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What is the exact meaning of $1/0$? Does that mean a number that is very large, a number that cannot be expressed as the one we know, infinite numbers of number so that giving one particular value is not feasible, or really undefined?

Just to check, consider functions $f_1(x) = 1/x$ and $f_2(x) = 1 - 1/x$. These two are "not defined" at $x = 0$. But their sum is. What should be interpreted from it? Can we say that $1/x$ was trying to get a value at $x = 0$, but $-1/x$ just cancelled it out and we get $1$? Or something else?

Ricky
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KeSHAW
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  • The sum of the function is $1$ where they are defined. We cannot sum undefined numbers , so the sum is also undefined at $x=0$. – Peter Oct 22 '23 at 07:55

2 Answers2

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It's undefined. Remember that division is reverse multiplication, so $8 \div 2$ is asking "what number times 2 equals 8?". $8 \div 0$ is asking "what number times 0 equals 8?". However, every number times 0 is 0, so there is no number that when multiplied by 0 equals 8 (or any other nonzero number). Therefore, any nonzero number divided by 0 is undefined. $0 \div 0$ is indeterminate since it is asking "what number times 0 is 0?". Since every number times 0 is 0, we cannot "determine" a correct value.

John
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Cogito ...

Taking a cue from the so-called imaginary number $i = \sqrt {-1}$, we could define $\frac{1}{0} = I$.

The value of a function like $f(x) = \frac{x + 3}{1 - x}$ at $x = 1$ would then be $f(1) = 4 \cdot \frac{1}{0} = 4I$

Special rules may be required to prevent pandemonium!!

Agent Smith
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  • Wouldn't work. By the rules of arithmetic, if $\frac{1}{0}=I$, then $1=0\cdot I = 0$, which is absurd. You'd essentially need to have two kinds of zero, one that behaved normally and one which as a divisor would give your result. But then how would the latter behave under multiplication and addition? This second kind of zero would have to give a nonzero result when multiplied by something, but unless you are saying that $I=1$ (which you aren't, by definition) then that wouldn't make sense either. – H. sapiens rex Oct 22 '23 at 08:39
  • $x^2 = -1$ also doesn't make sense (imaginary numbers). We need special rules and invertibility need not be a criterion for a definition (many functions, I hear, are noninvertible and yet defined over all real numbers). We hardly ever mix the reals with the imaginary (they remain distinct entities, perhaps exemplified by $a + bi$). The same may be done with $I = \frac{1}{0}$. We could obtain results like $9 + 3I$ for example. If no real satisfies $\frac{1}{0} = x$ then, $x$ must be "imaginary", oui? – Agent Smith Oct 22 '23 at 09:15
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    okay, then you should at least be able to tell me this: what would be the result of $I^2$? Another new kind of number? – H. sapiens rex Oct 22 '23 at 09:35
  • $I^2 = \frac{1}{0} \times \frac{1}{0} = \frac{1 \times 1}{0 \times 0} = \frac{1}{0} = I$ – Agent Smith Oct 22 '23 at 12:12
  • But that can't be right, can it? If $I^2=I$, then $\frac{I^2}{I}=\frac{I}{I}$, which implies that $I=1$. But $I=\frac{1}{0}$ as per your definition, which is decidedly not equal to $1$. – H. sapiens rex Oct 22 '23 at 15:21
  • Basically, what I'm trying to get at is that you cannot assert the existence of an entity with certain properties without first either a) constructing an entirely new theory to handle it or b) extending an existing theory to include it in such a way that it doesn't render past results invalid. Both ways require much more rigour than merely saying "let there be a number that kinda sorta acts similarly to the imaginary unit". In your case, I see no evidence of you constructing a new theory, so one is forced to assume that you intend an extension of the theory of complex numbers, but this results – H. sapiens rex Oct 22 '23 at 15:28
  • in absurd statements (with respect to the rest of the theory of complex numbers), which I have shown. – H. sapiens rex Oct 22 '23 at 15:29
  • $\frac{x}{x} = 1 \implies x = 1$??! – Agent Smith Oct 22 '23 at 15:52