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

Question

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 $b=0$

Answer 1

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.

Answer 2

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.

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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 $b = 0$.