Question

Prove that root 3-root2 is irrational.

Answer 1

Let √3-√2 be a rational number.

A rational number can be written in the form of p/q.

√3-√2 = p/q

Squaring on both sides,

(√3-√2)² = (p/q)²

√3²+√2²-2(√3)(√2) = p²/q²

3+2-2√6 = p²/q²

5-2√6 = p²/q²

2√6 = 5-p²/q²

2√6 = (5q²-p²)/q²

√6 = (5q²-p²)/2q²

p,q are integers then (5q²-p²)/q² is a rational number.

Then √6 is also a rational number.

But this contradicts the fact that √6 is an irrational number.

So,our supposition is false.

Therefore,√3-√2 is an irrational number.

Hence proved

Answer 2

Answer:

To Prove,

√3 - √2 is an irrational number.

Recall the concept:

A rational number is a number that can be expressed in the form, where p and q are integers and q≠0

Solution:

Let us assume the contrary, √3 - √2 is a rational number.

Then there exist two relatively prime integers p and q, with q≠0, such that

√3 - √2 = $\frac{p}{q}$

⇒√3 - $\frac{p}{q}$ =√2

⇒(√3 - $\frac{p}{q}$)² =2

⇒3 -2√3 $\frac{p}{q}$  + $\frac{p^2}{q^2}$ =2

2√3 $\frac{p}{q}$ = 1+ $\frac{p^2}{q^2}$ = $\frac{p^2+q^2}{q^2}$

⇒√3 = $\frac{p^2+q^2}{2pq}$

Since p and q are integers, p²+q² and 2pq are also integers.

So, as per the definition of rational numbers, √3 is a rational number.

Since √3 is not a rational number, our assumption is wrong

That is, √3 - √2 is an irrational number.

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