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Prove that $x \neq 0,y \neq 0 \Rightarrow xy \neq 0$.

Suppose $xy = 0$. Then $\frac{xy}{xy} = 1$. Can we say that $\frac{xy}{xy} = 0$ and hence $1 = 0$ which is a contradiction? I thought $\frac{0}{0}$ was undefined.

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Proofs that have 0 in the denominator are usually suspect! –  GEdgar Aug 20 '11 at 15:05
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This proof completely depends on the axioms you choose to use, as in some universes this implication does not hold (e.g. $\mathbb{Z}_n$, where n is not prime$) –  sxd Aug 20 '11 at 15:10
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As has been explained, the proposed argument is not right. To help you with an appropriate proof, one needs context. For what kind of numbers? What kind of course? –  André Nicolas Aug 20 '11 at 17:07
    
@Qia No, his hypothesis is $x,y\ne 0\:.$ It's a proof by contradiction - see my answer. –  Bill Dubuque Aug 20 '11 at 21:13
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@GEdgar The proof is correct - see my answer. –  Bill Dubuque Aug 20 '11 at 22:02

5 Answers 5

up vote 10 down vote accepted

The proof is valid in any field $\rm\: K$ (though it might be circular depending on the context). Namely, $\rm\:0\ne x,y\in K\:$ $\:\Rightarrow\:$ $\rm\: 1/x,1/y\in K\:$ $\:\Rightarrow\:$ $\rm\:(1/x)(1/y) = 1/(xy)\in K\:$ $\:\Rightarrow\:$ $\rm\:xy\ne 0\:.$ The OP's proof is simply this proof recast into a proof by contradiction. To be precise the OP's proof is as follows:

As above, $\rm\: x,y\ne 0\ \Rightarrow\ z := 1/(xy)\in K\:,\:$ i.e. $\rm\:xyz = 1\:.\:$ So $\rm\ xy=0\ \Rightarrow\ 0 = 1\:,\:$ a contradiction.

That's precisely the OP's proof, except I've replaced $\rm\:xy/(xy)\:$ by $\rm\:xyz\:$ to avoid possible confusion.

This is a valid proof. The confusion stems from the fact that it is a proof by contradiction. Such proofs - by their very nature - may encounter all sorts of strange looking mathematical objects, such as the above expression of the form $\rm\: 0/0 = 1\:.\:$ This is just $\rm\:1/1 = 1\:$ in the trivial ring $\:\{0\}\:$ where $\rm\:0 = 1\:.\:$ However, the trivial ring is not a field, since $\rm\:0\ne 1\:$ by the definition of a field (or integral domain). So, as above, $\rm\:0 = 1\:$ is a common target for proofs by contradiction in a field.

Proofs by contradiction often prove immensely confusing to students when first encountered. Learning to wrap one's mind around the bizarre contradictory objects encountered in such proofs is skill that comes with practice. A striking example of such confusion is Euclid's classical proof that there are infinitely many primes. Although Euclid's proof was constructive, it is widely presented as a proof by contradiction (and falsely claimed that this was Euclid's proof). When presented in contradictory form this proof often leads to much confusion. There are hundreds of threads on sci.math permeated by such confusion. One can reach all sorts of contradictions to terminate Euclid's proof, e.g. $\rm\:0 = 1\:$ or $\rm\: 1\:$ is prime, or some integer is both prime and composite, etc. Indeed, one can deduce anything in a contradictory theory such as the integers with finitely many primes. Such contradictions often prove too much to grasp for many beginners. Apparently this is because we have such strong intuition about integers that one contradiction easily implies many others, and this quickly grows too much to handle intuitively. This does not occur to the same degree when one works with more abstract structures, where real-world intuition has less chance to restrain logical thought processes. Such is the strange nature of proofs by contradiction.

Note $\ $ The OP has revealed the source as Proposition 1.16 in Rudin's Principles of Mathematial Analysis. I've appended it below. It is essentially as I surmised above.enter image description here

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I don't think that's what OP had in mind... –  anon Aug 20 '11 at 17:33
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@anon Only the OP can say what they "had in mind". My point is that the above is one valid way of viewing it. –  Bill Dubuque Aug 20 '11 at 17:33
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@Bill, assuming $xy=0$ means that one cannot write $\frac{z}{xy}$ validly, since only numbers which are nonzero have an inverse. If you can write $\frac{z}{xy}$ then you assume that $xy\neq 0$ implicitly. –  Asaf Karagila Aug 20 '11 at 22:48
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The way I originally understood OP, the two cornerstones of his proof are that $a/a=1$ and $0/a=0$ unilaterally for all $a$ (these are not sound assumptions, as they don't make sense with $a=0$). I wouldn't be surprised if OP did, in fact, have your specific reasoning in mind, but I see no actual evidence of it in the question. And while your proof by contradiction is obviously valid, I think the confusion here is whether or not it's a faithful interpretation of the OP. Since @Numberguy hasn't responded I say the point is moot for the time being. –  anon Aug 20 '11 at 23:17
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This is why I think my interpretation is the most sensible and parsimonious: (1) The number of people aware of the basic algebraic rules $a/a=1$ and $0/a=0$ vastly outnumbers the number of people who can do a fleshed out proof by contradiction. (2) The OP includes every step that is from my interpretation, and includes 0 steps from alternative interpretations to distinguish otherwise. (3) A valid textbook proof by contradiction would be difficult to "paraphrase" in the way OP wrote the question, as the expression $\frac{xy}{xy}$ would not be present without any explanation or lead-in. –  anon Aug 20 '11 at 23:40

