Simple mathematical problem (?)

[tsg]

If you aren't convinced by the mathematical proofs, this isn't likely to help, but it does show that .9~ = 1 is at least mathematically consistent.

What is .999~ - .888~ ?

Now, what is 1 - .888~ ?

 

[geni]

No, it's not. Name a number between them. Giving a formula is not the same thing as naming it.

Fine then I name it fred.

2) .9=.99

huh?

It's irrelevant whether he's a legit professor. What matters is whether his argument is legit, and I don't see that it is.

Proveing that it isn't is tricky (as I understand it the problem is he is useing a non satandard defintion of reals).

How do you know that?

To start with the way atomic matter has a half-life of rather more a a second. Allowing the universe to be infinerly devisible causes serious issues with quantum physics.

 

[69dodge]

Fine then I name it fred.

Ha ha.

I think what he meant was this: You don't have to start with the assumption that .999... = 1. All you have to assume is that every real number has a decimal expansion. (The decimal expansion is its name.) Now, what's the decimal expansion of (1 + .999...) / 2 ?

 

[Art Vandelay]

Fine then I name it fred.

Facetious responses reaally don't add to the discussion.

huh?

.9 and .99 are, technically speaking, not the same expression, and it must be established that they represent the same quantity.

Proveing that it isn't is tricky (as I understand it the problem is he is useing a non satandard defintion of reals).

Then do it. But if he's using a "non satandard[sic] definition of reals", then he isn't talking about reals, he's talking about another number system which he is disingenuously labeling "reals".

To start with the way atomic matter has a half-life of rather more a a second. Allowing the universe to be infinerly devisible causes serious issues with quantum physics.

Not sure what you mean by the first statement. Pions, for instance, have a half-life of quite a bit less than a second, and furthermore I don't see what that has to do with anything. As for your second statement, you'll have to explain how that is so. In fact, standard formulation is based on calculus, and calculus depends on infinite divisibility.

 

[gnome]

Try this one out for size...

Let x=.999999...

Then x = .9 + .09 + .009 + .0009 + ... and so forth

If it is agreed that 1/9 = .1111111.... or
1/9 = .1 + .01 + .001 + ....

then x + 1/9 = (.9 + .1) + (.09 + .01) + (.009 + .001) + etc...

then x + 1/9 = 1.0 + 0.1 + 0.01 + 0.001 ...

then x + 1/9 = 1.0 + 1/9

then x = 1.0

 

[geni]

Facetious responses reaally don't add to the discussion.

square root 2 / sqare root of 3 can only really be represented in an eqation built around surds.

Then do it. But if he's using a "non satandard[sic] definition of reals", then he isn't talking about reals, he's talking about another number system which he is disingenuously labeling "reals".

You could try useing that argument. Again it is tricky to prove since first you have to figure out what the heck the guy is on about. It has been claimed that he is useing intuitionist version of the standard reals but that involves philosophy. Then again it has been claimed that he isn't so take your pick.

Not sure what you mean by the first statement. Pions, for instance, have a half-life of quite a bit less than a second, and furthermore I don't see what that has to do with anything.

If you have an infinetly divisable universe you run into the problem that electron orbital cease to be qantitiesed. This results in the election spiraling into the nucleas in fair short order (it also result in a load of other weird effects even if you ignore that one. Light bulbs would give out large amounts of gammer rays and the like).

As for your second statement, you'll have to explain how that is so. In fact, standard formulation is based on calculus, and calculus depends on infinite divisibility.

Since energy is quantitised so is space time. Planck length and all that.

 

[LW]

Oh dear, the Stars were Wrong and the Thread that Should Not Have Been Rose Again to Haunt the Mankind.

.0000...00001 IS 0

No. Because 0 is a real number and 0.000...01 is not. They can't be the same.

 

[jpowell]

If you have an infinetly divisable universe you run into the problem that electron orbital cease to be qantitiesed. This results in the election spiraling into the nucleas in fair short order (it also result in a load of other weird effects even if you ignore that one. Light bulbs would give out large amounts of gammer rays and the like).

Gah, as both a student of maths and physics this thread is really painful. You can apply quantum mechanics to atoms without requiring the quantisation of spacetime. I.e see the Schrodinger formulation, it requires continuous spacetime to work... otherwise you couldn't have differential operators in the wave equation.

