# 2+2= 8 ; is this proof correct?

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• Apr 21st 2012, 01:11 PM
Mytutoing
Re: 2+2= 8 ; is this proof correct?
I get that, but it doesn't explain why there is difference between "two square roots" and "the square root" appart from convention - is there a mathamatical difference.
• Apr 21st 2012, 01:20 PM
Plato
Re: 2+2= 8 ; is this proof correct?
Quote:

Originally Posted by Mytutoing
I get that, but it doesn't explain why there is difference between "two square roots" and "the square root" appart from convention - is there a mathamatical difference.

There are two square roots of 4, $2~\&~-2$.
The square root of 4 is $\sqrt{4}=2.$.
• Apr 21st 2012, 01:29 PM
Mytutoing
Re: 2+2= 8 ; is this proof correct?
OK by why is this just symantics and convention or is there a mathamatical difference
• Apr 21st 2012, 01:31 PM
Plato
Re: 2+2= 8 ; is this proof correct?
Quote:

Originally Posted by Mytutoing
OK by why is this just symantics and convention or is there a mathamatical difference

There a mathematical difference: $\sqrt{4}\ne -2$.
• Apr 21st 2012, 01:42 PM
Mytutoing
Re: 2+2= 8 ; is this proof correct?
so the opperator/ symbol for taking the square root of a number is differnet to the square root of the number?!

Sorry still dont get it - the mathamatical notation seems to just be rewriting the previous MOD version in another way.
• Apr 21st 2012, 01:45 PM
Mytutoing
Re: 2+2= 8 ; is this proof correct?
besides if I solve the problem to find the intersect of the two lines y=5 and y=x^2 I expect two answers although I would represent the solution as [root symbol]5=x. Your argument suggests that the answer is only the positive values.
• Apr 21st 2012, 01:52 PM
Plato
Re: 2+2= 8 ; is this proof correct?
Quote:

Originally Posted by Mytutoing
so the opperator/ symbol for taking the square root of a number is differnet to the square root of the number?! Sorry still dont get it - the mathamatical seems to just be rewriting the previous MOD version in another way.

If $x^2=4$ then $x=2\text{ or }x=-2$. There are two roots of that equation.

$\sqrt{4}=2$ and $-\sqrt{4}=-2$.

If $|x|=2$ then $x=2\text{ or }x=-2$.
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