Dividing a number by zero
- Thread starter chikis
- Start date
I thought it was equal to 0 = -1, in that case there is no solution for P, or it is no reals.
All false statements are equivalent.
Starting from anything false (like (0 * P) + 6 = 5), you can conclude anything at all (eg 6=5, 0=-1, or even "Your mama's not fat.")
Actually, I think P is -∞.
(0*(-∞))+6 = -1 + 6 = 5
Of course it can produce any other answer but its the only "value" that fits.
Ew, I need to scrub my eyes.
.
that's if you're trying to solve for P. you get P = -1/0
$$0 \times P = 5 -6 = -1$$
$$P = -1/0$$
///ERROR Will Robinson. ERROR!
Division by zero.
Does not compute.
Core dump follows: 0394048583743FAB43949CD3940390AAA3485401285405885...
Dave, my mind is going. I can feel it. I ... can ... feel ... it.
Would you like me to sing you a song, Dave?
Daisy, Daisy, give me your answer do...
xxx Blurk Kerching.
This thread fully account of the various division by zero issues contributed by many users. It also briefly talk about infinity. The only thing it does not account for is whether there is a proof on every conceivable number system involving division by zero must end with contradictions
Took ages to find the thread because the search function does not search any threads beyond a year ago (return only my latest two threads), which I am not sure whether its a bug or intended
Therefore there exist no P in (R or C or Q etc.) such that (0 * P) + 6 = 5 is true
Took ages to find the thread because the search function does not search any threads beyond a year ago (return only my latest two threads), which I am not sure whether its a bug or intended
Therefore there exist no P in (R or C or Q etc.) such that (0 * P) + 6 = 5 is true
KitemanSA
Registered Senior Member
Sorry, his equation was correct. Anything except zero, when divided by zero equals infinity with the appropriate sign. 0/0 and infinity*0(and their inverses) are undefined.
"Exist" in this context means "is a number in a logically well-defined system"
Even in systems with numbers bigger than any finite numbers, transfinite numbers, multiplying any number by zero is still zero.
Introducing a "reciprocal" of zero breaks arithmetic in many ways.
$$ \color{red} 1 = 0 \times \infty = 0^2 \times \infty = 0 \times ( 0 \times \infty ) = 0 \times 1 = 0
N = ( M - 1 ) + (N + 1 - M ) = [ ( (M - 1) \times \infty ) + ( ( N + 1 - M ) \times \infty ) ] \times 0 = ( ( M - 1 ) \times \infty ) \times 0 + 0 \times ( ( N + 1 - M ) \times \infty ) = ( ( M - 1 ) \times 1 ) + ( 0 \times \infty ) = M$$
// Edit -- the red in the above mathematics was used to show that the above trains of symbols are Contradictions and reasons why the reciprocal of zero is not a number
Even in systems with numbers bigger than any finite numbers, transfinite numbers, multiplying any number by zero is still zero.
Introducing a "reciprocal" of zero breaks arithmetic in many ways.
$$ \color{red} 1 = 0 \times \infty = 0^2 \times \infty = 0 \times ( 0 \times \infty ) = 0 \times 1 = 0
N = ( M - 1 ) + (N + 1 - M ) = [ ( (M - 1) \times \infty ) + ( ( N + 1 - M ) \times \infty ) ] \times 0 = ( ( M - 1 ) \times \infty ) \times 0 + 0 \times ( ( N + 1 - M ) \times \infty ) = ( ( M - 1 ) \times 1 ) + ( 0 \times \infty ) = M$$
// Edit -- the red in the above mathematics was used to show that the above trains of symbols are Contradictions and reasons why the reciprocal of zero is not a number
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So since you say $$\frac{-1}{0} = -\infty$$.
Does that mean $$ -\infty \times 0 = -1$$?
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$$\lim_{x\to0^-}\frac{-1}{x}=+\infty$$ whereas $$\lim_{x\to0^+}\frac{-1}{x}=-\infty$$ So $$\frac{-1}{x}$$ goes in different directions depending on which side of 0 we approach. Hence $$\lim_{x\to0}\frac{-1}{x}$$ does not exist.What do you mean by "exist"? How do you define it?
You are right in that $$0/0$$ is not defined. But $$-1/0$$ is not $$-\infty$$ and neither is $$1/0$$ equal to $$\infty$$. This is an all too common mistake – and what’s more, people should stop writing things like $$-1/0$$ and $$1/0$$ and treating them as numbers (which they’re not).
Perhaps you’re thinking of $$\lim_{x\to\infty}\frac1x=0$$ and $$\lim_{x\to-\infty}\frac{-1}x=0$$. These limits exist. However $$\lim_{x\to0}\frac{\pm 1}{x}$$ do not exist (see above). If you’d only avoid writing the first two limits as $$1/\infty=0$$ and $$-1/-\infty=0$$ (which really you shouldn’t) then you wouldn’t be fooled into thinking that $$1/0=\infty$$ and $$-1/0=-\infty$$.
Tex Tip:
In tex, "x" shows as a cursive letter x, and so shouldn't be used for a multiplication symbol. Use \times instead.
$$7 \times 6 = 42$$
$$7 \times 6 = 42$$