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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 $π‘Žπ‘=0$ and by (A6) $𝑏=1⋅𝑏=(π‘Ž^{βˆ’1}β‹…π‘Ž)⋅𝑏=π‘Ž^{βˆ’1}β‹…(π‘Žβ‹…π‘)$. Either $π‘Ž=0$ or it is not. If a does not equal $0$,then by (A8),there is a unique element $π‘Ž^{βˆ’1}=(1/π‘Ž)βˆˆβ„$ such that $π‘Ž^{βˆ’1}β‹…(π‘Žβ‹…π‘)=π‘Ž^{βˆ’1}β‹…0=0$ thus $b=0$


marked as duplicate by rschwieb abstract-algebra Feb 19 at 21:18

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  • $\begingroup$ How from $a \cdot a^{-1}=1$ you conclude that $b=0$ ? $\endgroup$ – Mauro ALLEGRANZA Feb 19 at 13:28
  • $\begingroup$ Because ab=0 so if a = 1, then b must be 0. Should I include that? @MauroALLEGRANZA $\endgroup$ – brucemcmc Feb 19 at 13:29
  • $\begingroup$ You have to use : $a^{-1} \cdot (a \cdot b) = a^{-1} \cdot 0=0$. $\endgroup$ – Mauro ALLEGRANZA Feb 19 at 13:30
  • $\begingroup$ So would this be efficient to say then? Suppose π‘Žπ‘=0. Either π‘Ž=0 or it is not. If a does not equal 0,then by (A8),there is a unique element π‘Ž^βˆ’1=(1/π‘Ž)βˆˆβ„ such that π‘Ž^βˆ’1β‹…(π‘Žβ‹…π‘)=π‘Ž^βˆ’1β‹…0=0 thus ab=0? $\endgroup$ – brucemcmc Feb 19 at 13:32
  • $\begingroup$ @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. $\endgroup$ – Arthur Feb 19 at 13:34

Suppose $ab = 0$ and $𝑏=1⋅𝑏=(π‘Ž^{βˆ’1}β‹…π‘Ž)⋅𝑏=π‘Ž^{βˆ’1}β‹…(π‘Žβ‹…π‘)$.

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

Either π‘Ž=0 or it is not. If a does not equal $0$,then by (A8),there is a unique element $π‘Ž^{βˆ’1}=(1/π‘Ž)βˆˆβ„$

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

such that $π‘Ž^{βˆ’1}β‹…(π‘Žβ‹…π‘)=π‘Ž^{βˆ’1}β‹…0=0$ 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.


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