Is Using Multi-Valued Operators for Proofs Risky?

[MatheusMkalo]

(-3)² = (\sqrt{9})² --> True

-3 = \sqrt{9} --> False

Wolfram Alpha result: False

 

[chiro]

Because -3 doesn't equal 3?

 

[micromass]

We had a thread about this recently. The point is that \sqrt{x^2}\neq x. The square root of x is defined as the unique POSITIVE number who's square is x. Thus \sqrt{(-3)^2} is the unique positive number who's square is (-3)^2=9, obviously, this number is 3.

Thus, in general, we have that \sqrt{x^2}=|x|.

So, your first equation:

(-3)^2=(\sqrt{9})^2

is true, since (-3)^2=9and \sqrt{9}=3 and 3^2=9

But when taking the square root of both sides we get

\sqrt{(-3)^2}=\sqrt{(\sqrt{9})^2}

which evaluates to 3=\sqrt{9}, which is perfectly true!

 

[chiro]

Building on what micromass just said, use the horizontal line test to find out that there is no unique inverse except for the case when x = 0.

 

[KingNothing]

When you "undo" a squared operation, it is not as simple as just removing the little "2" superscript.

The square root of x is defined as the unique POSITIVE number who's square is x.

Really? In my classes, we've always used \sqrt[]{x^{2}}=\pm x

 

[Mentallic]

When you "undo" a squared operation, it is not as simple as just removing the little "2" superscript.



Really? In my classes, we've always used \sqrt[]{x^{2}}=\pm x

No, this is wrong. The square root function is defined to be the unique positive square root of the number. \sqrt{9}=3 and if \sqrt{9}=\pm3 then I have no clue why we would write such answers as x^2=9 \rightarrow x=\pm \sqrt{9} when the square root is already \pm

 

[Char. Limit]

And this is why when you use a multi-valued operator to prove something, you always have to check and see if the answer fits the original equation. Or you'll end up with extraneous or wrong solutions like -3=3.