Is Using Multi-Valued Operators for Proofs Risky?

  • Thread starter MatheusMkalo
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In summary, the conversation discusses the concept of inverse operations, specifically the square root function and its relationship to squared operations. It is clarified that the square root of a number is only defined as the unique positive square root, and not the positive or negative square root. The conversation also points out the importance of checking solutions when using multi-valued operators.
  • #1
MatheusMkalo
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(-3)² = ([tex]\sqrt{9}[/tex])² --> True

-3 = [tex]\sqrt{9}[/tex] --> False

Wolfram Alpha result: False
 
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  • #2
Because -3 doesn't equal 3?
 
  • #3
We had a thread about this recently. The point is that [tex]\sqrt{x^2}\neq x[/tex]. The square root of x is defined as the unique POSITIVE number who's square is x. Thus [tex]\sqrt{(-3)^2}[/tex] is the unique positive number who's square is [tex](-3)^2=9[/tex], obviously, this number is 3.

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

So, your first equation:

[tex](-3)^2=(\sqrt{9})^2[/tex]

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

But when taking the square root of both sides we get

[tex]\sqrt{(-3)^2}=\sqrt{(\sqrt{9})^2}[/tex]

which evaluates to [tex]3=\sqrt{9}[/tex], which is perfectly true!
 
  • #4
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.
 
  • #5
When you "undo" a squared operation, it is not as simple as just removing the little "2" superscript.

micromass said:
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 [tex]\sqrt[]{x^{2}}=\pm x[/tex]
 
  • #6
KingNothing said:
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 [tex]\sqrt[]{x^{2}}=\pm x[/tex]

No, this is wrong. The square root function is defined to be the unique positive square root of the number. [tex]\sqrt{9}=3[/tex] and if [tex]\sqrt{9}=\pm3[/tex] then I have no clue why we would write such answers as [tex]x^2=9 \rightarrow x=\pm \sqrt{9}[/tex] when the square root is already [tex]\pm[/tex] :wink:
 
  • #7
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.
 

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