Friday, November 9, 2018

arithmetic - "Proof" that $sqrt{x^2} = x$ by using index laws. Why is this wrong?



This is a very short argument that I have often seen about the definition of $|x|=\sqrt{x^2}$ for $x\in \mathbb{R}$. I'm not quite sure what is the flaw in this argument below, can someone please illustrate at what line this goes wrong and reference what "restriction" was broken from the original index law.




Claim: $\sqrt{x^2} = x$ where $x\in\mathbb{R}$.



Proof: Note that $$\begin{align*} \sqrt{x^2} &= (x^2)^{\frac{1}{2}} \\ &= x^{\frac{2}{2}} \\ &= x^{1}=x.\end{align*}$$



What is wrong with this?


Answer



First the counterexample to this is $x = -1$.
Now to the error in your reasoning is the following $(a^b)^c = a^{bc}$ is only true if $a \geq 0$.


No comments:

Post a Comment

analysis - Injection, making bijection

I have injection $f \colon A \rightarrow B$ and I want to get bijection. Can I just resting codomain to $f(A)$? I know that every function i...