The fault occurs immediately: "Suppose $xy=0$. Then $\frac{xy}{xy}=1$." This is not true, as the property $\frac{a}{a} = 1$ holds only when $ a\neq 0 $, and we have just assumed that $xy$ is indeed $0$.

However, we can still go along the road of contraction. Suppose $xy=0$. By assumption $x\neq 0 $ so we may divide $xy$ by $x$ to obtain $y=0$, which is in contraction with the other assumption that $y\neq 0$.

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No, the proof is correct if viewed appropriately. See my answer. –  Bill Dubuque Aug 20 '11 at 19:53
    
I think the OP just wrote the "suppose" line too soon, which is the cause for the confusion. "It is given that $x \neq 0$, $y \neq 0$. So $\frac{xy}{xy} = 1$. Suppose $xy=0$. Then we get a contradiction <somehow> (e.g., because $\frac{xy}{xy}=1$ is not true for $xy=0$)." Something along these lines is the proof the OP seems to have been reaching for. –  ShreevatsaR Aug 21 '11 at 7:29
    
@Shr That's not the problem. Every proof by contradiction starts that way. Rather, the problem is that proofs by contradiction can be very difficult to comprehend when the negated statement contradicts facts that are deeply hardwired into our intuition. The large number of erroneous votes here are witness to that, along with many hundreds of analogous erroneous posts on other math forums, e.g. sci.math. –  Bill Dubuque Aug 21 '11 at 19:31

consider that xy=0 from the basic axioms we know that product is zero when at least one of the multiplier is zero so is means that x is zero or y is zero or both x and y is equal zero which contradicts to given condition that neither x or y is zero so it means that their product is also not equal to zero

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"$A$ when $B$" means "$B\Rightarrow A$", yet you are treating this as though it were "$A\Rightarrow B$," a logical fallacy. –  jwodder Aug 20 '11 at 22:10
    
@jwodder actually the original statement says "$x\neq 0 \land y\neq 0 \implies xy \neq 0$, and this post attempts to prove $xy = 0 \implies x = 0\lor y = 0$ which is perfectly equivalent (although he seems to be saying things about contradiction which is unnecessary) and hence not a logical fallacy –  Deven Ware Aug 21 '11 at 6:36

If xy=0, then x=0 or y=0. So, by contraposition, if is not the case that "x=0 OR y=0", then it is not the case that xy=0. By De Morgan's Laws, we have that "if it is not the case that x=0 AND it is not the case that y=0, then it is not the case that xy=0". Now we move the negations around as a "quantifier exchange" and we have "if x is not equal to 0 and y is not equal to 0, then xy is not equal to 0."

If xy=0, we can't say that xy/xy=1. Division doesn't exist in the rational numbers. It only exists in non-zero rational numbers, or if you prohibit 0 in any denominator. xy/xy=1 only on the condition that "/" is defined. So this "Suppose xy=0. Then xy/xy=1." isn't correct.

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Sorry the answer is a bit late, but I just saw this post and happened to have Rudin's book on me so I've taken a look at the proposition the OP was referring to.

1.16 (b) If $x\neq 0$ and $y\neq 0$ then $xy \neq 0$.

the proof essentially goes as follows

Suppose $x\neq 0$ and $y\neq 0$, and suppose for contradiction that $xy = 0$. Since $x, y\neq 0$ we know that $\frac{1}{x}, \frac{1}{y}\in F$, Hence $1 = (\frac{1}{x})(\frac{1}{y})xy = (\frac{1}{x})(\frac{1}{y})0 = 0$ a contradiction.

The fallacy given in the OP is that since we assume that $xy = 0$ we cannot write $\frac{xy}{xy}$ because we are clearly dividing by zero. However, we can write $(\frac{1}{x})(\frac{1}{y})$ because we know both are in the field (and hence their product is as well) since $x, y \neq 0$.

I hope this helps even though it's late

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There is no "fallacy". The expression $(xy)/(xy)$ denotes $xy(xy)^{-1} = xyz$. See my answer. –  Bill Dubuque Aug 21 '11 at 6:45
    
@Bill Dubuque I see you're answer and I agree that your proof is correct, I simply answered because I went to see the proof in Rudin and hoped another post may address confusion the OP had that may have stemmed from the proof he laid out there –  Deven Ware Aug 21 '11 at 6:59

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