a$, we have $9\mid a^2$, and we can divide both sides of $3b^2=a^2$ by 3.