2 under root3 prove that this no is 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
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
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
quest2:
But the question is 2√3
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