1/(1/0)=?

[little boy]

1/(1/0)=? (not limit)

 

Is it zero or undetermined?

[Shadow]

[math]\frac{1}{\frac{1}{0}}=\frac{1}{?}=?[/math]. Division by zero is always undefined.

Edited September 8, 2011 by Shadow

[rktpro]

1/(1/0)=? (not limit)

 

Is it zero or undetermined?

 

It's infinite.

[little boy]

It's infinite.

 

I said it is not a limit, and it is 1/(1/0), not 1/0.

[Hal.]

If this was part of a calculation I did I would probably apply the rule of turning the denominator upside down and multiplying , giving 0/1 , then finishing thinking that 0 divided by an integer is 0 .

 

0

[khaled]

normally [math]\frac{1}{(\frac{1}{0})} = \frac{1}{?} = ?[/math],

 

but if you apply limit, then we have: [math]\lim_{x \to 0} \frac{1}{ (\frac{1}{x}) } \approx \lim_{x \to \infty} \frac{1}{x} = 0[/math],

 

I don't know if using limits would help finding the answer ...

Edited September 8, 2011 by khaled

[Realitycheck]

1/(1/0) is the same as 1*(0/1).

[khaled]

That gave me an idea, if [math]\frac{1}{(\frac{1}{0})} = 1 \times \frac{0}{1} = 1 \times 0 = 0[/math],

 

then we have [math]\frac{1}{(\frac{1}{0})} = {(\frac{1}{0})}^{-1} = (\frac{1}{0})^{-r} = 0[/math], besides [math](\frac{1}{0})^{0} = \frac{1^0}{0^0} = \frac{1}{?} = ?[/math]

 

we still get [math]\frac{1}{0} = {(\frac{1}{0})}^{+r} = {((\frac{1}{0})^{-1})}^{-1} = {0}^{-1} = \frac{1}{0} = ?[/math]

 

for any [math]-r \in \mathcal{R}^{-}[/math], and [math]+r \in \mathcal{R}^{+}[/math] ...

Edited September 8, 2011 by khaled

[math-helper]

The denominator of the main fraction is undefined and and the answer should be zero, as 1/undefined = 0

[Shadow]

It's infinite.

 

No.

 

If this was part of a calculation I did I would probably apply the rule of turning the denominator upside down and multiplying , giving 0/1 , then finishing thinking that 0 divided by an integer is 0 .

 

0

No. The identity [math]\frac{\frac{a}{b}}{\frac{c}{d}} = \frac{ad}{bc}[/math] holds only for [math]b, c, d \neq 0[/math]. It is for this reason that, when simplifying an algebraic expression, the variable conditions you write out pertain to the original expression, not the one you end up with after simplification. Ie. [math]\frac{(x-2)(x-3)}{(x-4)(x-3)} = \frac{x-2}{x-4}, x \neq 4 \wedge x \neq 3[/math]. What you're saying by this is that [math]\frac{(x-2)(x-3)}{(x-4)(x-3)}[/math] is the same as [math]\frac{x-2}{x-4}[/math], but ONLY IF x isn't four or three. Because if it were equal to either three or four, the former expression wouldn't be the same as the latter. For example, if x=3, the former expression turns out to be [math]\frac{0}{0}[/math], while the latter -1. And I think we can agree that [math] \frac{0}{0} \neq -1[/math]. And it's the same with [math]\frac{\frac{a}{b}}{\frac{c}{d}} = \frac{ad}{bc}[/math]. The two are the same ONLY IF b, c and d aren't zero. If they are, the identity doesn't apply.

 

1/(1/0) is the same as 1*(0/1).

That gave me an idea, if [math]\frac{1}{(\frac{1}{0})} = 1 \times \frac{0}{1} = 1 \times 0 = 0[/math],

 

then we have [math]\frac{1}{(\frac{1}{0})} = {(\frac{1}{0})}^{-1} = (\frac{1}{0})^{-r} = 0[/math], besides [math](\frac{1}{0})^{0} = \frac{1^0}{0^0} = \frac{1}{?} = ?[/math]

 

we still get [math]\frac{1}{0} = {(\frac{1}{0})}^{+r} = {((\frac{1}{0})^{-1})}^{-1} = {0}^{-1} = \frac{1}{0} = ?[/math]

 

for any [math]-r \in \mathcal{R}^{-}[/math], and [math]+r \in \mathcal{R}^{+}[/math] ...

No. See above.

