$N×0=0$ what is the proof behind this?

Question

I Am new to this Site. My question is how we can prove $N×0=0N×0=0$.

Cause suppose for example:

$$2×3=6$$ then we can solve it like this:

$$2=\frac63$$

This makes LHS=RHS.

In this way:

$$2×3=6$$

$$3=\frac62$$

This also makes LHS=RHS.

That means if $M×N=Z$, then $M=\frac ZN$ as well as $N=\frac ZM$, so in above equation $$N×0=0$$ Then It Should Be The Same,

Like:

$$N=\frac00$$

$$0=\frac0N$$

which is mathematically wrong. So please help me to make it understand. According to my perceptions, zero is an unique no. And there should be difference between the single $00$ and a placeholder $00$.

Answer 1

What you wrote is wrong from the start. You wrote that, in order to prove that $2\times3=6$, what we do is $2=\dfrac63$. Really? And how do we know that $2=\dfrac63$?

Actually, since, in a sense, division is the opposite of multiplication, we use properties of multiplication in order to prove properties of division and not the other way around.

Being able to prove that $n\times0=0$ (in $\mathbb R$) depends upon what you accept as a starting point. For instance, you can prove it as follows:\begin{align}n\times0&=n\times(0+0)\\&=n\times0+n\times0\end{align}and therefore\begin{align}0&=n\times0-n\times0\\&=(n\times0+n\times0)-n\times0\\&=n\times0+(n\times0-n\times0)\\&=n\times0+0\\&=n\times0.\end{align}

Answer 2

Indeed. $xy=z\land x\ne 0$ implies $\frac{z}{x}=y$. You cannot get rid of the second condition because $\frac{z}{x}=y$ does become nonsense when $x=0$.

This is very important to keep in mind when your $x,y,z$ are not plain numbers, but expressions, e.g. in case you are solving an equation. Imagine you have the equation:

$$2x=5x$$

then, by blindly dividing by $x$ you would say

$$2=\frac{2x}{x}=\frac{5x}{x}=5$$

(which is false, $2\ne 5$). But, this just proves that the original equation does not have non-zero solutions. The case $x=0$ must be examined separately, and, indeed, $x=0$ is a solution.

In general, that is how things go: whenever you want to divide by something that might be $0$, your calculations splits into two cases: when it is $0$ and when it is not $0$.