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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.

Update $ $ Replying to comments, the source of the proof is Proposition $1.16$ in Rudin's textbook Principles of mathematical analysis. I was trying to prove it and the proof there was basically the same as mine.

Bill Dubuque
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Numberguy
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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
  • @André See my answer. – Bill Dubuque Aug 20 '11 at 22:49
  • Re-tagged, as I think this is more algebra than number theory. – Gerry Myerson Aug 20 '11 at 23:16
  • Bill Dubuques answer is the one I want. – Numberguy Aug 20 '11 at 23:31
  • Could you please tell us about the source of the proof. Was it a proof that you found somewhere or one that you were attempting to construct based on a hint, or perhaps something altogether different? – Bill Dubuque Aug 20 '11 at 23:54
  • @Bill Dubuque: It was in Rudin's principles of math analysis. I was trying to prove it and the proof there was basically the same as mine. – Numberguy Aug 20 '11 at 23:58
  • If it's not too much trouble to tell me precisely where it is in Rudin I'd be grateful to know. – Bill Dubuque Aug 21 '11 at 00:04
  • The way to indicate which answer you want is to click in the little check mark to the left of it. – Gerry Myerson Aug 21 '11 at 00:18
  • @Bill Dubuque: Your proof is very attractive. But if integers have just been introduced semi-axiomatically, a considerable amount of work needs to be done before your proof becomes available. But one can prove the result the OP wants without additional conceptual apparatus. – André Nicolas Aug 21 '11 at 03:13
  • @André The context is rudimentary properties of a field, Prop. 1.16 in Rudin's PMA, see below. No considerable work is needed, only a general notation for inverses - see my comments to Asaf below. I had no luck explaining it to Asaf there, so I'm curious if it is comprehensible to you. I'd like to improve the answer but I'm not sure what is the true source of confusion for many. If you could lend any insight on this I'd be most grateful. – Bill Dubuque Aug 24 '11 at 02:00
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    I think the clickbait title of the question needs to be edited. Rudin does not use $0/0$. The OP does. – Martin Brandenburg Dec 15 '24 at 21:35

4 Answers4

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[Update OP revealed the source as Rudin's PMA. Despite prior doubts in the comments below, my (surmised) analysis below turns out to be essentially what was intended - cf. Note below].

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

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

That's precisely the OP's proof, except I replaced $\,xy/(xy)\,$ by $\,xyz\,$ 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 $\, 0/0 = 1.\,$ This is just $\,1/1 = 1\,$ in this trivial ring $\,\{0\}\,$ where $\,0 = 1.\,$ However, the trivial ring is not a field, since $\,0\ne 1\,$ by the definition of a field (or integral domain). So, as above, $\,0 = 1\,$ 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 Euclid's proof is presented by contradiction it often leads to much confusion. We can reach all sorts of contradictions to terminate Euclid's proof, e.g. $\,1 = 0\,$ or $\, 1\,$ is prime, or some integer is both prime and composite, etc. Indeed, by explosion we can deduce anything in a contradictory theory such as the integers with finitely many primes. There are hundreds of threads on sci.math permeated by such confusion, e.g. thread$_1$ & thread$_2$ & thread$_3$.

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 we work 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 later 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

