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Fractional Indices

  1. Jul 20, 2009 #1


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    I've always just accepted that [tex]x^\frac{1}{2}=\sqrt{x}[/tex] but I don't understand why. I guess I'm looking for a proof for this, and possibly lead this onto:


  2. jcsd
  3. Jul 20, 2009 #2
    [x^(m/n)] ^ n = [x ^ (m/n * n)] = [x ^ m] for all n.

    root_n(x^m) ^ n = x^m for all n.

    Since raising something to a power is single-valued for real numbers, and since raising root_n(x^m) to the power n yields the same thing as raising [x^(m/n)] to the power n, it only makes sense that root_n(x^m) = x^(m/n).

    I don't think this counts as a proof, but it should probably help show that the notation is consistent.
  4. Jul 20, 2009 #3


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    I can't believe how I could have missed that. Of course, [tex](x^\frac{1}{2})^2=x[/tex]
    And considering the square-root's definition, thus [tex]x^\frac{1}{2}=\sqrt{x}[/tex]

    The same idea goes of the n-th root.

    Again, I can't believe I missed that.
  5. Jul 21, 2009 #4
    I was worried that this wouldn't be a sufficient answer to your question. You see, I think this is probably how it worked out...

    (1) People started with integer powers as a shorthand for multipling a number n times.
    (2) Somebody proved (x^n)^m = x^(nm).
    (3) Somebody realized that since sqrt(x)^2 = x = x^1, then sqrt(x) could be thought of as x^(1/2) (whatever that means) because (1/2)*2 = 1.
    (4) They then proved that it worked for any rational power, and from that it was probably pretty easy to generalize to irrational numbers.

    Maybe somebody with more knowledge of the history of mathematics can clear this up for me?

    I bet that's also how they got negative exponentials.

    x^2 * 1/(x^2) = 1 = x^0, so 1/(x^2) = x^(-2) since 2+(-2)=0. etc.

    Then again, perhaps the rules for manipulating exponents came later? Good questions.
  6. Jul 21, 2009 #5


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    Yes it is pretty interesting.

    If I remember correctly, my teacher told us that while [tex]x^0[/tex] doesn't make sense (multiplying x by itself 0 times), if you use the exponential proofs it works. Also by using limits as [tex]x \rightarrow 0[/tex] from both positive and negative ends, it does limit to 1. Thus [tex]x^0=1[/tex]

    This problem probably then came to be after the proof of [tex]x^ax^b=x^{a+b}[/tex] and possibly before the proof of all exponentials to any rational.
  7. Jul 21, 2009 #6


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    "Using limits as [itex]x\rightarrow 0[/itex]"? But x is the base in your formula. Did you mean [itex]\lim_{y\rightarrow 0} x^y[/itex]?

    That strikes me as invalid anyway. How do you define x to negative or fractional (or irrational) powers if you haven't already defined x0?

    Of course, we don't "prove" that a0= 1, that's a definition. You may be thinking of motivating that definition.

    It is easy to show that, for positive integers m and n, and any positive real a, that aman= am+n. Basically, that is just "counting".

    In order that that rule still hold for "0", we must have a0an= a0+n= an. Dividing both sides by an (which is why we need a non-zero) we get a0= 1.
    Now, if want that rule to hold for negative integers as well, we must have a-nan= a-n+ n= a0= 1. That is, in order for that rule to still hold for negative powers, we must define a-n= 1/an.

    We can also prove that, for any positive integers, m and n, and any positive a, that [itex](a^n)^m= a^{mn}[/itex]. If we want that to hold for fractional powers as well, we must have [itex](a^{1/n})^n= a^1= a[/itex]. That is [itex]a^{1/n}[/itex] must be an nth root of a. Those only exist (in the real numbers) if a is positive which is the reason for the "any positive a" restriction above. Also there two real nth roots of a positive number. In order that we can continue to take roots, we define a1/n to be the positive nth root of a. Of course, from [math](a^m)^n= a^{mn}[/math] it follows that am/n= (a1/n)m= (am)1/n.

    Finally, we define ax for irrational x by requiring that ax be a continuous function. If ax is continuous and [itex]\left{r_n\right}[/itex] is a sequence of rational numbers converging to x, we must define
    [tex]a^x= \lim_{n\rightarrow\infty}a^{r_n}[/itex].

    Note that none of those are proofs. We don't prove definitions. They are motivations for the particular definitions we use.
    Last edited by a moderator: Jul 21, 2009
  8. Jul 21, 2009 #7


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    Hallsofivy, thanks for your insights. And yes, I meant for the index to be approaching 0.

    Was this idea shown/proven only because it could not be proven for negative "a" as there have been contradictions with the complex numbers? Is there more reasoning behind this statement?
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