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Prove that 5=-5 $$ \sqrt{(-5)^2} = \sqrt{25} = 5 = \sqrt{(-5)^2} = \sqrt{(-5)\cdot(-5)} = \sqrt{(-5)} \cdot \sqrt{(-5)} = (i \sqrt{5})\cdot(i\sqrt{5}) = -5\,. $$

desos
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Karim Hmoud SY
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  • The function $ \ y \ = \ \sqrt{x^2} \ $ is equivalent to the absolute-value function $ \ y \ = \ | x | \ $ , which is not a one-to-one function, that is, two different values of $ \ x \ $ can be assigned to the same value $ \ y \ $ from the function. It does not follow for functions that are not one-to-one that $ \ f(x) \ = \ f(y) \ $ implies only that $ \ x \ = \ y \ $. – colormegone Apr 11 '15 at 23:37
  • the basic point here is that $x=1 \to x^2=1$ is true, $x^2=1 \to x=1$ is false – JMP Apr 12 '15 at 04:40

1 Answers1

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$f(x) =\sqrt{x}$ is the principle square root function, since 25 has two square roots the function is defined to only give the positive one otherwise it would be a relation not a function and it would be ambiguous.

The issue is this step:

$ \sqrt{(-5)(-5})= \sqrt{-5}\sqrt{-5}$

You tried to use this identity:

$\sqrt{ab}= \sqrt{a}\sqrt{b}$

But the rule only applies when $a,b\geq 0$

futurebird
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