Question

Porove that following irrational √6​

Answer 1

Answer:

Prove That Root 6 is Irrational by Contradiction Method

As we know a rational number can be expressed in p/q form, thus, we write, √6 = p/q, where p, q are the integers, and q is not equal to 0. The integers p and q are coprime numbers thus, HCF (p,q) = 1.

Answer 2

Answer:

$\sqrt{6}$ is irrational.

Step-by-step explanation:

Proof:

Suppose that $\sqrt{6}$ is rational, then we can write "$\sqrt{6} =\frac{p}{q}$, where p, q are integers with no common factors".

squaring $\sqrt{6} =\frac{p}{q}$, $6=\frac{p^{2} }{q^{2} }$

  implies          $6q^{2}=p^{2}$

 implies $p^{2}$ is a multiple of $6$ and hence p is a multiple of $6$, $p=6m$

that is, $p^{2} =(6m)^{2}$,  where m is any integer.

hence write $6q^{2}=(6m)^{2}$

    that is        $q^{2} =6m^{2}$

that means $q^{2}$ is a multiple of $6$ and hence $q$ is a multiple of $6$.

that is we get $6$ as a common factor for both $p$ and $q$.

which is a contradiction to our assumption. hence our assuption is wrong.

That is, $\sqrt{6}$ is rational.

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