Why is |p|=|q| the same as (p)^2=(q)^2?

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Kulla_9289

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Why is |p|=|q| the same as [imath](p)^2[/imath]=[imath](q)^2[/imath]? How do I prove this?
 
You could consider the four cases separately.
1. p and q are both positive
2. p is positive and q is negative
3. q is positive and p is negative
4. both p and q are negative
and also when both p and q are 0.
 
So, |p|=|q| is the same as [imath](p)^4[/imath]=[imath](q)^4[/imath]also, right?
 
Kulla_9289 said:
So, |p|=|q| is the same as [imath](p)^4[/imath]=[imath](q)^4[/imath]also, right?
Click to expand...
Once again, it is difficult to answer you because you do not use math notation or precise English. What do you mean “is the same as”?

Clearly, they are different statements rather than identical statements. They have equivalent meanings. In math notation,

[math]|p| = |q| \iff p^2 = q^2.[/math]
 
Yes. In fact |x|= |y| is equivalent to [math]x^{2n}= y^{2n}[/math] where n is any integer.

That is because, if |x|= |y| then either x= y or x= -y. Do you understand that?
 
Kulla_9289 said:
Okay. Is [imath]|p|=|q|[/imath] equivalent to [imath](p)^4=(q)^4[/imath], too?
Click to expand...
Please try using the ideas you've been given to answer this question yourself. How do those ideas apply? If you only ask questions, and never think for yourself, you can't learn mathematics.

Also, however, I need to point out that everything being said here depends on context. If we are talking about real numbers only, then these various equations are equivalent; if the variables can represent complex numbers, everything changes. I believe your context is the former, which everyone has been assuming, but it's worth mentioning.
 
Also if you want assuming we are in real numbers as Dr peterson said, you can use the fact that p ^ 2n where n is integer is the same thing as p^2^n so after proving p^2 = q^2 is equivilant to |p| = |q| all other even powers follow as p^2 will imply p^2^n
As if a = b a^n = b^n so p^2 = q^2
So p^2^n = q^2^n
 
AbdelRahmanShady said:
Also if you want assuming we are in real numbers as Dr peterson said, you can use the fact that p ^ 2n where n is integer is the same thing as p^2^n so after proving p^2 = q^2 is equivilant to |p| = |q| all other even powers follow as p^2 will imply p^2^n
As if a = b a^n = b^n so p^2 = q^2
So p^2^n = q^2^n
Click to expand...
p^2n = p^2*n
p^(2n) = (p^2)^n
 
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