Is Using Multi-Valued Operators for Proofs Risky?

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Discussion Overview

The discussion revolves around the implications of using multi-valued operators, specifically square roots, in mathematical proofs. Participants explore the definitions and properties of square roots, particularly in relation to squaring negative numbers and the potential for extraneous solutions.

Discussion Character

  • Debate/contested
  • Technical explanation
  • Conceptual clarification

Main Points Raised

  • Some participants assert that \sqrt{x^2} is defined as |x|, emphasizing that the square root function yields the unique positive root.
  • Others challenge this by stating that in their experience, \sqrt{x^2} is understood as \pm x, suggesting a different interpretation of square roots.
  • One participant highlights the importance of verifying solutions when using multi-valued operators to avoid incorrect conclusions, such as -3 equating to 3.
  • A participant mentions the horizontal line test in relation to finding unique inverses, indicating that uniqueness may not hold in all cases.
  • There is a discussion about the nature of squaring and square roots, with some participants emphasizing that undoing a squared operation is not straightforward.

Areas of Agreement / Disagreement

Participants express differing views on the definition of square roots and the implications of using multi-valued operators. There is no consensus on whether \sqrt{x^2} should be considered as |x| or \pm x, indicating a fundamental disagreement on this topic.

Contextual Notes

Limitations include varying definitions of the square root function and the implications of using multi-valued operators in proofs, which remain unresolved in the discussion.

MatheusMkalo
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(-3)² = ([tex]\sqrt{9}[/tex])² --> True

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

Wolfram Alpha result: False
 
Last edited:
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Because -3 doesn't equal 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!
 
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.
 
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]
 
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:
 
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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