Canceling Roots and Squares in Modulus (closed)

Question

I'm honestly a bit lost. So, consider this: $$x^2 = k$$ for some generic k. Then take square root giving: $$x = ±\sqrt k$$ And consider this: $$\sqrt x = k$$ Square both sides so: $$x=k^2$$ Right? Showing root cancels square and square cancels root. But now consider: $$ ∣x∣ = \sqrt{x^2}$$ thus: $$ ∣x∣ =(x^{1/2})^2$$ $$ ∣x∣ = x^{2/2}$$ $$ ∣x∣ = x^1$$ $$ ∣x∣ = x$$ But this does not make sense as modulus is always positive so i'm having a hard time grasping why the square and root do not cancel. If anyone could explain the specifics behind the restrictions of roots and powers canceling it would be much appreciated.

I've read that for root to cancel by square radicand must be ≥ 0. But this leads to a huge variety of questions of how are radical equations solved? As one of the first steps is eliminating root by raising to exponent = index of radical (cube root -> third power, nth root -> nth power). And the problem for me arises in the fact that in general math x is assumed to element of reals not ≥ 0 making radical equations impossible as wouldn't it be wrong to assume the domain of x to be [0,infinity)

Answer 1

The reason why you need to be careful about radicals lies in its very definition. It's tempting to define, for a positive $x$, that $\sqrt{x}$ is whatever the real number whose square is $x$; but this definition is problematic because if $a$ is such a number, so is $-a$, and therefore there is no "the" number satisfying this property.

Fortunately, you can prove (with some real analysis) that there is indeed a unique non-negative number $a$ satisfying the condition that $a^2 = x$, for every non-negative real number $x$. So, in this case, we define: $$ \sqrt{x} \text{ is the unique non-negative real number whose square is x} $$

Now, (assuming we are working in $\mathbb{R}$), we have the following remarks

• $\sqrt{x}$ is only defined for non-negative real number $x$.
• $\sqrt{x}$ is always non-negative (by the above definition).
• $\sqrt{x^2} = |x|$
• $(\sqrt{x})^2 = x$ but this formula is only defined for non-negative $x$, instead for all $x$.

To point out what is exactly problematic in your reasoning, just notice that, as you pointed out, both $\pm\sqrt{k}$ are mapped to $k$ as you take squares, but when you take square root, you only get $\sqrt{k}$. Therefore, they don't cancel each other in a one-to-one manner.

Hope this helps.

Answer 2

The rule $(x^a)^b=x^{ab}$ always works for positive values of $x$ but does fail for negative values of $x$ in some cases. You've identified one such case: $\sqrt{x^2} = (x^2)^\frac{1}{2} \neq x^{\frac{2}{2}} = x$ for $x < 0$. It's a bit tricky to comprehensively describe the cases in which that rule doesn't work, but to slightly generalize what you are describing, it is true for any value of $x$ that $\sqrt[a]{x^a} = (x^a)^{\frac{1}{a}} = |x|$ when $a$ is even.