# Thread: prove that root 6 is irrational

1. ## prove that root 6 is irrational

plzz prove that root 6 is irrational.

i have to show the whole prove not just say that 6 is not a perfect square so root 6 is irrational. i have to prove it wholly.

2. Suppose $\sqrt{2\cdot3} = n/m$. Then $2\cdot3\cdot m^2 = n^2$. Factor both sides into primes. Then the power of 2 on the RHS is even because it is a square, but the power of 2 on the LHS is odd. This contradicts unique factorization.

3. extremely sryy but i can't understand it properly plzz give me detail.

4. Originally Posted by saha.subham
plzz prove that root 6 is irrational.

i have to show the whole prove not just say that 6 is not a perfect square so root 6 is irrational. i have to prove it wholly.

Or note that $\sqrt{6}$ is a solution to $x^2-6=0$ and the only rational solutions to that are...

P.S. It follows that $\sqrt{2}+\sqrt{3}\notin\mathbb{Q}$

5. srryy but i can;t understand without details

6. But I thought you said you had prove it. What do you understand about proving that roots are not rational?

7. haha! If you square root anything but a perfect square then the answer will be irrational!

PROVEN!!

8. Originally Posted by jgv115
haha! If you square root anything but a perfect square then the answer will be irrational!

PROVEN!!
I have a feeling that if the OP is going to say that then that statement has to be proven in the process.

Assume that $\sqrt{6}$ is rational.

Then $\sqrt{6} = \frac{a}{b}$, where $a, b$ are integers, and $\frac{a}{b}$ is in simplest form (since every rational number can be reduced to simplest form).

Therefore $6 = \frac{a^2}{b^2}$

$6b^2 = a^2$

$2\cdot 3b^2 = a^2$.

Since $a^2$ is even, that means $a$ is even. So write it as $a = 2c$. Then you have

$2\cdot 3b^2 = (2c)^2$

$2\cdot 3b^2 = 4c^2$

$3b^2 = 2c^2$

$b^2 = 2\left(\frac{c^2}{3}\right)$.

Remembering that $b$ is an integer, so is $b^2$. Also, since 2 is a factor, $b^2$ is even, and so $b$ is even. Write it as $b = 2d$.

Remember that $\sqrt{6} = \frac{a}{b}$

but $a = 2c$ and $b = 2d$

So $\sqrt{6} = \frac{2c}{2d}$

$\sqrt{6} = \frac{c}{d}$.

But we said that $\frac{a}{b}$ was already in simplest form.

So we have a contradiction.

This means that $\sqrt{6}$ can not possibly be rational.

Therefore $\sqrt{6}$ is irrational.

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# square root of 6 is irrational

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