 

The denominator of the main fraction is undefined and and the answer should be zero, as 1/undefined = 0

No. Dividing one, or any number for that matter, by an undefined quantity results in yet another undefined quantity. Also, it's not just dividing; doing anything with an undefined quantity gives an undefined result.

Edited September 9, 2011 by Shadow

[Hal.]

shadow ,

 

If you were a math lecturer and you told me that I was to apply the rules your way , then I would . If you are correct , then I've learned a new exception or maybe forgotten an old one I shouldn't have forgotten , so thanks .

 

 

[Shadow]

I'm not a math teacher, but I'm as positive as I can be about this. [math]\frac{1}{\frac{1}{x}} = x \Leftrightarrow x \neq 0[/math]. And you can really ignore all this and just think of it this way. If you evaluate, for example [math]\frac{3}{\frac{3}{2}}[/math], by flipping the denominator, you get the same result as you would by first evaluating [math]\frac{3}{2}[/math] and then dividing 3 with it. However, if you evaluate [math]\frac{1}{\frac{1}{0}}[/math] by flipping the denominator, you would end up with zero, and if you do it the other way, you end up with undefined. This alone should set off alarm bells.

 

I'd be more comfortable if someone with a higher education than mine would confirm this, but I'm pretty confident I'm right.

Edited September 10, 2011 by Shadow

[khaled]

1: [math]\frac{1}{(\frac{1}{0})} = 1 \times \frac{0}{1} = 0[/math],

 

2: [math]\frac{1}{(\frac{1}{0})} \;\; \approx \;\; \lim_{x \to 0} \frac{1}{(\frac{1}{x})} \;\; = \;\; \lim_{x \to \infty} \frac{1}{x} = 0[/math] .. in terms of limits

 

what do you think ?

Edited September 10, 2011 by khaled

[Shadow]

Like I said before, I agree with 2, I disagree with 1. While [math] \lim_{x \to 0} \frac{1}{\frac{1}{x}} = 0[/math], this does not mean that [math]\frac{1}{\frac{1}{0}} = 0[/math], in the same way that [math]\lim_{x \to \infty} \frac{1}{x} = 0[/math] does not mean that [math]\frac{1}{\infty} = 0[/math].

[Realitycheck]

If you want to add these other conditions correlated with these more expansive formulas, you can, but it's not the same formula anymore.

 

1/0 does not equal infinity though, but I see where you're coming from. Chalk it up to semantics. Undefined is not a number, equals zero.

 

For a bit of humor, see:

 

http://www.undefined.net/1/0

Edited September 10, 2011 by Realitycheck

[little boy]

The denominator of the main fraction is undefined and and the answer should be zero, as 1/undefined = 0

 

"undefined" has no meaning, how does it result in 1/undefined = 0?

[DrakeCennedig]

(Just a disclaimer: I'm no mathematician. I have nought but the wisdom of fools.)

 

I've been thinking a lot about dividing by zero. It's lack of definition is one of the least sensible details of mathematics. It seems to me that the only way to give 1 / (1 / 0) any kind of rational meaning would be to call it zero. The very term undefined is, to my line of thinking, just an excuse for not explaining the mathematics.

 

Consider this. If you were to take a perfect geometric cone, how would you calculate the number of points around one of its circumferences? You would have to divide the finite circumference by an interval of zero for each point. This would be "undefined" : yet we know that there are an infinite number of points around the cone. Therefore, shouldn't 1 / 0 be infinity?

 

Likewise, what would you get if you multiplied that infinite number of points with a circumference of zero? You would get one point, the tip of the perfect geometric cone. To my mind, this would imply that infinity * 0 = 1. In fact, a whole multiplicative family could be concluded from these "invalid" arguments (which I will note are based on valid geometry) : 1 / 0 = infinity, 1 / infinity = 0, infinity * 0 = 1

 

Another, related topic is 0 / 0, the undefined unit. If we are to simply accept the statement that any number divided by itself is equal to one, then 0 / 0 would have to equal 1. Evidence for this is the self-divided identity function, f(x) = x / x. For all values x ("except" 0), x / x = 1. Basic calculus will tell you that as x approaches 0, the limit of f(x) is 1. For some reason, the common consensus of mathematicians is that lim f(x) x --> 0 = undefined, but it certainly seems more sensible to say that x / x = 1, period. I can't imagine why an awkward codicil to a sensible law should break an otherwise continuous function.

