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Can anyone explain why this not true?

9 Answers

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The error is assuming that x = |x| for all x, but that is only true if x ≥ 0. The square root as a function is taken as the positive root, and the square of any real number is nonnegative, so sqrt(x^(2)) ≥ 0.
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The last statement is true. That’s is how the absolute value is defined.
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Your error is in assuming that (x\^2)\^(1/2) = x; this only holds for positive x. There are a couple ways to see this - one is to simply note that sqrt(x\^2) = |x|, which is not equal to x for negative x.

A way that potentially gives more insight is that we can note that x\^2 can't admit an inverse on the whole real line. The graph fails the horizontal line test, so the "inverse" would not be a function (algebraically, this states the function x\^2 is not injective, so it cannot admit an inverse). However, if we restrict x\^2 to *only* the positive numbers, or *only* the negative numbers, then it is injective/it passes the verticla line test, so it does have an inverse on those restricted sets. For positive numbers, the inverse function of f(x) = x\^2 is g(x) = x\^{1/2}, and for negative numbers, the inverse function is h(x) = -x\^{1/2}, where we are taking x\^{1/2} to be the positive square root.

Because of this, when x is negative, we can't assume (x\^2)\^(1/2) = x, and so we don't get a contradiction that x = |x|.
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First, you wrote it backwords, which confused some other commenters. The convention is to start with a known fact, and move step by step to get to what you want to prove. So you have to switch the top and the bottom.

The property: `(a^b)^c = a^(b*c)` is not true for all real numbers. This rule holds if one of the following is true:

* a > 0; or
* b and c are integers

This condition is often left out when teaching. (There are probably other conditions that make the property work, but the two I listed are the most common cases.)
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I was reading that sqrt(x^2) = |x| and that got me thinking about this, which seems just plainly false, I just don't understand what's wrong with it.
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X is Not abs value of x
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When x=i you have

sqrt(i^2) =? |i|

sqrt(-1) =? 1

i =? 1
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The first line is not correct. Then, saying that x^1 = sqrt(x^2) is also incorrect. Remember, square root of positive x has two possible answers. By taking the sqrt of x^2, you implicitly pick the positive one.

However, sqrt(x^2) = abs(x) for reals.
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If the first statement is saying that for ALL x values that x is always positive it’d be wrong because square roots give two answers, however if the first line is just the X is positive, the it is correct

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