Given that under root 2 is irrational. Prove that 5+3root2 is an irrational
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Answered by
1
Heya !!
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Given that, √2 is rational.
Let 5+3√2 be irrational.
5+3√2 = a/b, where a and b are integers and (b≠0)
=> 3√2 = (a/b) – 5
=> √2 = (a–5b) / 3b
Therefore (a–5b) / 3b is irrational as √2 is irrational.
But this contradiction has arisen because of our incorrect consumption.
So, we conclude that 5+3√2 is irrational number.
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Hope my ans.'s satisfactory.☺
==================================
Given that, √2 is rational.
Let 5+3√2 be irrational.
5+3√2 = a/b, where a and b are integers and (b≠0)
=> 3√2 = (a/b) – 5
=> √2 = (a–5b) / 3b
Therefore (a–5b) / 3b is irrational as √2 is irrational.
But this contradiction has arisen because of our incorrect consumption.
So, we conclude that 5+3√2 is irrational number.
==================================
Hope my ans.'s satisfactory.☺
Answered by
1
here your answer
we can assume the given no. as rational and use contradiction method to solve it
HOPE IT WILL HELP YOU
we can assume the given no. as rational and use contradiction method to solve it
HOPE IT WILL HELP YOU
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