# 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.

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

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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

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?

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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 $x\cdot0=0$.

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