# A real number times zero is zero [closed]

Justify each step of the following direct proof, which shows that if $x$ is a real number, then $x\cdot0=0$. Assume that the following are previous theorems: If $a$, $b$, and $c$ are real numbers, then $b+0=b$, $a(b+c)= ab+ac$, and if $a+b=a+c$, then $b=c$.

Please show the steps as I'm confused here.

-
Please surround math with dollar signs, this makes the formulae stand out and makes them easier to read. For example $(a+b)c$ gives you $(a+b)c$. –  Jack M Nov 3 '13 at 22:22

## closed as off-topic by azimut, user1337, Daniel Fischer, Daniel Rust, Stefan HamckeNov 3 '13 at 23:04

This question appears to be off-topic. The users who voted to close gave this specific reason:

• "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – azimut, user1337, Daniel Fischer, Daniel Rust, Stefan Hamcke
If this question can be reworded to fit the rules in the help center, please edit the question.

Note that

\begin{align*} x \cdot 0 &= x\cdot (0 + 0) \\ &= x\cdot 0 + x\cdot 0 \end{align*}

Rearrange this to give

$$x \cdot 0 + 0 = x \cdot 0 + x \cdot 0$$

So what can you conclude from here? And can you justify each step?

-
I'm lost on the logic there, how did you translate the x*0=0 to x*0+)=x*0+x*0? –  johnny Nov 3 '13 at 22:49
Since $a(b+c)=ab+ac$ for all real numbers $a$, $b$, and $c$, we know that $x\cdot0=x\cdot(0+0)=x\cdot0+x\cdot0$. This implies that $x\cdot0=x\cdot0+x\cdot0$. Since $b+0=b$ for all real numbers $b$, we know that $x\cdot0+0=x\cdot0+x\cdot0$. Since $a+b=a+c$ implies that $b=c$ for all real numbers $a$, $b$, and $c$, we can see that $x\cdot0+0=x\cdot0+x\cdot0$ implies that $x\cdot0=0$. Thus $x\cdot0=0$.