Question

prove that
$\sqrt{6}$
is a irrational


​

Answer 1

Answer:

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Step-by-step explanation:

$\sqrt{6} is equal to \sqrt{2*3}\\and since \sqrt{2} and \sqrt{3} is irrational so \sqrt{6} is also irrational.$

Since Irrational when multiplied with irrational number is irrational so under root 6 is irrational number.

Answer 2

$\huge\mathfrak\pink{Question}$

$prove \: that$

$\sqrt{6}$

$is \: a \: irrational$

$\huge\mathfrak\purple{Solution}$

The following proof is a proof by contradiction.

$Let \: us \: assume \: that$

$\sqrt{6}$

$is\: rational \: number.$

Then it can be represented as fraction of two integers.

Let the lowest terms representation be:

$\sqrt{6}$

$= \frac{a}{b} \: where \: b \: not \: = 0$

Note that this representation is in lowest terms and hence, a and b have no common factors

$a² =6b²$

From above a²is even. If a² is even, then a should also be even.

$⟹a=2c$

$4c² =6b²$

$2c²=3b²$

From above 3b² is even. If 3b² is even, then b² should also be even and again b is even.

But a and b were in lowest form and both cannot be even. Hence, assumption was wrong and hence,

$\sqrt{6}$

is an irrational number.

Hope it's helpful for you .