Physics also has absolutely no input into mathematics. Maths is the study of axiomatic systems, mathematical statements are either true, false, or we don't know. There isn't an in between where something is half true and half false. Physics is the study of the universe... while mathematics can help with our insight into the universe, the universe can't tell us anything about mathematical theorems.

I'm also firmly on the side of the 0.9999... = 1 they are both two different representations of the same real number, in much the same way that the square root of 4 is another way of representing the real number 2.

 

[Art Vandelay]

square root 2 / sqare root of 3 can only really be represented in an eqation built around surds.

The whole issue is not whether (1+.9999...)/2 is a number, but whether it's between .9999... and 1. If I ask you to name a number that is between .81 and .82, and you give an answer like "sqrt(2/3)", that's not enough; you should give the decimal expansion up to at least three places to show that it satisfies the requirement. Similarly, it's not enough to simply declare that (1+.9999...)/2 is between .9999... and 1; you have to actually show it to be so.

You could try useing that argument. Again it is tricky to prove since first you have to figure out what the heck the guy is on about.

No, I don't. It's been proven that .9999... = 1, therefore anyone who disagrees is wrong.

Since energy is quantitised so is space time. Planck length and all that.

Except energy is not quantized. If energy were quantized, then we'd run into problems with the uncertainty principle.

 

[drkitten]

The whole issue is not whether (1+.9999...)/2 is a number, but whether it's between .9999... and 1. If I ask you to name a number that is between .81 and .82, and you give an answer like "sqrt(2/3)", that's not enough; you should give the decimal expansion up to at least three places to show that it satisfies the requirement.

No, I don't.

I can equally well show that 0.81 squared is 0.6561, which is less than 2/3, and that 0.82 squared is 0.6724, which is more than 2/3.

I can even do it by showing that 0.81 is 81/100, which is 6561/10000, and I don't need to use "decimal" representations at all.

Similarly, I can easily prove that, in general, for any two real numbers x < y, there is a number (x+y)/2 such that x < (x+y)/2 < y. (In fact, that's a standard proof in any elementary topology class.) IF you assume that 0.99999... < 1, then I simply declare x to be 0.999.... and y to be 1, and the proof is complete.

Similarly, it's not enough to simply declare that (1+.9999...)/2 is between .9999... and 1; you have to actually show it to be so.

Shown. It's a more general case of the previous theorem.

Of course, this also assumes that .9999 < 1, a statement with which I disagree. But that just shows the utter vacuity of the "show me a number in between" counterargument.

 

[69dodge]

[...] and I don't need to use "decimal" representations at all.

You don't need to, but you always can.

Similarly, I can easily prove that, in general, for any two real numbers x < y, there is a number (x+y)/2 such that x < (x+y)/2 < y. (In fact, that's a standard proof in any elementary topology class.) IF you assume that 0.99999... < 1, then I simply declare x to be 0.999.... and y to be 1, and the proof is complete.

What good is a proof that assumes its conclusion?

The question is precisely whether 0.999... < 1.

But that just shows the utter vacuity of the "show me a number in between" counterargument.

It seems a reasonable argument to me. If 0.999... differs from 1, then there's a real number between them, which has a decimal expansion, as every real number does. So, what is that decimal expansion?

 

[drkitten]

It seems a reasonable argument to me. If 0.999... differs from 1, then there's a real number between them, which has a decimal expansion, as every real number does. So, what is that decimal expansion?

"Obviously," 0.99999....5, where the .... represents an infinite string of nines.

Or perhaps our decimal notation itself is inadequate, precisely because it doesn't allow for the representation of infinitesimals.

There are any number of pseudoanswers that could be presented.

 

[Art Vandelay]

No, I don't.

You don't what?

I can equally well show that 0.81 squared is 0.6561, which is less than 2/3, and that 0.82 squared is 0.6724, which is more than 2/3.

Yes, that works (with a few more details, such as establishing that the squaring function is monotonically increasing on the positive reals). My point is that if I say "there are no real numbers between .81 and .82", then "sqrt(2/3)" is not a full counterexample to that claim. Giving a counterexample is not merely a matter of giving a instance which contradicts the claim; you need to prove that it does indeed contradict the claim.

