1

I've come across a certain proof by contradiction at many places, textbooks and online likewise, which seems faulty to me. I'm no mathematician, but I just can't wrap my head around this issue.

The story goes like this:

Suppose dividing some integer by $0$ has some value:

$$a/0 = b$$

Now, first assume that $a$ is a non-zero integer. Then by multiplying both sides by $0$ we get:

$$a = b \cdot 0$$

but given the rules of arithmetic about multiplication with $0$ we arrive at a contradiction with our assumption that $a$ is non-zero.

Afterwards, assume that $a = 0$. Then, using the same step we get:

$$0 = b \cdot 0$$

which is unsatisfactory because it doesn't give a unique value, because for every $n$ it holds that $n \cdot 0 = 0$.

The problem I have with this argument is that it somehow presupposes the very thing it argues against, i.e. that division by zero is defined (at least when looking at the case when zero is divided by zero). What else could justify the move from

$$a/0 = b$$

to

$$(a/0) \cdot 0 = b \cdot 0$$

to

$$a\cdot 0/0 = b \cdot 0$$

to

$$a = b \cdot 0$$

if not a presupposition that $0/0 = 1$? In no other way can we leave $a$ sitting there alone on the left-hand side.

On the other hand, the same applies for the second assumption. Namely, we can get

$$0 = b \cdot 0$$

only if we suppose that

$$0/0 \cdot 0 = b \cdot 0$$

resolves to

$$0 = b \cdot 0$$

which in this case works even if we don't set its value to $1$, because any $n$ would suffice, but for the sake of the argument suppose it's still set to that.

It seems to me that discarding this presupposition the two steps of the argument would collapse into the same

$$0/0 = b \cdot 0$$

which opens a door to an infinite regress because we're back at the starting point again, leaving us with multiplying by $0$ again and having

$$0 \cdot 0/0 = b \cdot 0 \cdot 0$$

which is again the same old left-hand

$$0/0 = b \cdot 0 \cdot 0$$

Therefore, the presented argument is invalid in its aim. Leaving

$$0/0 = b \cdot 0 \cdot 0$$

to be standardly resolved as

$$0/0 = 0$$

Now, where am I wrong and what to make of this whole business?

Karlo Mikić
  • 37
  • You have shown which consequences it has if we divide by $0$. This is the reason that it is forbidden , $\frac{1}{0}$ is undefined as well as $\frac{0}{0}$ is. – Peter Nov 22 '22 at 17:34
  • It seems to me that I've shown that a certain kind of proof for why the result of division by zero can't be any integer, either zero or non-zero, isn't a proof, but a fallacious use of a common algebraic technique. And I can live with that. The real problem is that the implication of my remark here is that any number divided by zero is also a zero. Where did I err in my reasoning? – Karlo Mikić Nov 22 '22 at 17:40
  • Say , $\frac{a}{0}=b$ for real numbers $a,b$. Then, $a=0\cdot b=0$ , so $a\ne 0$ is impossible. The problem with $\frac{0}{0}=a$ is that $a\cdot 0=0$ holds whatever $a$ is. Since $\frac{0}{0}$ could be anything , there is no way to reasonably define it. – Peter Nov 22 '22 at 17:45
  • And that is the exact argument I presented here, only written more concisely. My issue with it is that I can't fathom what sanctions the move from the first to the second equation, i.e. from $a/0=b$ to $a = 0 \cdot b$. What leaves us with the $a$ on the left-hand side after we've multiplied both sides by $0$? How are we suppose to cancel out $a/0 \cdot 0$ to be left with only $a$? – Karlo Mikić Nov 22 '22 at 17:51
  • You should rephrase the entire argument in terms of the existence of a multiplicative inverse: "division by $n$" is the same as multiplication by the multiplicative inverse of $n$ (i.e. the number $n^{-1}$ such that $n^{-1} n = 1$). So when we prove "you can't divide by $0$", what we are actually saying is "there is no multiplicative inverse of $0$". The proof is then by contradiction: Suppose there is some multiplicative inverse $1/0$ of $0$, then it must be unique (this can be proven in sufficient generality), and it must satisfy $0(1/0) = 0/0 = 1$. Then your arguments show a contradiction. – stillconfused Nov 22 '22 at 17:58
  • Fair enough, and I can't really point any objections to such way of framing the issue. But, is my first claim - that the above algebraic "proof" isn't really a proof - a sound one then? Seems to me that it's ill-concieved then trying to prove this by that manner of algebraic manipulation. – Karlo Mikić Nov 22 '22 at 18:13
  • I think it also depends on what your definition and prerequisites are. Because $a\cdot 0=0$ is also something that may not be a prerequisite and first needs to be proven with the axioms of addition and multiplication. – LegNaiB Nov 22 '22 at 18:15

