if m and n are prime positive integer prove that root M + root N is a irrational number
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Given:-
M and N are positive primes
To prove:-
√M + √N is an irrational number.
Proof :-
Let √M + √N be a Rational number
Therefore, √M + √N = p/q
==> ( √M + N ) ² = p ²/q ²
[ Squaring on both sides ]
==> M + N + 2 √MN = p ²/q ²
==> 2 √MN = p ²/q ² - M - N
==> 2 √MN = p ² - Mq ² - Nq ²/q ²
==> √MN = p ² - Mq ² - Nq ²/2q ²
Now, √MN which is an irrational number as M and N are primes is equal to a Rational number where ( p ≠ 0, q ≠ 0 , M ≠ 0, N ≠ 0 ) is a contradiction.
This contradiction has arisen because of our assumption that √M + √N is a rational number, which is incorrect .
Hence, √M + √N is an irrational number [ proof by contradiction ].
M and N are positive primes
To prove:-
√M + √N is an irrational number.
Proof :-
Let √M + √N be a Rational number
Therefore, √M + √N = p/q
==> ( √M + N ) ² = p ²/q ²
[ Squaring on both sides ]
==> M + N + 2 √MN = p ²/q ²
==> 2 √MN = p ²/q ² - M - N
==> 2 √MN = p ² - Mq ² - Nq ²/q ²
==> √MN = p ² - Mq ² - Nq ²/2q ²
Now, √MN which is an irrational number as M and N are primes is equal to a Rational number where ( p ≠ 0, q ≠ 0 , M ≠ 0, N ≠ 0 ) is a contradiction.
This contradiction has arisen because of our assumption that √M + √N is a rational number, which is incorrect .
Hence, √M + √N is an irrational number [ proof by contradiction ].
Answered by
2
Answer:
Answer. √M + √N is an irrational number.
If n is a perfect square then √n is a an integer and therefore rational, so it suffices to prove that if n is not a perfect square, then √n is irrational
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