Bill Dubuque
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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
  • The second proof you have isn't like the first one nor the OP. In the first one, you can start with the hypothesis that x does not equal 0 and y does not equal 0, which basically implies the rest. In the second one, if x, y does not equal 0, then xy is not equal to 0, and thus xy=0 is false (given a two-valued logic). Though such a proof ends up valid, if xy=0 gets taken as a hypothesis along with that x, y does not equal 0, it simply doesn't work out as consistent throughout, while just having a hypothesis of x, y not equal to 0 does work out as consistent (or at least not inconsistent). – Doug Spoonwood Aug 20 '11 at 21:12
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    @Doug What you wrote above shows confusion about the nature of proofs by contradiction. Comments are too short for a tutorial on such. Perhaps you might consider posing a new question about such. – Bill Dubuque Aug 20 '11 at 22:20
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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
  • @Asaf Not true. The above proof is correct. If you do in fact believe that the proof is incorrect then please state precisely which inference you think is incorrect and I'll be happy to set you straight. Beware that proofs by contradiction, by their nature, require extra care to comprehend correctly. – Bill Dubuque Aug 20 '11 at 22:58
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    @Bill The proof in the OP has xy=0. This happens as a supposition. It doesn't get inferred. Your proof infers to the conditional "xy=0 => 0=1". The antecedent of that conditional isn't true in the presence of x and y both not equal to 0. The supposition of xy=0 works fine in the OP. However, from that you can't infer that for a consistent field xy/(xy)=1, as the OP tries to do, since division by 0 is not defined in a consistent field. You haven't done that. You don't have that xy/xy=1 for all elements of the field as the OP suggests. You only have that for the non-zero elements. – Doug Spoonwood Aug 20 '11 at 23:09
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    @Bill Asaf didn't say either of the above proof wasn't correct. I haven't claimed either incorrect also. However, the first proof still comes as dissimilar to the first proof you gave. It does hold that xy=0 means you can't write z/(xy) validly from the assumption of consistency. A field does not define division for all elements of the field, but only nonzero elements. This "x,y≠0 ⇒ z:=1/(xy)∈K, i.e. xyz=1" doesn't claim otherwise, since the antecedent implies xy as a nonzero element. – Doug Spoonwood Aug 20 '11 at 23:13
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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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    @Bill You're simply wrong that I'm confused here, and at best even if I were you've only guessed at that instead of providing evidence or a discussion as to where my confusion lies. In truth, you're the one here who comes as confused, and as I've pointed out your confusion lies in thinking the proofs fundamentally similar and confusion a supposition and an inference, with a conditional. In logic, this is like confusing "p|-q" with "|- Cpq" which compares to confusing the object language with the metalanguage. This can get remedied by comments and discussion. – Doug Spoonwood Aug 20 '11 at 23:18
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    @anon This is a common proof by contradiction in field theory - one might even say a prototypical proof. I suspect that the OP was puzzled by one of these proofs, and paraphrased it here in his question. That fits perfectly with how the question is presented above. I see no other sensical way to interpret the question. – Bill Dubuque Aug 20 '11 at 23:26
  • @Doug There is nothing wrong with my proof. The only possible point of contention is the metamathematical matter of whether or not the proof the OP intends to discuss is the one I present or something else. I don't think the latter alternative is plausible given the way the question is presented. – Bill Dubuque Aug 20 '11 at 23:29
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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
  • @anon The first two sentences of the OP's question make it absolutely clear that it is a proof by contradiction. I'm baffled why anyone would think otherwise. Given that it is a proof by contradiction, I can see no other logical way to interpret the proposed proof other than as I explained above. – Bill Dubuque Aug 20 '11 at 23:43
  • @anon (3) is easily explained. The OP could be attempting to understand a textbook proof such as that I presented above. To attempt to understand it, he eliminates $z$ by replacing it by $1/(xy)$ and notices that this yields $xy/(xy)$ leading to the $0/0$ confusion. It makes perfect sense. – Bill Dubuque Aug 20 '11 at 23:45
  • Let me clarify: I do not believe OP's proof is not a proof by contradiction, I believe it is one that is not as fleshed out as yours. Specifically, my interpretation of OP's proof is $$(1)\quad a/a=1 \implies \frac{xy}{xy}=1,$$ $$(2)\quad 0/a\implies \frac{xy}{xy}=\frac{0}{xy}=0,$$ $$(3)\quad 1=0.$$ Both our interpretations agree with OP's on the initial hypotheses and (3), but only my interpretation explicitly includes what OP said for (1) and (2). – anon Aug 20 '11 at 23:53
  • @Bill Dubuque: It is proposition 1.16 in rudin. Can someone please convert this into a comment? Also I would like to accept Bill's answer but can't seem to check it. – Numberguy Aug 21 '11 at 00:39
  • @Numberguy: I have merged your duplicate account with the one you used to ask the question. You should now be able to accept Bill's answer by clicking the green check mark that is on the upper left of his answer. – Zev Chonoles Aug 21 '11 at 04:21
  • @Asaf Do you still believe your above comment is correct? If so, I'd be grateful if you could please elaborate. I've always been fascinated by cognitive hurdles encountered in comprehending proofs by contradiction. – Bill Dubuque Aug 21 '11 at 19:22