1 Answers1

1

The proper definition of $a/b$ is that it is the unique number you multiply $b$ by to get $a$.

Try to read that when $b=0$. Then you want the number that you multiply $0$ by to get $a$.

If $a \ne 0$ there is no such number.

If $a = 0$ any number will do for $0/0$. There is no unique number to define it as.

There is no need to invoke the rule that says $$ a \times \frac{b}{c} = \frac{ab}{c}. $$ (That rule follows from the definition of division.)

Ethan Bolker
  • 95,224
  • 7
  • 108
  • 199
  • Exactly, the point here is that division is defined as the reverse operation of multiplication, i.e. $\frac a b \cdot b = a$ for all $a,b$ by definition – LegNaiB Nov 22 '22 at 17:58
  • Of course, that makes sense. But still when looking the above argument in a step by step fashion I can't lose the impression it's wrong. Could you analyze it through the mentioned steps and point out what my mistake is? – Karlo Mikić Nov 22 '22 at 17:58
  • The thing is that you don't really "suppose" that $0/0 \cdot 0 = b\cdot 0$ resolves to $0 = b\cdot 0$. It is just the definition of multiplicative inverses. And writing $/0$ is nothing else than (supposing that a multiplicative inverse exists for $0$) multiplying with the theoretical value $0^{-1}$. – LegNaiB Nov 22 '22 at 18:00
  • Nor am I in fact. Only thing I did was using the non-controversial $b \cdot 0 = 0$ and substituting $0$ for $b \cdot 0$ in $0/0 \cdot 0 = b \cdot 0$. But I'm not really interested here in claiming that dividing by $0$ is somehow sensible. I'm interested in understanding is the algebraic step that is commonly invoked for proving that it's not sensible, the one from $a/0 = b$ to $a = b \cdot 0$ sensible itself. Because as far as I see it's not - because we can't cancel $0/0$ if we assume it's undefined, and then we can even less get $a = b \cdot 0$. – Karlo Mikić Nov 22 '22 at 18:06
  • Is that algebraic proof then merely a trick then? Should one resort to talking about multiplicative inverses to really show that there is no sensible way to define division by zero? – Karlo Mikić Nov 22 '22 at 18:09
  • 1
    It's difficult to first grab this as a non-mathematician: You said the "only" thing you did was substituting something in the $0/0 \cdot 0 = b\cdot 0$ equation. However the problem is in $0/0$, more generally in the division $a/b$ itself! There is no algebraic step between $a/0=b$ and $a=b\cdot 0$ as this is the exact definition of division!

    In math you generally have only addition and multiplication (which is also some sort of addition). Subtraction and division are no operations that exist by itself, you could replace them by additive/multiplicative inverse.

     – LegNaiB Nov 22 '22 at 18:10
  • So your question boils down to: "What is the algebraic step between multiplying $a$ with the multiplicative inverse of $0$ to get $b$ and getting $a$ by multiplying $b$ with $0$". There is none, as it is the same, just written differently. The multiplicative inverse of $a$ is defined as the unique number $c$ so that $a\cdot c = 1$. We then write $c$ as $a^{-1}$ or as $/a$ for short. – LegNaiB Nov 22 '22 at 18:11
  • Maybe it's clearer by using different notation: Assume that there exists a multiplicative inverse $c$ of $0$. Then $c\cdot 0 =1$ which is a contradiction to the known fact that $0\cdot a=0$ for all $a$. You are writing $0/0$ instead of $c\cdot 0$ which seems to confuse you – LegNaiB Nov 22 '22 at 18:13