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Is the following proof by contradiction valid?

Theorem: Let $a$ and $b$ be real numbers. If $ab=0,$ then $a=0$ or $b=0$ (or both).

Proof: We will prove this by contradiction. Assume that $ab=0$ and $a\neq 0$ and $b\neq 0.$ Since $a \neq 0$ and $b \neq 0,$ there exist $a^{-1}$ and $b^{-1}$ such that $aa^{-1}=1$ and $bb^{-1}=1.$ Thus, since $ab=0,$ we can multiply both sides of this equation by $a^{-1}b^{-1}:$ $$a^{-1}b^{-1}ab=a^{-1}b^{-1}0.$$ Thus, rearranging the terms gives us $$aa^{-1}bb^{-1}=0,$$ so that we have $$1=0.$$ But this is absurd. So it must be the case that $a = 0$ or $b = 0.$ This completes the proof. $\square$

Where does this proof contradict the hypothesis? Does it have to? Is the conclusion that $1=0$ sufficient because it's absurd? Any insight is appreciated. Thank you.

Bill Dubuque
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nomadvagabond
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    Any contradiction is enough. In a sense, all contradictions are the same, by the “principle of explosion” – Malady Dec 15 '24 at 16:09
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    I mean... $1=0$ looks absurd enough to me – Davide Masi Dec 15 '24 at 16:11
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    @DavideMasi Agreed. I'm asking if this is enough, or if we have to contradict the hypothesis, namely that $ab=0.$ – nomadvagabond Dec 15 '24 at 16:12
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    Now I get what you meant. There is no need to contradict the hypothesis. Every contradiction with other axioms or theorems is fine – Davide Masi Dec 15 '24 at 16:13
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    @DavideMasi Awesome - that's what I assumed as well, but I wanted to verify. Thank you! – nomadvagabond Dec 15 '24 at 16:14
  • I suggest that you read https://math.stackexchange.com/questions/4236804/validity-of-the-law-of-excluded-middle – Davide Masi Dec 15 '24 at 16:32
  • It contradicts the hypothesis that $1\neq 0$ in a field (or, equivalently, that fields/domains are not trivial rings, i.e. not ${0}).,$ See here in the linked dupe for elaboration (include a more extreme form of this contradiciton). – Bill Dubuque Dec 15 '24 at 18:20

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At face value, something like that should be enough to make a proof by contradiction. However, if this is some sort of introductory mathematics class you are following where you are discussing stuff like axioms, it could be that you would need to add more onto that proof.

As you may know, the field of Real Numbers (and any other field) has 6 axioms:

  1. Closure
  2. Multiplicative + Additive Identity
  3. Multiplicative + Additive Inverse
  4. Commutativity
  5. Associativity
  6. Distributivity

So if you are strictly using axioms, you would also have to prove that:

• Anything times 0 remains zero (I am sure you can arrive at that using axioms)

And

• Either the additive or the multiplicative identity is unique (then you can for sure logically say that 0=1 is absurd).

For this one, you can go by contradiction too:

Suppose for contradiction that the additive inverse is NOT unique. In other words, there exists a certain $y,z$, where $y≠z$ such that $$x+y=0=x+z$$ Hence,

$$x+y=x+z$$

Adding y to both sides, we get

$$y + (x+y) = y + (x+z) $$ $$(y+x) + y = (y+x) + z$$

And since $$y+x = 0$$ Then

$$0 + y = 0 + z$$ $$y=z$$

Which contradicts our initial statement that $y≠z$. Therefore, the additive identity is unique. So 0 is unique, and 0=1 is impossible as 0 is a unique value.

Therefore, that also means that your theorem holds true, i.e. If $ab=0$, then $a=0 $ and/or $b=0$.

Honestly I don't think you need to go this much in depth if you are proving something like that. But sometimes it is important that in maths we go to the very fundamentals in order to submit a 100% correct proof even if it's as trivial as this one!

Kris Nenezic
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