Question

Show that √3 is a irrational.​

Answer 1

Answer:

Then r2 is odd and 3r2 is odd which implies that q2 is odd and so q is odd. Since both q and r are odd, we can write q=2m−1 and r=2n−1 for some m,n∈N. ... Therefore there exists no rational number r such that r2=3. Hence the root of 3 is an irrational number.

Step-by-step explanation:

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Answer 2

$\red{\bf {Given:}}$

$\sqrt {3}$

$\red{\bf {To\: find:}}$

$\sqrt {3}$

$\small\bf\color{purple}{is \:an\: irrational\: number.}$

$\red{\bf {Solution:}}$

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

(Were p and q is a co-prime)

$\sqrt {3 \times q}$

Squaring both the side in above equation

$3 \times q^{2} = p^{2}$

if 3 is a factor of$p^{2}$

Then, 3 will also be a factor of p

$\Rightarrow Let\quad p = 3 \times m$

{where m is a integer}

Squaring both sides we get

$\color{darkblue}p^{2} = {3 \times m}^{2}$

$\color{darkblue}p^{2} = 9 \times m^{2}$

$\small\textbf\color{darkgreen}{Substitute the value of}$ $p^{2}$

$\small\textbf\color{magenta}{in the equation}$

$3 \times q^{2} = p^{2}$

$3 \times q^{2} = 9 \times m^{2}$

$q^{2} = 3 \times m^{2}$

If 3 is a factor of $q^{2}$

Then, 3 will also be factor of q

Hence, 3 is a factor of p & q both

$\bold{So,}$

our assumption that p & q are co-prime is wrong.

$\bold{So,}$

$\sqrt {3}$

$\small\textbf{is an "irrational number".}$$\sf\color{red}{Hence\: proved.}$

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