Prove that √3-√2 is a irrational
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Let √3-√2 be a rational number
A rational number can be written in the form of p/q.
√3 - √2 = p/q
√3 = p/q + √2
(√3)² = (p/q + √2)²
3 = (p/q)² + √2² + 2(p/q)(√2)
3 = p²/q² + 2 + 2√2p/q
3 - 2 = p²/q² + 2√2p/q
1 = p²/q² + 2√2p/q
2√2p/q = 1 - p²/q²
2√2p/q = (q²-p²)/q²
√2 = (q²-p²)/q² × q/2p
√2 = (q²-p²)/2pq
p,q are integers then (q²-p²)/2pq is a rational number.
Then √2 is also a rational number.
But this contradicts the fact that √2 is an irrational number.
So,our supposition is false.
Therefore, √3-√2 is an irrational number.
A rational number can be written in the form of p/q.
√3 - √2 = p/q
√3 = p/q + √2
(√3)² = (p/q + √2)²
3 = (p/q)² + √2² + 2(p/q)(√2)
3 = p²/q² + 2 + 2√2p/q
3 - 2 = p²/q² + 2√2p/q
1 = p²/q² + 2√2p/q
2√2p/q = 1 - p²/q²
2√2p/q = (q²-p²)/q²
√2 = (q²-p²)/q² × q/2p
√2 = (q²-p²)/2pq
p,q are integers then (q²-p²)/2pq is a rational number.
Then √2 is also a rational number.
But this contradicts the fact that √2 is an irrational number.
So,our supposition is false.
Therefore, √3-√2 is an irrational number.
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