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if $b^{1/n}$ where $b$ is a real number and $n$ even is defined as the positive real solution to the equation $x^n=b,$ how did they suddenly decide to define the value of $b^{m/n}$ where $m$ and $n$ are integers (rational exponent). did they calculate $(b^{1/n})^m$ and then used the multiplicative property of exponents assuming its still applies even when $m\ne n.$

Thanks

Michael Hardy
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James Doherty
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  • Who is "they"? And what is the question, exactly? Note a similar recent question: https://math.stackexchange.com/questions/2844960/irrational-exponent-understanding/2844963#2844963 – Arnaud Mortier Jul 10 '18 at 16:06
  • What do you mean by m=/=n? Perhaps $m\neq n$? – Crosby Jul 10 '18 at 16:11
  • when m and n are equal you can see from the definition of b^(1/n) that the multiplicative property applies, oh yeah thats what i meant – James Doherty Jul 10 '18 at 16:17
  • For example how was the value of 2^(2/3) decided, was it just (2^(1/3)) and then square that value? In other words was the multiplicative property forced to work here – James Doherty Jul 10 '18 at 16:22

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When $m,n$ are positive integers we have $x^m\cdot x^n=x^{(m+n)}$ and $(x^m)^n=x^{(mn)}$ for every real $x.$

It was found interesting and useful to extend this property to all integers $m,n.$ To be consistent, this requires $x^0=1$ when $x\ne 0,$ and $x^{(-m)}\cdot x^m=1$ when $x\ne 0.$

It was found to be interesting and useful to further extend this property to all rational $m,n.$ To be consistent, this requires $(x^{(1/m)})^m=x$ when $x>0$ and $m\in \Bbb Z^+,$ and $(x^{(1/n)})^m=x^{(m/n)}$ when $x>0.$

It was found interesting and useful to further extend this property to all $m,n \in \Bbb R^+$ by defining $x^y=\lim_{k\to \infty}x^{(q_k)}$ for any sequence $(q_k)_{k\in \Bbb N}$ of rationals converging to $y,$ when $x>0.$

Some restrictions are necessary. For example, there is no real number $x^{(1/2n)}$ when $x<0$ and $n \in \Bbb Z^+.$ And $0^y$ is not defined when $y< 0.$

It is sometimes convenient to define $0^0=1,$ because $x^x$ converges to $1$ when $x$ converges to $0$ through positive values. This definition is often implicitly assumed : When an author writes "Consider a polynomial $p(x)=\sum_{j=1}^nA_jx^j$ " you may very well ask what this means when $x=0$ because the term $A_0x^0$ is $A_0\cdot 0^0$ when $x=0.$ The author is defining $0^0=1$.

DanielWainfleet
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  • Great answer, but I want to note that the extension to all of $\Bbb R$ requires proving that the limit exists; the definition of $x^{m/n}$ requires checking that for any two quotient-of-integer representations of the same rational (e.g., $r = 2/3 = 4/6$), the two different definitions, i.e., $ (x^{1/3})^2$ and $(x^{1/6})^4$ produce the same answer; I say this so that OP will realize that you cannot simply say "it'd be convenient if ..." and then define something. That attitude has led to a million "Why can't I define $0/0$ to be $1$?" questions, and others like them. – John Hughes Jul 10 '18 at 17:05
  • @JohnHughes. I appreciate your comment. I was not in the mood to write a complete textbook-style essay, but merely to touch the main points. – DanielWainfleet Jul 11 '18 at 00:11
  • I figured as much. :) I like your repeated use of "to be consistent." – John Hughes Jul 11 '18 at 11:27
  • Thanks for the answer but just to clarify (not that it needs it, I'm just a bit inferior) , after positive integer exponents were defined on the reals and theorems proved about this definition like the multiplicative property, was the definition of the 0 power, negative integer powers and rational powers created in such a way so that the properties hold for these too? – James Doherty Jul 11 '18 at 12:24
  • Yes. Precisely. Although people knew of such things as square roots, cube roots, etc., from very ancient times, a general theory of exponents came about well after the discovery of logarithms by John Napier. – DanielWainfleet Jul 11 '18 at 15:53
  • "because $x^x$ converges to $1$ when $x$ converges to $0$" That's not the reason. The reason is that $1$ is the only sensible value to use whenever you find yourself needing $0^0$. In fact, it can be defined like that always. This leads to not a single issue that I have ever heard of, and I find it strange that it is not universal. – Arthur Jul 11 '22 at 17:50