is 0 even or odd?

[drummond13]

I said good day, sir! That's that.

Is that what your video said? I didn't watch it. And if you had REALLY said good day, then you would have responded again.

Go do your own research. At this point I don't think any facts quoted on this board will convince you. Have a good one.

 

[Denamic]

People seriously need to stop using that shit now.
If you're able to understand this kind of explanation, you should also realize the unreasonable jump in logic in it.

People seriously need to stop saying there are flaws without actually pointing them out....

I shouldn't need to.

 

[drummond13]

People seriously need to stop using that shit now.
If you're able to understand this kind of explanation, you should also realize the unreasonable jump in logic in it.

People seriously need to stop saying there are flaws without actually pointing them out....

I shouldn't need to.

Well, you DO need to in this case. There are multiple mathematical proofs of this that have no jumps in logic at all. Some of them have even been posted on this thread. If you disagree, that's fine. But you need to actually back it up with something. You're disagreeing with an established mathematical fact, albeit a nonintuitive one.

 

[Dominic Corner]

I shouldn't need to.

Forgetting for a moment that you absolutely do need to point out what the flaws are in order for it to be a valid argument; there is no flaw, it was 100% correct and there was no leap in logic whatsoever... Just because you don't follow the proof, doesn't mean it is a false one.

For x = 0.9*

10x = 9.9*
9x = 9
x = 1

0.9* == 1

The logic is sound, complete, and requires no assumptions...

You're disagreeing with an established mathematical fact, albeit a nonintuitive one.

I'd say it was entirely intuitive. It even works as a thought experiment, without getting all that pesky maths involved (all three lines of it)...

In order for the two numbers to be different, there needs to be a difference between them (something you can add to the smaller number to make the larger). No matter how far down the row of 9s you go, there is never any point where a difference will ever present itself, therefore there is no difference and they must be the same number.

 

[blankedboy]

Okay, and you give me the fucking square root of -1. "i" may not be physically definable but it's still a number >.>
Yeesh.

Precisely my point. (and -1 is me taking an apple away from you, welcome to how these numbers came to be in the first place).

By your logic, neither of us giving apples will result in a 0, so I can just do that twice. :/

 

[Denamic]

I shouldn't need to.

Forgetting for a moment that you absolutely do need to point out what the flaws are in order for it to be a valid argument; there is no flaw, it was 100% correct and there was no leap in logic whatsoever... Just because you don't follow the proof, doesn't mean it is a false one.

For x = 0.9*

10x = 9.9*
9x = 9
x = 1

0.9* == 1

The logic is sound, complete, and requires no assumptions...

You're disagreeing with an established mathematical fact, albeit a nonintuitive one.

I'd say it was entirely intuitive. It even works as a thought experiment, without getting all that pesky maths involved (all three lines of it)...

In order for the two numbers to be different, there needs to be a difference between them (something you can add to the smaller number to make the larger). No matter how far down the row of 9s you go, there is never any point where a difference will ever present itself, therefore there is no difference and they must be the same number.

The fact remains that you multiplied a number with decimals.
Those decimals are no longer the same as the original number.
You've introduced change to the decimals, and you cannot just assert that they are the same because they 'look' the same.
That is not 'intuitive'.

 

[bojac6]

I shouldn't need to.

Forgetting for a moment that you absolutely do need to point out what the flaws are in order for it to be a valid argument; there is no flaw, it was 100% correct and there was no leap in logic whatsoever... Just because you don't follow the proof, doesn't mean it is a false one.

For x = 0.9*

10x = 9.9*
9x = 9
x = 1

0.9* == 1

The logic is sound, complete, and requires no assumptions...

You're disagreeing with an established mathematical fact, albeit a nonintuitive one.

I'd say it was entirely intuitive. It even works as a thought experiment, without getting all that pesky maths involved (all three lines of it)...

In order for the two numbers to be different, there needs to be a difference between them (something you can add to the smaller number to make the larger). No matter how far down the row of 9s you go, there is never any point where a difference will ever present itself, therefore there is no difference and they must be the same number.

The fact remains that you multiplied a number with decimals.
Those decimals are no longer the same as the original number.
You've introduced change to the decimals, and you cannot just assert that they are the same because they 'look' the same.
That is not 'intuitive'.

So your argument is that 10 multiplied by .9* is not 9.9* but something else? I just don't understand. Multiplication is simple arithmetic, and keeping both sides of the equation balanced is basic algebra. What part of that is not the same or not intuitive?