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    @Bill: Yes. I still think my comment as stated is true. The expression $\frac{0}{0}$ is invalid in any context. The same way I could write "Proof that $\pi=3$ by contradiction: Assume $\frac{0}{0}=1$, therefore: $$\pi=\pi\cdot 1 =\frac{\pi\cdot 0}{0} = \frac{0}{0}=\frac{3\cdot 0}{0}=3\cdot 1=3$$ The point is that if you assume that $xy=0$ then you cannot write $\frac{xy}{xy}$ in a meaningful way without intentionally introducing a false statement into your proof. You can, however, write $$xy=0\implies x^{-1}xy=x^{-1}\cdot 0\implies y=0$$ – Asaf Karagila Aug 21 '11 at 19:32
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    @Asaf $xy/(xy) = xy(1/x)(1/y)$ is well-defined: by hypothesis, $0 \ne x,y\in K$ so, by field axioms, they have inverses $1/x, 1/y$. So $xy(1/x)(1/y)$ is a product of $4$ elements of $K$. It's well-defined because multiplication is a total function. The proof by contradiction proceeds by hypothesizing additionally that $xy=0$. That one can now deduce myriad contradictions is not a bug, but rather a feature of proof by contradiction. In this case we even have a model where the proof may be interpreted, viz. the zero ring ${0}$ where $0 = 1$ so $ 0/0 = 1/1 = 1$ – Bill Dubuque Aug 21 '11 at 20:15
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    @Bill: Note that Rudin does not write $\frac{xy}{xy}$. Was he trying to extend his proof to consume more paper and damage the rain forests? I doubt that. Yes, the contradiction is indeed in the fact that one derives $0=1$. The point is that one derives it from writing $1=xy\frac{1}{x}\frac{1}{y}=0$, not from writing an invalid (except in the zero ring) expression as $\frac{0}{0}$. – Asaf Karagila Aug 21 '11 at 20:28
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    @Asaf At issue isn't Rudin's proof but, rather, the OP's reinterpretation in terms of $xy/(xy):.:$ To interpret the OP's proof I employ a notation for inverses that works universally, i.e. in all rings, including the zero ring, viz. in any commutative ring if $\ rs = 1\ $ then $1/r$ denotes $s$, the inverse of $r$ (necessarily unique). So in the proof since $xy(1/y)(1/x) = 1$ we may write $1/(xy)$ for $1/y\ 1/x$ in Rudin's proof, whch yields the OP's reinterpretation, after using the definition that $a/b = a(1/b)$ so $xy/(xy) = xy\ 1/(xy)$. There is nothing "invalid" about that. – Bill Dubuque Aug 23 '11 at 00:09
  • @Asaf That $0/0$ is invalid in rings except $0$ isn't relevant since the ring in the proof is $0$. It's akin to this. Consider this form of Euclid's classical proof of infinitely many primes: if $p_1,\ldots,p_n$ are all the primes then $1 < 1 + p_1\cdots p_n$ has no smaller prime factor so is prime, contra hypothesis. Some claim this is invalid since $1+2\cdot 3\cdots 13 = 59 \cdot 509$, so the proof does not construct a prime. That is not a valid objection since the proof does not work in N. Rather, it works in the hypothetical object $\ll$ N with a finite number of primes $\gg$. – Bill Dubuque Aug 23 '11 at 01:52
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    @Bill: The issue here that you appeal to the Rudin proof and claim the two are the same. They are not. Please do not interpret my lack of further response here as agreement with any of the points you have made so far. – Asaf Karagila Aug 23 '11 at 05:43
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    @Asaf The "sameness" of the proofs is not what we were discussing. Rather, we were discussing your (false) claims in your first two comments that my proof uses expressions "invalid in any context". Hopefully what I wrote above makes it clear that this is not true. The point of my proof is simply that the OP's reinterpretation of Rudin's proof can be made rigorous - in a manner that is quite natural algebraically, contra the many unfounded objections elsewhere on this web page. – Bill Dubuque Aug 23 '11 at 14:26
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    @Bill: While my interpretation seemed to me to have the better backing, and public support, it turns out in the end that you are in the right. Kudos. – anon Aug 24 '11 at 06:29
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    To me the main point here is that not enough care is taken to write exactly the right expressions; in the very early stage of theory we cannot write one expression for another because they can (later!) be proven equivalent (and we have the habit of normalising the form immediately). Notable, after "To be precise", one cannot replace $(\frac1x)(\frac1y)$ from Rudin's proof by $\frac1{xy}$. The field axioms allow writing $a^{-1}$ for any nonzero $a$ (and $b/a$ means $ba^{-1}$). So writing $\frac1{xy}$ requires having $xy\ne0$ as fact or assumption; not the case here. But Rudin's form is OK. – Marc van Leeuwen May 07 '16 at 07:11
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    Similarly, the second line of this answer would be fine if $(1/x)(1/y) = 1/(xy)$ (which again suffers from taking the inverse of a number not known to be nonzero) were replaced by $(1/x)(1/y)xy=1$ (which has no such problem, and contradicts with $xy=0$ as desired). – Marc van Leeuwen May 07 '16 at 07:17
  • See also my comments here. $\ \ $ – Bill Dubuque Dec 15 '24 at 22:36
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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 $y\neq 0$.

Ragib Zaman
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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 (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 07:29
  • @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
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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

Deven Ware
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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 06:45
  • @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 06:59
  • This should be the accepted answer. – Martin Brandenburg Dec 15 '24 at 21:30
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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.

Doug Spoonwood
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