$x^2$ cannot equal $0$, where $x$ is a real number.

Question

Hey can someone help me figure this out, because I just don't know how to get started.

if $x^2$ cannot equal $0$, where $x$ is a real number, then $x$ cannot equal $0$. Let $a$ be a real number with $a^2$ cannot equal $0$, then $a$ cannot equal $0$.

What I think

If we take a number for $x$, lets say 2^2 = 4 which does not equal 0.. turning the statement to be true.

That is all I can think of right now..

I need to know if this argument is valid or not..and how if they are valid. I think they are valid. but i just dont know the HOW part.

Answer 1

Let's proceed by a proof by contradiction. Assume that $x\neq 0$ and that $x^2=0$, but then $x^{-1}$, the unique real number such that $xx^{-1}=x^{-1}x=1$, exists as $x\neq 0$ and so multiplying the second equality by $x^{-1}$ we have that $$0=x^{-1}0=x^{-1}x^2=(x^{-1}x)x=1x=x$$ which gives the desired contradiction. Hence, by reductio ad absurdum, $x^2\neq 0$ whenever $x\neq 0$ as you want. Note that for any $y$, $y0=0$ since, by the distributive law, $y0=y(0+0)=y0+y0$ and so $y0=0$ after substracting $y0$ from both sides.

Using a similar argument, you can show that the product of two non-zero real numbers is non-zero. The crucial point of the proof is that if a real number $x$ is non-zero, i.e., $x\neq 0$, then you can divide by it, i.e., there exists its inverse $x^{-1}$.

EDIT: In view of the changes of the question, the version I saw I requested to prove that if $x\neq 0$, then $x^2$, I provide a prove of the fact that $x^2\neq 0$ implies $x\neq 0$. I let the above proof showing that $x\neq 0$ implies $x^2\neq 0$ just in case.

For showing that $x^2\neq 0$ implies $x\neq 0$, we apply again a proof by contradiction. So assume $x^2\neq 0$ and $x=0$, then $x^2=x0=0$ as noted in the above proof. This is a contradiction and thus the desired implication is proven. Note that in this case the essential property used is that any number multiplied by zero is zero.

Answer 2

Let $x^2\not=0$. Then $x\cdot x\not=0$.

Now, assume $x=0$. Then we have $0\cdot 0=0^2=0$.

But $x^2\not=0$, contradiction!

Therefore, $x\not=0$.

Answer 3

Let's try to use the following law:

Trichotomy Law: If $a\in\mathbb{R}$ then exactly one of the following holds

$$(i)a>0\quad (ii)a=0\quad (iii)a<0$$

Let us use direct proof to your question. Assume that $x\neq 0$. Using Trichotomy Law, either $x>0$ or $x<0$. If $x>0$, then $$x^2=x\cdot x>0.$$ If $x<0$ then $-x>0$ and so $$x^2=-x\cdot -x>0.$$ In either case, we have shown that $x^2>0$. Again, using the Trichotomy Law, we conclude that $x^2\neq 0$.

Answer 4

Let $x y = 0$ in a field like $\mathbb{R}$. This implies that $x$, $y$ or both must be $0$. Otherwise, they would be invertible, and multiplying by the inverse we would get $1 = 0$. (More generally, this property holds for integral domains).

Therefore, if $x^2$ were $0$, it would mean $x = 0$, but $x \not= 0$, so $x^2 \not= 0$.

Answer 5

If we had an $x \neq 0$ with $x^2 =0$, then every real number $r$ would satisfy that $r^2=0$. For since $x\neq 0$ we have $r= x \cdot \frac{r}{x}$. But then $r^2 = x^2 \cdot (\frac{r}{2})^2 = 0 \cdot (\frac{r}{2})^2 = 0$.

However, $1^2 = 1 \neq 0$.

Answer 6

Since x=0 is the only solution to x^2=0 and x^2=0 is true if x=0 just use the contrapositive.