Prove that the only solution of $|x| + \left(y-1\right)^2 = 0$ is $x=0$ and $y=1$.

Question

My teacher in general mathematics added another problem in the problem set I need to answer. I need to prove that the only solution of $|x| + \left(y-1\right)^2 = 0$ is $x=0$ and $y=1$.

I am told, and know for a fact, that the only possible solutions for both $|x|$ and $\left(y-1\right)^2$ are non-negative numbers. That is,

$$\forall x,y \in \Bbb R , |x| \geq 0 \; and \; \left(y-1\right)^2 \geq 0$$

And that the sum of two non-negative terms will only be $0$, if and only if they are both $0's$.

That's all off my progress and I don't know how to write it as a proper mathematical proof. Thanks in advance for anyone who will gladly help.

P.S. Please do correct me if I have stated something wrong.

Answer 1

Let $a$ and $b$ be two non negative real numbers such that $a+b=0$. Assume that $a>0$. Then $a+b>b\geq 0$, which is absurd.

So in your case here, you may immediately infer that $|x|=(y-1)^2=0$ and hence, that $x=0$ and $y=1$.

Answer 2

If $a,b\geq 0$ then $a+b=0$ if and only if $a=b=0$. Therefore, let $a=|x|$ and $b=(y-1)^2$, then $$a=b=0\implies |x|=0,(y-1)^2=0\implies x=0,y=1.$$

Answer 3

As the question is related to proof writing, I'll propose you a redaction for another equation: $x^4+\exp(\theta)=0$ which has no real solution.

Notice that, for any real $x$, $x^4$ is non-negative and for any real $\theta$, $\exp(\theta)$ is also non-negative. As the equation we consider is "sum of these two terms equal zero", we have: $x^4=0$ and $\exp(\theta)=0$.

However, $\exp(\theta)$ is positive, hence the equation $\exp(\theta)=0$ has no real solution.

As a solution $(x,\theta)$ of our equation must satisfy $\exp(\theta)=0$, we conclude that our equation has no real solution.

Your task is to adapt this kind of redaction to your actual problem

P.S.: As I'm not fluent in English, this redaction may have some grammatical or spelling error. Please take time to flag them in the comments or edit

Answer 4

$|x|=-(y-1)^2\le0$ is only possible with $x=0$ and the equation reduces to $(y-1)^2=0$.

Answer 5

$|x|+|(y-1)^2|=0$

$\iff $

$|x|=0$ and $|(y-1)^2|= 0.$

Hence: $x=0$, and $y-1=0$.