prove that root n is not a rational number, if n is perfect square
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If n is not a perfect square then is irrational
Let on the contrary say it is rational .
Then
where p and q are coprime integers.
so n =p2/q2
p2 =nq2
This shows p divides q
which is a contradiction.
Hence is irrational if n is not a perfect square
divya326:
but the question is when n is a perfect square
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If a root n is a perfect square such as 4, 9, 16, 25, etc., then it evaluating the square root quickly proves it is rational. The square root of the perfect square 25 is 5, which is clearly a rational number. To prove a root is irrational, you must prove that it is inexpressible in terms of a fraction a/b, where a and b are whole numbers. For the nth root of x to be rational: nth root of x must equal (a^n)/(b^n), where a and b are integers and a/b is in lowest terms. So the square root of 3 for example would be done like this: 2nd root of 3 must equal (a^2)/(b^2) 3 = (a^2)/(b^2) 3(b^2) = (a^2) So we know a^2 is divisible by three. (b^2) = (a^2)/3 So we know b^2 is divisble by three. Since this is neither of these are in lowest terms, the sqrt(3) is an irrational number.
Hope this will help you :-))
Biee ^_^
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Hope this will help you :-))
Biee ^_^
❤❤
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