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Prove that for all $a,b \in\mathbb{R}$, $a·b=0$ 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 $=0$ and by (A6) $=1⋅=(^{−1}⋅)⋅=^{−1}⋅(⋅)$. Either $=0$ or it is not. If a does not equal $0$,then by (A8),there is a unique element $^{−1}=(1/)∈ℝ$ such that $^{−1}⋅(⋅)=^{−1}⋅0=0$ thus $b=0$

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brucemcmc
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  • How from $a \cdot a^{-1}=1$ you conclude that $b=0$ ? – Mauro ALLEGRANZA Feb 19 '19 at 13:28
  • Because ab=0 so if a = 1, then b must be 0. Should I include that? @MauroALLEGRANZA – brucemcmc Feb 19 '19 at 13:29
  • You have to use : $a^{-1} \cdot (a \cdot b) = a^{-1} \cdot 0=0$. – Mauro ALLEGRANZA Feb 19 '19 at 13:30
  • So would this be efficient to say then? Suppose =0. Either =0 or it is not. If a does not equal 0,then by (A8),there is a unique element ^−1=(1/)∈ℝ such that ^−1⋅(⋅)=^−1⋅0=0 thus ab=0? – brucemcmc Feb 19 '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 '19 at 13:34
  • @Arthur Suppose =0 and =1⋅=(^−1⋅)⋅=^−1⋅(⋅). Either =0 or it is not. If a does not equal 0,then by (A8),there is a unique element ^−1=(1/)∈ℝ such that ^−1⋅(⋅)=^−1⋅0=0 thus ab=0? – brucemcmc Feb 19 '19 at 13:36
  • Why are you concluding with "thus $ab = 0$? That's unnecessary. We already assumed that, and it's not what we're after. And don't use $a^{-1}$ until you have assumed that $a\neq 0$. – Arthur Feb 19 '19 at 13:38
  • @Arthur didn't I assume that a=0 or it is not? And should it then conclude as... thus a or b = 0? Sorry I am lost with my proof now :( – brucemcmc Feb 19 '19 at 13:40
  • @brucemcmc I suggest you write your entire proof, as you have it right now, from start to finish, in your question post. I think that's easier than trying to fit bits and pieces of it in comments. – Arthur Feb 19 '19 at 13:44
  • @Arthur I just did. By the way, thank you for all your help thus far. – brucemcmc Feb 19 '19 at 13:50
  • @Arthur thank you for the help again. I was wondering, what literature do you recommend to enhance my abilities in proof writing/math? – brucemcmc Feb 19 '19 at 14:41
  • @brucemcmc Do not repost the same question as you did: edit the old one. Also, please take time to write more useful titles. The original title of this question and the original duplicate are useful to nobody. – rschwieb Feb 19 '19 at 15:09
  • @rschwieb okay will do :( – brucemcmc Feb 19 '19 at 19:46

2 Answers2

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Suppose $ab = 0$ and $=1⋅=(^{−1}⋅)⋅=^{−1}⋅(⋅)$.

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

Either =0 or it is not. If a does not equal $0$,then by (A8),there is a unique element $^{−1}=(1/)∈ℝ$

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

such that $^{−1}⋅(⋅)=^{−1}⋅0=0$ 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 $b = 0$.

Arthur
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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.

YiFan Tey
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