# Prove that for all $a,b \in\mathbb{R}$, $a·b=0$ if and only if $a=0$ or $b=0$. [duplicate]

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Prove that for all $$a,b \in\mathbb{R}$$, $$aΒ·b=0$$ if and only if $$a=0$$ or $$b=0$$.

Using these field axioms:

• (A1) Addition is commutative
• (A2) Addition is associative
• (A3) Addition has a neutral element $$0$$
• (A4) Any element has an additive inverse
• (A5) Multiplication is commutative
• (A6) Multiplication is associative
• (A7) Multiplication has a neutral element $$1$$
• (A8) Any non-zero element has a multiplicative inverse
• (A9) Multiplication distributes over addition

My answer: Suppose $$ππ=0$$ and by (A6) $$π=1β π=(π^{β1}β π)β π=π^{β1}β (πβ π)$$. Either $$π=0$$ or it is not. If a does not equal $$0$$,then by (A8),there is a unique element $$π^{β1}=(1/π)ββ$$ such that $$π^{β1}β (πβ π)=π^{β1}β 0=0$$ thus $$b=0$$

## marked as duplicate by rschwieb abstract-algebra StackExchange.ready(function() { if (StackExchange.options.isMobile) return; $('.dupe-hammer-message-hover:not(.hover-bound)').each(function() { var$hover = $(this).addClass('hover-bound'),$msg = $hover.siblings('.dupe-hammer-message');$hover.hover( function() { $hover.showInfoMessage('', { messageElement:$msg.clone().show(), transient: false, position: { my: 'bottom left', at: 'top center', offsetTop: -7 }, dismissable: false, relativeToBody: true }); }, function() { StackExchange.helpers.removeMessages(); } ); }); }); Feb 19 at 21:18

• How from $a \cdot a^{-1}=1$ you conclude that $b=0$ ? – Mauro ALLEGRANZA Feb 19 at 13:28
• Because ab=0 so if a = 1, then b must be 0. Should I include that? @MauroALLEGRANZA – brucemcmc Feb 19 at 13:29
• You have to use : $a^{-1} \cdot (a \cdot b) = a^{-1} \cdot 0=0$. – Mauro ALLEGRANZA Feb 19 at 13:30
• So would this be efficient to say then? Suppose ππ=0. Either π=0 or it is not. If a does not equal 0,then by (A8),there is a unique element π^β1=(1/π)ββ such that π^β1β(πβπ)=π^β1β0=0 thus ab=0? – brucemcmc Feb 19 at 13:32
• @brucemcmc If you add $b = 1\cdot b = (a^{-1}\cdot a)\cdot b = a^{-1}\cdot (a\cdot b)$ to the start of that sequnce of equalities, you're good to go, because that gives you $b = 0$ if you compare the leftmost end to the rightmost end. – Arthur Feb 19 at 13:34

Suppose $$ab = 0$$ and $$π=1β π=(π^{β1}β π)β π=π^{β1}β (πβ π)$$.

Here you get into trouble because you haven't assumed $$a\neq 0$$. Thus $$a^{-1}$$ might not exist.

Either π=0 or it is not. If a does not equal $$0$$,then by (A8),there is a unique element $$π^{β1}=(1/π)ββ$$

Now you have established that $$a^{-1}$$ is valid.

such that $$π^{β1}β (πβ π)=π^{β1}β 0=0$$ thus $$b=0$$

and this is where I meant you should put the hting you put right at the start.

Finished proof, in the right order:

Either $$a = 0$$ or $$a\neq 0$$. If $$a = 0$$ we're done. If not, then by $$A8$$ there is an $$a^{-1}$$ such that $$a\cdot a^{-1} = 1$$. By $$A5$$, we also have $$a^{-1}\cdot a = 1$$.

Thus if $$a\neq 0$$, we get $$b \stackrel{A7}= 1\cdot b \stackrel{A8+A5}= (a^{-1}\cdot a)\cdot b \stackrel{A6}= a^{-1}\cdot(a\cdot b)\stackrel{Assumption}=a^{-1}\cdot 0 = 0$$ showing that $$b = 0$$.

Suppose neither $$a$$ nor $$b$$ is zero. Then, since $$\mathbb R$$ is a field, all nonzero elements are invertible, hence, $$a^{-1},b^{-1}$$ exist. Now $$a^{-1}abb^{-1}=1=0$$, a contradiction.