 

Returning to the cone, we know that any two circumferences of a cone will have different sizes. One circumference might be one-pi long, and another might be 2-pi. To my uneducated mind, one circumference twice as big as another should have twice as many points, even if that means a larger version of infinity. A circumference of two divided by an interval of zero is what I would call 2-infinity, or two-times-infinity. It makes no logical sense to say that the number of points is the same across two unequal distances. Yes, both are infinite, but could not one remain larger if multiplied by zero?

 

Finally, this logic could also extend to zero itself : One empty box contains zero items, and a hundred empty boxes also contain zero items. We know that in each case, we have the same nothing: but just try storing a hundred empty boxes versus one, and you will find that a hundred boxes actually contain a lot more nothing. In all mathematical seriousness, I would say that in one box, you would have 1(0), or one-times-zero, and in a hundred boxes, you would have 100(0), or one-hundred-times-zero. Both values are nothing, but one can become larger. For what would you have if the postman comes to multiply the amount within each box by a single infinite amount, 1-infinity? If you only had one box, you would have one item, but if you had a hundred boxes, you would have a hundred items, for a hundred empty boxes, though they contain nothing, can contain much more than a single empty box.

 

If we were to accept the above arguments as true, the function 1 / (1 / 0) would follow order of operations and rationalize as 1 / 1-infinity, which would come out as zero (one-zero to be specific). Incidentally, calculus tells us that this is the limit of the function, as described in previous posts. This mathematics would work the same way as the limit of any polynomial equation. For example, if you find the limit of f(x) = 2x as x approaches 4, you get 8. This is exactly the same as if you had solved for f(4).

 

 

Some words about the idea that infinity is not a number:

 

We are taught from grade school that 1 + 1 = 2, 2 + 3 = 5, and 4 - 4 = 0. Zero is a very basic concept in mathematics, so basic that even gradeschoolers will eventually try dividing by zero whenever they start fiddling with their calculator. At its core, the concept of zero is of something that is infinitely small; infinity, by extension, is the concept of something infinitely large. It seems to me like petty semantics to say that infinity is "merely a concept", while zero is a "useful placeholder". After all, multiplying by zero gives us follies like 1 * 0 = 0; 2 * 0 = 0; 0 = 0, therefore 1 * 0 = 2* 0 and 1 = 2. Furthermore, when students are taught that any number divided by itself equals one, and they think to treat zero the same way, it seems fruitless to add a useless codicil about how zero is an exception to the rules. It doesn't seem particularly useful to say that some numbers defy the laws of mathematics; many curious people would consider such an answer to be of the caliber as "because I said so."

 

I would define infinity to be as much a placeholder as zero, I would extend both their jobs to include multiplication terms. 1 does not equal 2, thus, 1 * 0 does not equal 2 * 0, even though both are forms of zero. 1 * 0 = 1(0), just like 1 * x = 1x. 2 * 0 likewise equals 2(0), just like 2 * x = 2x. 1 * 0 = 1(0), which does not equal 2(0), and thus, 1 * 0 does not equal 2 * 0, and neither does 1 = 2. The practical usefulness of answers like these is in all honesty, trivial or nonexistent; but such answers do give us solutions to problems like, (1 * 0 + 2 *0) / 0. We could say that (1(0) + 2(0)) / 0 = (3(0)) / 0 = 3 (0/0) = 3(1) = 3. Just use a few infinite values and you come out with 3, just like limit theorems have suggested time and time again.

 

One final thought experiment, this time using the old standard, pies. One pie can be broken into two servings, creating servings of one half a pie (for one divided by two equals one half). If one whole pie is only half a serving of pie, then we obviously have very small pies, and two whole pies must equal one serving of pie (for one divided by one half equals two). If we were to break our pie into an infinite number of servings, each serving would be so small, it would quite literally be nothing (for one divided by infinity must equal zero). And finally, if our pie is infinitely small, it must quite literally have no size at all, and it would take an infinite number of them to make any serving (for I would say that one divided by zero must equal infinity).

 

Again, I am not a mathematician. I am not one who has devoted my life to the study of numbers and their relationships. I am merely a high-schooler learning calculus for the first time, and hence truly do have nothing but the wisdom of fools. But without some theorem proving the exceptionable quality of zero, I hardly think it reasonable to accept it as true. Science and math, after all, should require proof before acceptance as truth, and I, a humble student, have found none.

Edited September 29, 2011 by DrakeCennedig

[questionposter]

A number divided by 0 is always undefined because there is nothing determining how many times something 0 actually goes into something or how. It's a permanent gray area. So you'd basically have one over undefined.

Edited September 30, 2011 by questionposter