Article by Dr p

[Cubist]

... We use functions if we want to remain sane, which these threads are making me doubt whether I still qualify.

??

((-1)^2)^1/2 =(1^2)^1/2
here we first square the Value inside,
1^ 1/2 = 1^1/2
Now to only way we can bring -1 on the lhs ...

Why would you want to "bring -1 on the lhs"? When you plug specific numbers in for x and y then the number of possible solutions is reduced to ONE (assuming that the numbers fit the equation). The only other possibility is that the numbers don't fit the equation. Typically you'd approach this type of problem in the following manner...

Question: Prove that if x=-1 and y=1 then the following equation is satisfied:- x^2=y^2
Answer: (-1)^2 =1^2 which implies that 1=1, and therefore the equation x^2=y^2 is satisfied when x=-1 and y=1
We don't even need to bother with taking square roots.

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In the context of the original article, I think that I didn't read down far enough when I wrote my response #8.

The taking square roots step was deliberately flawed to illustrate a common mistake. And it says further down that you need use an identity such as sqrt(x^2) = |x|

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BTW: Another method is to use difference of squares...

[math]x^2=y^2[/math]
[math]x^2-y^2=0[/math]
[math]\left(x-y\right)\left(x+y\right)=0[/math]

 

[Saumyojit]

(-1)^2 =1^2 which implies that 1=1

x = y => x^2 = y^2
But not the other way around especially if X and y are different sign.
Also , for the square function to be invertible it's Domain has to be either greater than equal zero or less than equal to zero.

We don't even need to bother with taking square roots.

To get from x^2 = y^2 to x and y taking the positive square root is obvious although we lose information about the original value of X and Y

Why would you want to "bring -1 on the lhs"?

Dr p intentionally considered sqrt to be as double valued function to show that sqrt is actually a function giving the absolute value of X : √x^2 = | x|

When the base is negative and Exponent are like this( ( -4)^2)^1/2 if we cancel the square with the square root then we would get back -4 .
Or if we square inside first (16)^1/2
We need to consider Exponention as multivalued then to get -4 .

But that's why negative bases are avoided.

Also check ( although I should give this in "complex post" ) https://math.stackexchange.com/ques...compute-negative-numbers-to-fractional-powers