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.
thanks in advance
Suppose $\displaystyle \sqrt{2\cdot3} = n/m$. Then $\displaystyle 2\cdot3\cdot m^2 = n^2$. 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.
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 $\displaystyle \sqrt{6}$ is rational.
Then $\displaystyle \sqrt{6} = \frac{a}{b}$, where $\displaystyle a, b$ are integers, and $\displaystyle \frac{a}{b}$ is in simplest form (since every rational number can be reduced to simplest form).
Therefore $\displaystyle 6 = \frac{a^2}{b^2}$
$\displaystyle 6b^2 = a^2$
$\displaystyle 2\cdot 3b^2 = a^2$.
Since $\displaystyle a^2$ is even, that means $\displaystyle a$ is even. So write it as $\displaystyle a = 2c$. Then you have
$\displaystyle 2\cdot 3b^2 = (2c)^2$
$\displaystyle 2\cdot 3b^2 = 4c^2$
$\displaystyle 3b^2 = 2c^2$
$\displaystyle b^2 = 2\left(\frac{c^2}{3}\right)$.
Remembering that $\displaystyle b$ is an integer, so is $\displaystyle b^2$. Also, since 2 is a factor, $\displaystyle b^2$ is even, and so $\displaystyle b$ is even. Write it as $\displaystyle b = 2d$.
Remember that $\displaystyle \sqrt{6} = \frac{a}{b}$
but $\displaystyle a = 2c$ and $\displaystyle b = 2d$
So $\displaystyle \sqrt{6} = \frac{2c}{2d}$
$\displaystyle \sqrt{6} = \frac{c}{d}$.
But we said that $\displaystyle \frac{a}{b}$ was already in simplest form.
So we have a contradiction.
This means that $\displaystyle \sqrt{6}$ can not possibly be rational.
Therefore $\displaystyle \sqrt{6}$ is irrational.