IF you assume that 0.99999... < 1, then I simply declare x to be 0.999.... and y to be 1, and the proof is complete.

But that's just stupid. You're engaging in circular reasoning. The question of whether .999... is less than 1 is the very question that is at issue.

Of course, this also assumes that .9999 < 1, a statement with which I disagree. But that just shows the utter vacuity of the "show me a number in between" counterargument.

It is your response that is vacuous. If we assume that .999... doesn't equal 1, then we can prove than .999... doesn't equal 1. Whoop-to-freaking-do.

(.999...+1)/2=(1.999...)/2, right? So what is that equal to? Can it be anything but .999...? It's hard to imagine that it's not. So (1+x)/2=x, therefore x=1. It's not a rigorous argument, but it's hard to argue with it.

 

[gnome]

Something inside me rejects the idea of keeping a decent fellow mathematics enthusiast on ignore. So I've changed it back. I hope we don't butt heads so much on other topics.

 

[69dodge]

"Obviously," 0.99999....5, where the .... represents an infinite string of nines.

Or perhaps our decimal notation itself is inadequate, precisely because it doesn't allow for the representation of infinitesimals.

There are any number of pseudoanswers that could be presented.

As long as there aren't any real answers, we're doing ok.

Yes, I'm assuming that every real number has a decimal expansion. Someone who doesn't accept that isn't talking about the same real numbers as everyone else is. A decimal expansion is an infinite sequence of digits. Each digit has a position which is a natural number: there's a first digit, a second digit, a third digit, etc. There is a hundredth digit, and a thousandth digit, and a millionth. Your 0.999...5 is not a decimal expansion: what is the position of the 5? Infinity? That's not a natural number.

This decimal notation is inadequate for representing infinitesimals. That's fine. It just means that infinitesimals aren't real numbers.

I believe (not entirely sure) that one characterization of the surreal numbers is that they're decimal expansions where the digits are indexed by all ordinals (less than any given one), not just by the finite ordinals. So then you could have something like 0.999...5. All the finite positions are occupied by 9s, and there's a 5 at position omega. (The rest are 0s, I guess. I'm not really very familiar with surreal numbers.)

 

[jpowell]

No, I don't.

I can equally well show that 0.81 squared is 0.6561, which is less than 2/3, and that 0.82 squared is 0.6724, which is more than 2/3.

I can even do it by showing that 0.81 is 81/100, which is 6561/10000, and I don't need to use "decimal" representations at all.

Why would you say 6561/10000 is the same as 81/100? It is the square of 81/100.

 

[Vampyric]

I will now prove that 999 = 1000.

x = 999
10x = 9999
10x -x = 9999-999
9x = 9000
x = 1000

I don't post much as you can tell but...you stupid morons. This is not a valid proof. You can't subtract x from one side and 999 from the other. Your math should be:
x = 999
10x = 9999
10x -x = 9999 - x (NOT - 999)
9x = 9999 -x
9 = (9999-x)/x
9 = 9999/x -1
10 = 9999/x
10x = 9999
x = 999

or however else you want to go around in cirlces. If you understand this that you understand why the original proof is bogus. How math defines .999~ not withstanding.

 

[jpowell]

I will now prove that 999 = 1000.

x = 999
10x = 9999
10x -x = 9999-999
9x = 9000
x = 1000

I don't post much as you can tell but...you stupid morons. This is not a valid proof. You can't subtract x from one side and 999 from the other. Your math should be:
x = 999
10x = 9999
10x -x = 9999 - x (NOT - 999)
9x = 9999 -x
9 = (9999-x)/x
9 = 9999/x -1
10 = 9999/x
10x = 9999
x = 999

or however else you want to go around in cirlces. If you understand this that you understand why the original proof is bogus. How math defines .999~ not withstanding.

10 times 999 is 9990.

Therefore

x = 999
10x = 9990
10x -x = 9990-999
9x = 8991
x = 999

It is perfectly valid to use your method, but you need to know how to multiply numbers together in the first instance to get the right answer .

 

[Art Vandelay]

I don't post much as you can tell but...you stupid morons. This is not a valid proof. You can't subtract x from one side and 999 from the other.

If they're the same thing, then you can. If you don't get that, then you're the moron.

 

[T'ai Chi]

limit _k->oo 1-10^-k = 1.

Anyone object?