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Given $a, b \in \mathbb{R},$ prove $ab = 0 \implies (a=0 \lor b=0).$

Is the following valid?

Proceed by contraposition. The contrapositive of $ab = 0 \implies (a=0 \lor b=0)$ is $(a \neq 0 \land b \neq 0) \implies ab \neq 0.$

Proceed by contradiction. Suppose $a \neq 0 \land b \neq 0$ and, for contradiction, assume that $ab = 0.$ Taking $(ab = 0) \times \frac1a$ gives $b = 0$, which contradicts the supposition that $b \neq 0.$ Therefore $ab \neq 0,$ thus $(a \neq 0 \land b \neq 0) \implies ab \neq 0.$

Therefore, by contraposition, $ab = 0 \implies (a=0 \lor b=0).$

ryang
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weatherisheinous
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    Looks fine to me! Doing the contrapositive isn't necessary; I think you can make the same argument by contradiction on the original statement. – Pavan C. May 02 '21 at 06:09
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    Spot on valid. Minor nuance: your use of $(1/a)$ is perfectly valid. You might mention to the reader that this is permitted where you have used it, because it is being assumed that $a \neq 0.$ – user2661923 May 02 '21 at 06:10
  • Great, thanks @PavanC. for catching that – weatherisheinous May 02 '21 at 06:35
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    When proving such a basic statement as this, whether your proof is valid depends very heavily on what facts you already know and are allowed to use in your proof. So we can't tell you whether your proof is valid without more information about your context. – Eric Wofsey May 02 '21 at 14:08
  • It seems excessively compicated to me. A more direct expression of the same idea goes like this: Suppose $ab=0$. Either $a=0$ or $a≠0$. In the former case there is nothing to prove. In the latter cass multiply by $ab=0$ by $\frac1a$ as you did, to obtain $b=0$. – MJD Jun 07 '21 at 09:03

2 Answers2

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Doing it as a proof by contradiction is completely unnecessary. You never use the assumption that $b\neq 0$ except to contradict your conclusion that $b=0$. In essence, you are doing a "fake proof by contradiction" of the contrapositive, by doing a contrapositive proof of the contrapositive. In fact, you are doing a direct proof in the first place!

Why do I say that? You start from $ab=0$. Then you say: "if $a\neq 0$, then multiplying by $\frac{1}{a}$ we conclude $b=0$"; fine up to here, but then you just use that conclusion to contradict your assumption that $b\neq 0$. Why bother? Just conclude that if $ab=0$ and $a\neq 0$ , then $b=0$...

And then you are done! Why? Because $P\vee Q$ is equivalent to $\neg P\implies Q$; so by proving that $\neg(a=0)\implies b=0$, you've actually proven that $(a=0)\vee (b=0)$... which is what you wanted to prove in the first place...

Or alternatively: assume $ab=0$. If $a=0$, you are done. If $a\neq 0$, then... and you put down your argument. So you've proven that either $a=0$ or $b=0$. No need for contrapositives, contradictions, or a double secret probation reversal...

Arturo Magidin
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Your proposed solution has a small internal error in the proof-by-contradiction portion: the negation of the contrapositive$$(a \neq 0 \land b \neq 0) \implies ab \neq 0,\tag{*}$$ is $$a \neq 0 \land b \neq 0 \land ab=0,$$ which you use as your supposition; when the contradiction later surfaces via $b$ turning out to be $0,$ the correct deduction is then that $$(a \neq 0 \land b \neq 0) \implies ab \neq 0,\tag{*}$$ but not necessarily because $$ab\neq0.$$

In any case, as pointed out by Pavan, contraposition unnecessarily convolutes the proof; here's a shorter one (although, as pointed out by Arturo, a direct proof suffices and contradiction is unnecessary either):

  • Let $a,b\in\mathbb R,$ suppose that $ab=0,$ and assume that $a,b\neq0.$

    Then $a=\frac{ab}b=\frac0b=0;$ this is a contradiction; therefore, the assumption is false; i.e., $a=0 \lor b=0.$

    Hence, for each $a,b\in\mathbb R,\quad ab=0\implies(a=0 \lor b=0).$

ryang
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