I need to prove the following statement: if $xy=0$ then $x=0$ $or$ $y=0$.
My first idea was to prove the negation of the original problem if $x\neq0$ and $y\neq0$ then $xy\neq0$ but I still don't know how to prove it rigorously.
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2The tags suggest you're in the context of real numbers. I'm guessing you were given a list of axioms. Were you? Please share it. – Git Gud Aug 31 '13 at 21:56
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To prove it rigorously, show that a contradiction arises when you assume $x \neq 0$ and $y \neq 0$ and $xy = 0$. – abiessu Aug 31 '13 at 21:57
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One way to do this is to assume that $xy=0$ and $x\not=0$, and then show that $y=0$. – user84413 Aug 31 '13 at 22:00
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1you mean "contrapositive", not "negation" – Stefan Smith Sep 01 '13 at 01:41
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Hint: Suppose that $x\ne 0$ and $xy=0.$ Since $x\ne 0,$ then there is a very special real number closely tied to $x$ with regard to multiplication. Check your axioms, and see if you can figure out what it could be. What will this real number allow us to do, since $1y=y$ and $z\cdot 0=0$ for all $z$?
Cameron Buie
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@Yossi: Indeed! Can you see what we can do with it? (Bear in mind that we assumed $x\ne 0,$ so what are we trying to show?) – Cameron Buie Aug 31 '13 at 22:10
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Suppose $xy=0$ and suppose $x\ne0$ and $y\ne 0$. Then there exists $x^{-1}$ and $y^{-1}$. Now, you should be able to derive a contradiction.
Alraxite
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Break it up into cases:
Case 1: x equals 0.
You are done proof in this case.
Case 2: x does not equal 0.
Since x does not equal 0, 1/x exists.
Take equation xy=0 and multiply both sides by 1/x:
(1/x)*xy = (1/x)*0
.. thus, y=0
So in this case we are also done.
General comments: Say you have a theorem of the form... If P then A or B
The way to prove this "directly" (meaning without contradiction) is to split it up into cases either using A or B.
That is if you choose to split it up on cases using A, the cases will be: 1) Assume A true. ...... Boom, we already proved that one of A or B is true (i.e., A).
2) Assume A not true. .... need to do some work to prove that B is true.
So when you become "pro" you usually don't show case 1), you usually just show the proof of 2), like:
Assume x does not equal 0. Then 1/x exists, so .... , boom y=0. Done.
do-leap
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How do you go from $(1/x)xy = (1/x)0 \implies y = (1/x)0 $ to $y=0$? – Stand with Gaza Dec 27 '18 at 08:21