Question

show that root 3-4 is ir rational number

Answer 1

Hey!

Let us assume that √3 - 4 is a rational number.

and a rational number can be written in the form of p/q where q ≠ 0.

So,

√3 - 4 = p/q { Where p and q are co- prime number and q ≠ 0 }

√3 = p/q + 4

$\sqrt{3} = \frac{p + 4q}{q}$


Since , √3 is an irrational number
and ( p + 4q )/q is a rational number

So, it's a contradiction
[ A rational number can't be equal to an irrational number ]

Hence , √3 - 4 is an irrational number !!

Answer 2

$Heya$‼

Let $\sqrt{3}-4$ is a rational number.

$\sqrt{3}-4=\frac{a}{b}$ ,where a and b are integers and (b ≠ 0) .

$\sqrt{3}=\frac{a}{b}-4$

$\sqrt{3}=\frac{a-4b}{b}$

Since $\frac{a-4b}{b}$ is rational so, $\sqrt{3}$ is also rational.

But this contradicts the fact that $\sqrt{3}$ is an irrational number.

Therefore, we conclude that $\sqrt{3}-4$ is an irrational number.
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