Prove this number fact

Question

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.

Answer 1

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.

Answer 2

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 $y\neq 0$.

Answer 3

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.

Answer 4

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

Answer 5

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