Question

prove that√3 is irrational number​

Answer 1

Answer:

Given:

\sqrt {3}

3

To find:

\sqrt {3}

3

is an irrational number.

Solution:

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

3

=

q

p

(Were p and q is a co-prime)

\sqrt {3 \times q} = p

3×q

=p

Squaring both the side in above equation

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

3 \times q^{2} = p^{2}3×q

2

=p

2

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

2

Then, 3 will also be a factor of p

\Rightarrow Let\quad p = 3 \times m⇒Letp=3×m {where m is a integer}

Squaring both sides we get

p^{2} = {3 \times m}^{2}p

2

=3×m

2

p^{2} = 9 \times m^{2}p

2

=9×m

2

Substitute the value of p^{2}p

2

in the equation

3 \times q^{2} = p^{2}3×q

2

=p

2

3 \times q^{2} = 9 \times m^{2}3×q

2

=9×m

2

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

2

=3×m

2

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

2

Then, 3 will also be factor of q

Hence, 3 is a factor of p & q both

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

So, \sqrt {3}

3

is an "irrational number". Hence proved.

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