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How can we prove that the property of exponents a^m × a^n =a^(m+n) (where "^" this sign denotes the power a is raised to)holds true even if m, n and a are fractions? Like I can clearly see that it will be true if m and n are integers but not when these are fractions.

Bill Dubuque
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Shyam
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2 Answers2

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Let $p,r\in\mathbb{Z}$ and $q,s\in\mathbb{N}$.

Write $x=a^{p/q}=a^{ps/qs}$ and $y=a^{r/s}=a^{qr/qs}$. Then $x^{qs}=a^{ps}$ and $y^{qs}=a^{qr}$. Then $(xy)^{qs}=x^{qs}y^{qs}=a^{ps}a^{qr}=a^{ps+qr}$.

Then $xy=a^{(ps+qr)/qs}=a^{p/q+r/s}$.

J. Chapman
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    Just adding that $p, q, r, s \in \mathbb {N}$. – Simon Jun 19 '24 at 21:09
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    @Simon Edited based on your suggestion. $p$ and $r$ could be 0 or negative as well. Then we are using only that the property $a^m a^n = a^{m+n}$ is true for integers $m$ and $n$. – J. Chapman Jun 19 '24 at 21:13
  • @Shyam: Based on what you've written (question and comments), I suspect both Antony Theo.'s and J. Chapman's answers are pitched a bit more advanced than you are currently ready for. I recommend going through J. Chapman's answer with the specific values $p=1,$ $q=3,$ $r=2,$ $s=5.$ This will show that $a^{\frac{1}{3}} \times a^{\frac{2}{5}} = a^{\frac{1}{3} + \frac{2}{5}}.$ And perhaps try some other possibilities, such as $p=2,$ $q=3,$ $r=4,$ $s=7.$ Doing this should make the "algebra letter-forest" a bit more concrete and comprehensible. – Dave L. Renfro Jun 19 '24 at 21:39
  • @J.Chapman so, I think when we are dealing with fractional exponents we have to assume that atleast one of the properties holds true to get a sense of what those exponents mean to say and then we can derive other properties using that. In your answer too you have assumed that the multiplication property i.e (a^m)^n =a^(m×n) holds true even when m and n are fractions that's why you have equated them in the first step?? – Shyam Jun 20 '24 at 14:15
  • Sorry not in the 1st step but the last step when you move qs inside the bracket – Shyam Jun 20 '24 at 14:44
  • If I'm looking at the correct step, I believe all that is being used is the equivalence $a^{m/n}=(a^m)^{1/n}$ for $m\in\mathbb{Z}$ and $n\in\mathbb{N}$. I would take this as the definition of what we mean by $a^{m/n}$ since we know what $a^m$ means for an integer $m$, and we know what an $n$th root is. Note that $ps+qr\in\mathbb{Z}$ and $qs\in\mathbb{N}$. There is no reason we should understand what the symbol $a^{m/n}$ means until we define it in some way. – J. Chapman Jun 20 '24 at 18:21
  • Hmm I understand, so just by the way we define fractional exponents. We can derive that all of these properties works for them as well – Shyam Jun 21 '24 at 06:14
  • @J.Chapman so when you move qs inside the bracket in last step. If you wouldn't that would mean that what no. On multiplying by itself qs times gives a^(ps+qr) and by DEFINITION it means (ps+qr)/rs so it means rs will divide both of them right – Shyam Jun 21 '24 at 06:18
  • @Shyam Right, by definition, $(a^{(ps+qr)})^{1/qs}=a^{(ps+qr)/qs}$. And at that point $(ps+qr)/qs$ can be simplified/distributed as usual because these are integers we are talking about. – J. Chapman Jun 21 '24 at 14:15
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$$x=a^ma^n$$

Where $m,n\in\mathbb{Q}$ therefore $x=a^{p/q}a^{r/s}$

$$\forall a\in\mathbb{R},x>0\implies\log(x)=\log(a^{p/q} )+\log(a^{r/s})=\frac{p}{q}\log(a)+\frac{r}{s}\log(a)=\log(a)(\frac{p}{q}+\frac{r}{s})=\log\left(a^{\frac{p}{q}+\frac{r}{s}}\right)\ \implies x=a^{p/q}a^{r/s}=a^{\frac{p}{q}+\frac{r}{s}} \text{ QED}$$

Antony Theo.
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    You are using logarithmic property for proving the property of exponents but the logarithmic property itself is a result of property of exponents. So it doesn't make any sense. – Shyam Jun 19 '24 at 20:46
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    There are many ways to prove things, given different frameworks we operate under. Some people define logarithms first and derive exponentials and their properties from them. Other people define exponentials first and then derive logs and their properties from them. If you don't like one proof, maybe you're just in a different framework. I think Antony's proof is pretty much needed if you are proving the property for real exponents, not just rational. – J. Chapman Jun 19 '24 at 21:02
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    @J.Chapman There is an alternate proof for real exponents. If $q_n\to p$ as $n\to\infty$ then define the function $x^p$ as the limit of the functions $x^{q_n}$. – Kepler's Triangle Jun 19 '24 at 21:34