Question

prove that (√2+√3) is irrational​

Answer 1

Answer:

ANSWER

Let us assume that

2

+

3

is a rational number

Then. there exist coprime integers p, q,q



=0 such that

2

+

3

=

q

p

=>

q

p

−

3

=

2

Squaring on both sides, we get

=>(

q

p

−

3

)

2

=(

2

)

2

=>

q

2

p

2

−2

q

p

3

+(

3

)

2

=2

=>

q

2

p

2

−2

q

p

3

+3=2

=>

q

2

p

2

+1=2

q

p

3

=>

q

2

p

2

+q

2

×

2p

q

=

3

=>

2pq

p

2

+q

2

=

3

Since, p,q are integers,

2pq

p

2

+q

2

is a rational number.

=>

3

is a rational number.

This contradicts the fact that

3

is irrational.

Thus, our assumption is incorrect.

Therefore,

2

+

3

is a irrational.

Answer 2

Answer:

Let as assume that √2 + √3 is a rational number .

Let as assume that √2 + √3 is a rational number .Then , there exists co - prime positive integers p and q such that

$\sqrt{2 + \sqrt{3} }$

⬆Is equals to p÷q

➡p÷q-root 3-root 2

Squaring on both sides.

So,

we get

(p÷q-root 3)

$(p \div q - \sqrt{3} ) {}^{2} = ( \sqrt{2) ^{2} } \\ {p {2}^\div {q}^{2} } - 2pq \sqrt{3} + 3 = 2 \\ {p}^{2} \div {q}^{2} + 3 - 2 =2 \sqrt{3} \: p \div q \\ p {2} \div {q}^{2} + 1 = 2 \sqrt{3} \: p \div q \\ \sqrt {p}^{2} \div {q}^{2} + 1 = 2 \sqrt{3} p \div q \\ ( {p}^{2} \ + {q}^{2} ) \div ({q}^{2}) \times q \div 2p \\ = \sqrt{3} \\ {p}^{2} + {q}^{2} \div 2pq \\ = \sqrt{3} \\ therefore \\ \sqrt{3} is \: a \: rational \: number$

I hope it helps you