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Name: Alex L.
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My students have asked me to explain the reason for the link between fractional exponents and roots (square root, cube root, etc). I know that x^(1/2) equals the square root of x, but I do not know why. The best explanation I have found is, "If we want to make any sense of, say, 9^(1/2), and have the laws of exponents continue to work, we are forced to define it as the square root of 9." Is there a better explanation or a step-by-step derivation from x^(1/2) to the square root of x?

The way my math books answer this problem is just like you state: In order for the laws of exponents (i.e. a^m*a^n=a^(m+n); (a^m)/(a^n) = a^(m-n); (a^m)^n = a^(m*n))to work, the following definitions are required

a^(p/q)= qth root of(a^p)

I know that does not seem to make it any clearer, but you may want to show them the relationship between exponents and logarithms.

If x=a^y then y=Log(base a)of x.

Run them through some numbers like the square root of 10 and the cube root of 10 because their calculators have the log base 10 function on them. You can at least show them that the numbers fall out. The one problem you might have with this avenue is that you might now have trouble explaining logarithms ;-).

Christopher Murphy, P.E.


For integral exponents, we know: (a^m)^n = a^(mn). There is no underlying rule. It is derived through observation. a^m is m a's multiplied together. (a^m)^n is n sets of m a's multiplied together, or (mn) a's multiplied together. There is no reason why this relation should not hold for real numbers as exponents. Without really knowing what a non-integer exponent represents, the relation is defined to be true for real exponents rather than just integer exponents. (a^(1/n))^n=a^((1/n)n)=a. Also, the (nth root of a)^n=a. As a result, we can discover that a^(1/n)=nth root of a.

Dr. Ken Mellendorf
Physics Professor
Illinois Central College

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