Can we discuss this?

[Leonhard]

The contradiction is do to using indeterminate values to make unequal answers.

Just a terminology pickle, its indeterminate form, not value.

0 x 1 = 0. 0 x 2 = 0. 0 x 100 = 0. So 0/0 is indeterminate. In an equation should 0/0 occur, its value is defined by the equality. Otherwise 0/0 by itself is indeterminate.

You weren't talking about indeterminate forms 37818, you were talking about '1/0', which is clearly not an indeterminate form.

As I've said, its relatively straightforward to prove that if you can divide 1 by 0, then you're forced to accept that '1 = 0'. There is no way around that.

The symbol for infinity has to be treated with care and can't be used as either a cardinal number or even a real number. However you used aleph_0. And the multiplication of zero and aleph_0 is zero. So if you try to decide by fiat of definition, that one divided by zero is equal to aleph zero, then by multiplying both sides of this equality with zero, you'd get 'one is equal to zero'. Clearly a contradiction! Ergo one divided by zero is not equal to aleph_0.

[Leonhard]

Yeah, I 'get' the concept but my brain really hates that equation since they are different. It reads as wrong to me - even though I understand why it isn't.

I get that, and that's why its sort of a hard equality to swallow. But its just two ways of writing the same number, '1' obviously being far more elegant than '0.999...'.

[37818]

. . .

You weren't talking about indeterminate forms 37818, you were talking about '1/0', which is clearly not an indeterminate form.

As I've said, its relatively straightforward to prove that if you can divide 1 by 0, then you're forced to accept that '1 = 0'. There is no way around that.
.

According to the accepted math which AFIK does not follow that indeterminate equality rule, as I personally understand it.

[Leonhard]

According to the accepted math which AFIK does not follow that indeterminate equality rule, as I personally understand it.

Its okay if you have your own rules, though I'm wondering whether they make as much sense as you hope. Again you seem to be inventing your own technical terminology, something I warned you about in another thread. You're free to call it that, but so far I sincerely have no idea what the rule is... and how it justifies '1/0 = a' without ending up with contradictions.

And again '1/0' is not an indeterminate form.

[Boxing Pythagoras]

It still looks way wrong. 1 = 2 looks wrong for the same reason. But I do get why one makes some sense...

0.999... equals 1 for exactly the same reasons that:

3-2 = 1
x/x = 1
x^0 = 1
e^(2iπ) = 1
Sum(n=2→∞) 1/n = 1

Et cetera, et cetera.

[37818]

Its okay if you have your own rules, though I'm wondering whether they make as much sense as you hope. Again you seem to be inventing your own technical terminology, something I warned you about in another thread. You're free to call it that, but so far I sincerely have no idea what the rule is... and how it justifies '1/0 = a' without ending up with contradictions.

And again '1/0' is not an indeterminate form.

If you graph y = 2x - 20
And graph 5 + y/(x - 10) = 7
You will obtain the same graph, except at y=0 and x=10. Which is the same at tangency. Where y/(x - 10) is 0/0 is equal to 2. In my view it is the same. In the standard view, I understand that it is not.

[Leonhard]

If you graph y = 2x - 20
And graph 5 + y/(x - 10) = 7
You will obtain the same graph, except at y=0 and x=10. Which is the same at tangency. Where y/(x - 10) is 0/0 is equal to 2. In my view it is the same. In the standard view, I understand that it is not.

You have to understand what exactly it is you're doing here. I know that you can plug both equations into a grapher, and it'll produce a neat line, and that neat line, will touch at the point you've listed. Seemingly without a hole. However, what's done is that you have a computer algorithm, checking whether a given set of points closely enough solves the equation, and if yes plots that point.

This can give rise to problems depending on what you're trying to draw, and if it was a perfect system, it should be able to recognise (at least occasionally) something mathematicians call a removable singularity, which is basically what you're pointing out here. That's what the graph is... you've just given two different definitions of it, that are close, but not completely equivalent.

In reality, '0/0' is an indefinite form, and so there should be hole in the graph right there. And if you've got very sophisticated graphing software, it might recognise this and draw a tiny circle at that point, to indicate that this point is undefined.

However, the hole can be removed, by simple adding in the value.

This doesn't mean though, that you can divide by zero. Nor does it mean that '0/0 = 1' or '0/0 = 0' or whatever. The fact is that it can have any value you want without contradiction if you approach it as the limit of f(x)/g(x) where both f(x) and g(x) go towards zero for x going towards c. That's why its indeterminate.

I've told, if you can divide one by zero, then inescapably '0 = 1'. When I asked you about this you replied that this was merely because 'standard' math didn't follow your own indeterminate equality rule. So far you haven't explained it, or shown how it avoids the contradiction I've given you.

[Leonhard]

Sum(n=2→∞) 1/n = 1

Actually the limit of the sum of 1/n tends towards infinity.

[Boxing Pythagoras]

Actually the limit of the sum of 1/n tends towards infinity.

D'oh! Brain fart, on that one. You're absolutely right. Should have been:

Sum(n=1→∞) 1/(2^n) = 1

[Leonhard]

Here's a quick one line proof.

[Teallaura]

0.999... equals 1 for exactly the same reasons that:

3-2 = 1
x/x = 1
x^0 = 1
e^(2iπ) = 1
Sum(n=2→∞) 1/n = 1

Et cetera, et cetera.

No, it doesn't. Don't be silly.

[Boxing Pythagoras]

No, it doesn't. Don't be silly.

Actually, it does. Any time we write "A = B," that statement is true if and only if the expression A represents the same value as the expression B represents. So, the value of (3-2) is the same as the value of 1. The value of (x^0) is the same as the value of 1. The value of [e^(2iπ)] is the same as the value of 1.

In exactly the same way, the value of 0.999... is the same as the value of 1.

[Teallaura]

No, it isn't. One is not an approximation of one.

[Leonhard]

No, it isn't. One is not an approximation of one.

0.999... is not an approximation of one, 0.999... is equivalent with 1. There's no difference, except how its written. I know its counter intuitive, but its just a quirk of repeating decimal fractions that you can write 1 in two different ways both as '1.0' and as '0.999...'

[Teallaura]

0.999... is not an approximation of one, 0.999... is equivalent with 1. There's no difference, except how its written. I know its counter intuitive, but its just a quirk of repeating decimal fractions that you can write 1 in two different ways both as '1.0' and as '0.999...'

Really? So you're saying the third place has no significance? That's not counter-intuitive - it's counter historical. Wasn't one of the main problems with Newtonian physics that it begins to fail as you get out beyond four or five (more? It's been a long time...) places?

I get that they have no practical difference (not sure about machining, actually...) and don't have issue with the concept. But X and Y aren't the same thing. And I don't even take issue with you can write 1 as 0.999 since there is no practical (machining?) difference - but my little brain ain't gonna go with 'same' when I see a difference.

It makes me a great Hocus-Pocus player whether or not I'd ever make much of a mathematician. (Actually, given the utter incompetence of mathematicians at buying groceries, this doesn't bother me at all. Can't keep a running tab without paper or a calculator.. )