 

[AlexMBrennan]

Forgetting for a moment that you absolutely do need to point out what the flaws are in order for it to be a valid argument; there is no flaw, it was 100% correct and there was no leap in logic whatsoever... Just because you don't follow the proof, doesn't mean it is a false one.

For x = 0.9*

10x = 9.9*
9x = 9
x = 1

0.9* == 1

The logic is sound, complete, and requires no assumptions...

You are interchanging multiplication and infinite summation, so you need to justify why you can do that here.

The fact remains that you multiplied a number with decimals.
Those decimals are no longer the same as the original number.
You've introduced change to the decimals, and you cannot just assert that they are the same because they 'look' the same.
That is not 'intuitive'.

OK, what does that even mean?

 

[drummond13]

Forgetting for a moment that you absolutely do need to point out what the flaws are in order for it to be a valid argument; there is no flaw, it was 100% correct and there was no leap in logic whatsoever... Just because you don't follow the proof, doesn't mean it is a false one.

For x = 0.9*

10x = 9.9*
9x = 9
x = 1

0.9* == 1

The logic is sound, complete, and requires no assumptions...

You are interchanging multiplication and infinite summation, so you need to justify why you can do that here.

Because just because a number goes on forever doesn't mean you can't multiply it by 10. .9999... times 10 is 9.9999... As with any other number, when you multiply it by ten, you simply move the decimal point one space to the right.

 

[bojac6]

Forgetting for a moment that you absolutely do need to point out what the flaws are in order for it to be a valid argument; there is no flaw, it was 100% correct and there was no leap in logic whatsoever... Just because you don't follow the proof, doesn't mean it is a false one.

For x = 0.9*

10x = 9.9*
9x = 9
x = 1

0.9* == 1

The logic is sound, complete, and requires no assumptions...

You are interchanging multiplication and infinite summation, so you need to justify why you can do that here.

Infinite Summation has nothing to do with this. 0.9* is simply a repeating decimal, not a series.

Here is the proof with all the steps put into it, I think this helps clear this up.

Let x = 0.9*
10x = 9.9* (because we multiplied both sides by 10)
10x - x = 9.9* - x (subtract x from both sides)
9x = 9.9* - 0.9* (because 10x-x = 9x and x = .9*)
9x = 9 (because 9.9* - 0.9* = 9)
9x/9 = 9/9 (divide both sides by 9)
x = 1 (Finish the last step)
0.9* = 1 (because x = 0.9*)

 

[ThunderDumpling]

I can't believe something like this has 12 pages of replies... I am disappoint...

 

[Dominic Corner]

You are interchanging multiplication and infinite summation, so you need to justify why you can do that here.

What does that mean?

0.9* is 0 and a decimal point with an infinite number of 9s... multiplying by 10 moves the decimal point to give 9 and a decimal point with an infinite number of 9s...

This is pretty basic mathematics (I was taught it at GCSE level as a way to deal with recurring decimals - not in such basic terms, though), so I would have thought everyone would know it.

In the form of a mathematical expression:

10 * 0.9* = 9 + 0.9*

Unless you're saying that the original 0.9* is different to the new 0.9* (and by inference that infinity - 1 != infinity...)?

 

[AlexMBrennan]

Of course that's true but you still need to prove it. Your reply is an appeal to common sense and some vigorous hand waving which is not a valid technique of proof.

 

[AlexMBrennan]

Infinite Summation has nothing to do with this. 0.9*

But what is 0.9*? How do you define it if not as sum(9x0.1^n) from 1 to infinity?

This is pretty basic mathematics (I was taught it at GCSE level as a way to deal with recurring decimals - not in such basic terms, though)

You're taught a lot of lies-to-children at GCSE, and then at A-level. Yes, you can exchange the order of infinite summation if certain conditions are met (they usually are, which is why students are taught this) - but you still need to show that that's the case here;

 

[Dominic Corner]

OK, lets try a different approach:

1/9 == 0.1*

0.9* == 9 * 0.1*

(1/9)*9 == 9/9

9/9 == 1

QED

Just let it go... Whichever way you look at it, 0.9 recurring is equal to 1. Always has been, always will be.

Also, why is this 12 pages when it was answered correctly in the 2nd reply (post 3 "Even.").

 

[SamuelT]

I love how such an simple question has spawned such debate and yet the guy who posted it still gets a low content post warning.

Regardless... well, I thought it was even before but reading some of these answers has put me off. If 0 isn't a number, is a negative number also not a number but the absence of value? I always figured it was simple in the fact that it goes even(0), odd(1), even(2), odd(3) ect..

YOU DON'T KNOW SCIENCE, YOU SHUSH.

