prove that root n is not a rational no. , if n is not 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.
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.
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