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If $a^{p}\cdot b^{p}= (a\cdot b)^{p}$ then why $$-1^{2}\cdot 3^{2}\neq (-1\cdot 3)^{2}\\ -1\cdot 9\neq (-3)^{2}\\ -9\neq 9$$

I'm sorry, I don't know how to put latex code.

Thomas Andrews
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Jose
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  • You presumably mean $a^2\cdot b^2=(a\cdot b)^2$? – Thomas Andrews Jun 09 '15 at 01:04
  • There is a difference between $(-1)^2$ and $-1^2$, when you plug in a value you technically need parentheses, but normally it doesnt matter, when it is a negatice value however, it does matter – nosyarg Jun 09 '15 at 01:20
  • @nosyarg I didn't know that, is that a rule? if so, could you tell me where to find it so I can read about it? – Jose Jun 09 '15 at 01:29
  • This is the page of a math teacher who wrote about it: http://www.angrymath.com/2014/08/when-are-parentheses-required-for.html?m=1 – nosyarg Jun 09 '15 at 01:32
  • Thank you, @nosyarg, a very interesting read – Jose Jun 09 '15 at 02:06

2 Answers2

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Because if $a=-1$ then $a^2=(-1)^2\neq -(1^2)$. Yo are treating $a^2=-(1^2)$, which is wrong.

Thomas Andrews
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  • But -1^2 is -1, what if a were -1? – Jose Jun 09 '15 at 01:17
  • Umm, did you even read my answer? It's all about if $a=-1$. @Jose – Thomas Andrews Jun 09 '15 at 01:18
  • I did, but I don't understand why if a is -1 why $a^2=(-1)^2$ and not $a^2=-1^2=-1$ I mean, sure I could put -1 in brackets, raise it to the power of 2 and it would give me just 1. But my question is much like $-2^2=-4$ as opposed to $(-2)^2=4$ – Jose Jun 09 '15 at 01:24
  • $a^2=a\cdot a = (-1)\cdot (-1)$. Sorry if you don't understand how parentheses work, but $a^2$ is multiplying the number by itself. You get two minus signs, not one. Substitution isn't in a vacuum. If $a=2+3$ then $3\cdot a\neq 3\cdot 2+3=9$. Rather $3\cdot a=3\cdot (2+3)$. When we substitute, we often add parens because otherwise we get ambiguities. – Thomas Andrews Jun 09 '15 at 01:25
  • @Jose: I am assuming here that $p$ is prime correct? The way you are going about this is $-1^2 = -(1)^2 = -1^{1}$. Now do you see why this doesn't hold? The exponent $1 \not \in {\textbf{primes}}$. – Mr.Fry Jun 09 '15 at 01:27
  • @Mr.Fry In this question $p=2$. All other questions about $p$ are irrelevant. – Thomas Andrews Jun 09 '15 at 01:29
  • If $a=1+1$ is $a^2=1+1^2=1+1=2$? @Jose No, $a^2=(1+1)^2$. See, we've again added parentheses when making a substitution. – Thomas Andrews Jun 09 '15 at 01:32
  • Basically, you are suggesting that $a^2\neq (a)^2$ for some values of $a$. That is going to make mathematics nonsensical. @Jose – Thomas Andrews Jun 09 '15 at 01:35
  • @ThomasAndrews. Yes. Please ready what I wrote carefully. I am showing that the way he is treating the exponent is equivalent to my comment. And I was suggesting that typically when someone uses $p$ it means prime. So as the statement stands it is relevant because this is what I use to draw a contradiction. – Mr.Fry Jun 09 '15 at 01:36
  • There is no reason at all to believe he wants $p$ is prime, rather than being just a variable for "power," considering the nature of his question. @Mr.Fry And it really does not affect his question at all, except for that we probably want $p$ to be an integer. – Thomas Andrews Jun 09 '15 at 01:38
  • @ThomasAndrews Yes, that is what I'm suggesting, I should have been clearer in expressing myself. I know it makes it nosensical to "defy" rules, but it's fun, just like trying to prove that 1+1 doesn't equal 2. I appreciate your input. – Jose Jun 09 '15 at 02:04
  • @Mr.Fry That's not what I meant, as Thomas said, p is just a variable for power, but what you said is really interesting too, so I thank you for bringing that up, more stuff to think about is always good :p – Jose Jun 09 '15 at 02:05
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You are wrong here.

$a^{p} b^{p}=(ab)^{p}$

And your math on that example is wrong. You forgot the brackets.

$(-1)^2(3)^2=(-3)^2=+9$.

Hope this helps

Aleksandar
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  • Thanks. But I didn't forget the brackets, considering -1^2 is -1 (not (-1)^2 which I know is 1), then if a or b is a negative number $a^{p}\cdot b^{p}\neq(a\cdot b)^{p}$, I want to know why. – Jose Jun 09 '15 at 01:19
  • You did, $-1^2=-1$ $(-1)^2=1$. – Aleksandar Jun 09 '15 at 01:21
  • @Jose, what you are doing with $-1^2$ is that you are only squaring the one, but leaving the $-$ untouched, the parentheses ensure that you capture all of $(-1)$ – nosyarg Jun 09 '15 at 01:24
  • @Aleksandar $-1(-1)^2=-1(1)=-1$, I don't understand how I'm wrong here. – Jose Jun 09 '15 at 01:27
  • Ya you're right but what you have there is $(-1)^3=(-1)(-1)(-1)=(-1)(-1)^2$. – Aleksandar Jun 09 '15 at 01:30
  • Hmmmm, I see, now that's interesting, I'll give it some thought. Thanks for the input, sorry if I come across as a hardheaded person :p – Jose Jun 09 '15 at 01:32