OT: 0 can be divided by an even number and not get a decimal result. As such; it is even.

 

[AlexMBrennan]

That's exactly the same problem - you need to show that 0.9* = 9x 0.1*. I'd expect something like the following:

I'll use bold N, R for natural numbers, real numbers, etc
Axiom 1.1: Every non-empty bounded set of real numbers has a least upper bound

Theorem 1.2: Given a r in R, there exists a n in N such that n>r
Proof: If it is false then r is an upper bound for N. By Axiom 1.1 N has least upper bound s. Then s-1 is not an upper bound for N so there is n in N with n>s-1. But then n+1>s, s in N contradicting s = least upper bound for N. QED

Corollary 1.3: greatest lower bound { 1/n : n in N } = 0
Proof: The set is non-empty and bounded below by 0, so t = greatest lower bound = inf {1/n : n in N} exists. Certainly t >= 0.
If t > 0 then 1/t < n for all n in N, contradicting Theorem 1.2. Hence t=0 QED

Corollary 1.4: 0.999... = 1
Proof: Let s = least upper bound = sup {a_n : n in N} where a_n = 0.99...9=1-10^-n (n nines)
Then a_n =< 1 for all n in N, so s exists by Axiom 1.1, and s =< 1. But if s<1 then by Corollary 1.3. 1-s>1-1/n for some n, so s<a_n. This is a contradiction, and hence s=1 QED

[Lecture notes for Mathematical Analysis by Prof Johnstone for the Mathematics Tripos at the University of Cambridge]

 

[Dominic Corner]

No, I don't need to show that at all. It has already been shown by other people.

If you're set on it not being equal to 1 then you can go on believing it.

The fact still remains that 0.9 recurring == 1 and that there are many, many proofs of this out there.

I shouldn't be required to prove all of mathematics every time I supply a proof... Nobody is expected to do that in academic circles, so why should it be any different here?

I'm out.

 

[Jake0fTrades]

Nothingness cannot be classified into any category or class. It simply isn't.

But Zero isn't nothingness.

And wouldn't the fact that nothingness has attributes such as being unclassifiable, make it have it's own special class of being?

Zero, by definition, is nothing.

 

[SeaCalMaster]

No, I don't need to show that at all. It has already been shown by other people.

If you're set on it not being equal to 1 then you can go on believing it.

The fact still remains that 0.9 recurring == 1 and that there are many, many proofs of this out there.

I shouldn't be required to prove all of mathematics every time I supply a proof... Nobody is expected to do that in academic circles, so why should it be any different here?

I'm out.

Um... he was agreeing that 0.999... = 1. Did you not read it at all?

That's exactly the same problem - you need to show that 0.9* = 9x 0.1*. I'd expect something like the following:

I'll use bold N, R for natural numbers, real numbers, etc
Axiom 1.1: Every non-empty bounded set of real numbers has a least upper bound

Theorem 1.2: Given a r in R, there exists a n in N such that n>r
Proof: If it is false then r is an upper bound for N. By Axiom 1.1 N has least upper bound s. Then s-1 is not an upper bound for N so there is n in N with n>s-1. But then n+1>s, s in N contradicting s = least upper bound for N. QED

Corollary 1.3: greatest lower bound { 1/n : n in N } = 0
Proof: The set is non-empty and bounded below by 0, so t = greatest lower bound = inf {1/n : n in N} exists. Certainly t >= 0.
If t > 0 then 1/t < n for all n in N, contradicting Theorem 1.2. Hence t=0 QED

Corollary 1.4: 0.999... = 1
Proof: Let s = least upper bound = sup {a_n : n in N} where a_n = 0.99...9=1-10^-n (n nines)
Then a_n =< 1 for all n in N, so s exists by Axiom 1.1, and s =< 1. But if s<1 then by Corollary 1.3. 1-s>1-1/n for some n, so s<a_n. This is a contradiction, and hence s=1 QED

[Lecture notes for Mathematical Analysis by Prof Johnstone for the Mathematics Tripos at the University of Cambridge]

I think you want 1/t > n in 1.3. (That should be a greater-than sign; someone needs to fix this forum.) Also, there's a slight issue with your proof: stating that the supremum of the set is equal to 1 is not quite the same as showing that the limit of a_n as n -> infinity is equal to 1, although pointing out that the series increases monotonically clears this up.

As for your comment about rearranging terms in the infinite series, I'm not sure why it comes up, as the standard proof doesn't rearrange terms at all. It merely distributes multiplication over the infinite sum (and if you like, you can demand a proof that that works, too). In any case, all the summands are positive, so you can rearrange them all you like.