$$ -1=(-1)^1=(-1)^{2\frac{1}{2}}=((-1)^2)^{\frac12}=1^\frac12=1 $$ I can't find a mistake. I think it might be that $$a^{mn}=(a^m)^n$$ doesn't hold for all $$a,m,n$$ but I don't know exactly. Can anyone help?
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5That's exactly it. It only holds for non-negative $a$. – Arthur Jul 19 '18 at 17:38
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The square root function does not preserve sign. – DynamoBlaze Jul 19 '18 at 17:39
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@MalayTheDynamo Yes it does, it is defined for non-negative numbers only and outputs non-negative numbers only. – Arnaud Mortier Jul 19 '18 at 17:40
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@ArnaudMortier True. But it doesn't here. You'd have to write $1^{\frac12}=\pm1$ – DynamoBlaze Jul 19 '18 at 17:41
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3@MalayTheDynamo This makes no sense. It does, period. Otherwise you're not talking about the square root function. Please avoid comments that can only confuse the OP. – Arnaud Mortier Jul 19 '18 at 17:44
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1$(a^m)^n=a^{mn}$ if i) $a$ is a non-negative real number ii) $m,n$ are integers or iii)(a more general case of ii) $m,n$ are rational numbers so that when written in fractions in the lowest terms the denominator is odd. If $m,n$ are integers and $a^m$ is $a$ multiplied by itself $m$ times then $(a^m)^n=a^{mn}$ is easily shown. If we think "well is $(a^{\frac 1m})^m=a^{\frac 1mm}=a^1=a$ so $a^{\frac 1m}$ must be whatever $x$ is so that $x^m = a$" we have two problems: what if there is no such number? What if there is more* than one such number? ... to be continued... – fleablood Jul 19 '18 at 18:25
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1"what if there is no such number? What if there is more than one such number?" If $a>0$ there is always exactly one such positive number that is true. So we define $a^{\frac 1m}$ as being the positive square root but only on a positive base. If you look at that proof $(-a)^2=a^2$ and $(a)^2=a^2$ so to go backwards from $a^2\to ???$ we have two choices. By definition we go back to the positive one. And... that simply was not the one we started from. So $(a^2)^{\frac 12} = a$ is simply not a rule we can have. But $(a^2)^\frac 12=|a|$ so a rule we can have. – fleablood Jul 19 '18 at 18:31
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Square roots are tricky, there are always two solutions:
$$1^{\frac{1}{2}} = \pm 1$$
So the part of the proof which goes $(-1)^1 = (-1)^{2 \frac{1}{2}}$, introduces $1$ as a second solution.
More generally, over the reals, $$(a^2)^{\frac{1}{2}} = \pm a$$
packetpacket
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This is intuitively true. Unfortunately, the range of the square root function is dogmatically defined as positive only. – David Diaz Jul 19 '18 at 17:44
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1More generally, over the reals, $\sqrt{a^2}=|a|$. The square root function outputs only one output at a time, as is the case for all functions. While true that $\sqrt{a^2}=a$ or $\sqrt{a^2}=-a$ for certain $a$, it is only ever equal to one of those at a time and never both (except when $a=0$). The square root of $1$ is equal to $1$ and $1$ only. The square root of $1$ is not equal to $-1$. You seem to be confusing the square root function with the operator which returns the set of roots, a different thing and notated differently. – JMoravitz Jul 19 '18 at 17:45
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@DavidDiaz My mistake, see my edit. I believe the answer still applies. – packetpacket Jul 19 '18 at 17:45
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@DavidDiaz The square root function is a multivalued function. In fact, $z^w$ is defined as $e^{w\log(z)}$ where the logarithm function is defined as $\log(z)=\text{Log}(|z|)+i(\text{Arg}(z)+2n\pi)$ and $-\pi<\text{Arg}(z)\le \pi$ is the principal argument of $z$ and $\text{Log}(|z|)$ is the logarithm for real numbers. So, for $z=1$ and $w=1/2$, we find the two values for $\sqrt{z}$ are $\pm 1$. – Mark Viola Jul 19 '18 at 17:51
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@JMoravitz The notation should be clear, but if you wish to distinguish the complex logarithm from the logarithm for the reals, your notation suffices. – Mark Viola Jul 19 '18 at 17:53
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@MarkViola you seem to misunderstand my comment. $\log$ is multivalued over complex numbers, sure. $\text{Log}$ on the other hand is single-valued, even over the complex numbers. I.e. force $n$ to be equal to zero in what you wrote above. – JMoravitz Jul 19 '18 at 17:54
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@MarkViola then I do not understand how you still claim that the square root function is multivalued when it returns only a single unambiguous output given an input. I was saying to change your definition of exponentiation to $z^w:=e^{w\text{Log}(z)}$. – JMoravitz Jul 19 '18 at 17:57
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@JMoravitz $z^w\equiv e^{w\log(z)}=e^{w\text{Log}(|z|) +i(\text{Arg}(z)+2n\pi )w}$. That is generally multivalued and clearly has two values for $w=1/2$ since $e^{in\pi}=(-1)^n$. – Mark Viola Jul 19 '18 at 18:01
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@MarkViola using the definition $z^w:=e^{w\text{Log}(z)}$ where $\text{Log}(z)$ denotes the principal logarithm results in $z^w=e^{w\text{Log}(|z|)+i\text{Arg}(z)w}$ which for $z=1,w=\frac{1}{2}$ results in $e^{0}=1$. By completely removing $n$ from the picture, we do get unambiguous results, which gives us the nice properties that taking the $n$'th root is well-defined giving a single output for all numbers. It does admittedly lead to frustrations like the OP has experienced where otherwise nice properties have specific requirements, however that is a small price to pay. – JMoravitz Jul 19 '18 at 18:13
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@Sebastiano I am making no further edits. Please read the other comments to see why it depends on the definition of $z^w$. – packetpacket Jul 19 '18 at 20:23
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Sorry for the hairsplitting :-), but $\sqrt{0}$ has only one solution. :-) – peterh Jul 19 '18 at 21:52
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@peterh it doesn't make sense mathematically to write that $\sqrt x$ "has a solution". An equation has a solution. A number doesn't. $\sqrt x$, for non-negative real $x$, is a number, and it is one number. – Arnaud Mortier Jul 20 '18 at 00:03
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@peterh this is why I flagged this post for low-quality, it is so misleading. People who upvote it feel like they gain in knowledge, while the truth is they will keep writing inaccurate equalities and be confused about what the square root function is. – Arnaud Mortier Jul 20 '18 at 00:21