Question

prove that root 3 plus root 2 is irrational​

Answer 1

Answer:

so root 3 plus root 2 is irrational.

Answer 2

ANSWER:

• (√3+√2) is an irrational number.

GIVEN:

• Number = √3 + √2

TO PROVE:

• (√3+√2) is an irrational number.

SOLUTION:

Let (√3+√2) be a rational number which can be expressed in the form of p/q where p and q have no other common factor than 1.

$\implies \: \sqrt{3} + \sqrt{2} = \dfrac{p}{q} \\ \\ \implies \: \sqrt{3} = \dfrac{p}{q} - \sqrt{2} \\ \\ \: \: \: \: squaring \: both \: sides \: we \: get. \\ \\ \implies \: ( \sqrt{3} ) {}^{2} = ( \dfrac{p}{q} - \sqrt{2} ) {}^{2} \\ \\ \implies \: 3 = \dfrac{p {}^{2} }{q {}^{2} } + 2 - \frac{2 \sqrt{2}p }{q} \\ \\ \implies \: 3 - 2 - \frac{p {}^{2} }{q {}^{2} } = - \frac{2 \sqrt{2}p }{q} \\ \\ \implies \: 1 - \frac{p {}^{2} }{q {}^{2} } = - \frac{2 \sqrt{2} p}{q} \\ \\ \implies \: \frac{q {}^{2} - p {}^{2} }{q {}^{2} } = - \frac{2 \sqrt{2}p }{q} \\ \\ \implies \: \frac{ - (p {}^{2} - q {}^{2}) }{q {}^{2} } \times q \: = - 2 \sqrt{2} p \\ \\ \implies \: \frac{p {}^{2} - q {}^{2} }{q} = 2 \sqrt{2} p \\ \\ \implies \: \sqrt{2} = \frac{p {}^{2} - q {}^{2} }{2pq}$

• Here (p²-q²)/2pq is rational but √2 is Irrational.
• Thus our contradiction is wrong .
• So (√3+√2) is an irrational number.