-4
  1. $1 = 1$
  2. $ 1^2 = 1$ and $(-1)^2 = 1$
  3. Therefore, $1^2 = (-1)^2$
  4. Square root both sides $\sqrt{1^2} = \sqrt{(-1)^2}$
  5. Therefore, $-1 =1$

This is an obvious paradox, but I don't know how to approach solving it

potapeno
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    Because you computed the square root incorrectly. –  May 05 '16 at 02:46
  • You made a mistake in your calculation on the RHS in line 4 – Xylius May 05 '16 at 02:47
  • $\sqrt{(-1)^2}=1$. – Michael Burr May 05 '16 at 02:47
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    Generally for real $a$ it's $\sqrt{a^2}=|a|$ not just $a$ --- the absolute value sign is crucial. – coffeemath May 05 '16 at 02:49
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    do you understand what means the notation $f : a \to \sqrt{a}$ ? it means that $f$ is a FUNCTION , which for a number $a$ outputs "its square root", and clearly $f(1)$ cannot output two different values. sometimes we define the square root of $1$ to be $1$, sometimes to be $-1$, but once we chose the value of $f(1)$, we cannot change anymore (without defining another function) – reuns May 05 '16 at 02:57

1 Answers1

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This is not obvious and wrong. You applied that "if $f(x)=f(y)$, then $x=y$ for all $x,y$ in $\operatorname{dom} f$" between step 4 and 5. However, it is true if and only if $f$ is one-to-one. $f(x)=\sqrt{x^2}$ is not one-to-one.

choco_addicted
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