Math makes me happy

[Flying Mullet]

How does it show why 1.9 != 2? You can show it algebraically.

1.9 = 1 + 0.9

Let 0.9 = X, then

10X = 9.9
9X = 9.9 - 0.9 = 9
X = 1

We know 1.9 = 1 + 0.9, so

1.9 = 1 + X = 1 + 1 = 2

Alright, I'm going to start over on this proof and look at it two ways:
The first says that we haven't proved that 0.9 = 1

So 0.9 = X

We don't prove that X = 1 until after these series of statements so we have to use 0.9 for all instances of X
10X = 9.9
9X - 0.9 = 9.9 - 0.9
9X - 0.9 = 9
9X = 9 + 0.9

9X = 9.9
X = 9.9 / 9

Okay, now we're dividing into an infinite number so this is a lost cause.

Now, I will take a second approach:
Let's assume that 0.9 = 1 from a previous proof.
So 0.9 = 1 = X (we can use these interchangably.
10X = 9.9
9X - 1 = 9.9 - 1
9X = 8.9

X = 8.9 / 9

And we're back to where we started, dividing into an infinite number.

It doesn't matter which way you look at it, you can counter-proof the proof.

Remember, the burden of the proof lies on the proof itself. It has to be able to stand up to all counter-proofs/attacks. If even one causes it to fail it's not a proof.

[crono_logical]

Alright, I'm going to start over on this proof and look at it two ways:
The first says that we haven't proved that 0.9 = 1

So 0.9 = X

We don't prove that X = 1 until after these series of statements so we have to use 0.9 for all instances of X
10X = 9.9
9X - 0.9 = 9.9 - 0.9

This step's incorrect, so this counter proof is invalid. YOu've taken 0.9 from the right, but you've taken 0.9 and X from the left.

Now, I will take a second approach:
Let's assume that 0.9 = 1 from a previous proof.
So 0.9 = 1 = X (we can use these interchangably.
10X = 9.9

This step's also invalid, though I think it's a typo judging from your next line, so I'll continue on this one

9X - 1 = 9.9 - 1
9X = 8.9

X = 8.9 / 9

And we're back to where we started, dividing into an infinite number.

Yes, we're back to dividing by an infinite number, but you've stopped midway. You can then see 8.9 = 8 + 0.9 = 8 + 1 = 9, so that resolves the dividing by an infinite number, if you don't like dividing such things. Stopping somewhere just because you can't see a way to continue doesn't provide a counter-proof.



Let's try writing this proof out even longer so it's easier to see what's going on

We want to proove 0.9 = 1.

Let's start with 0.9.
Let's define X = 0.9.

Now 10X = 9.9, because we've shifted the decimal place as we're multiplying by the base.

Next we subtract X from both sides.

10X - X = 9.9 - X

For the LHS, 10X - X = 9X, as you should know from algebra.
For the RHS, the only value of X we know is 0.9 since that's what we defined earlier, so the RHS becomes 9.9 - 0.9, so we now have

9X = 9.9 - 0.9

You should be able to see the RHS simplifies to 9, giving us

9X = 9

Divide both sides by 9, and we get

X = 1.

Nowhere earlier have I used X = 1 in obtaining this value.

EDIT: Yes, a proof has to stand up to potential counter-proofs and attaks, but they have to be valid counter-proofs or attacks and those cannot have flaws in themselves

[Flying Mullet]

We're not proving that 0.9 = 1, we're proving whether or not 1.9 = 2.

[Flying Mullet]

Fine, how's this?
The first says that we haven't proved that 0.9 = 1

So 0.9 = X

10X = 9.9
10X - 0.9 = 9.9 - 0.9
10X - 0.9 = 9
10X/10 - 0.9/10 = 9/10
X - 0.09 = .9
X = .9 + 0.09
X = .99

That just proved that 0.9 is equal to itself, not that 0.9 is equal to 1.

[Dr Unne]

Neat, overlines.

Okay, now we're dividing into an infinite number so this is a lost cause. --Flying Mullet

No, you can divide an infinite series by a number, or multiply it by a number, etc. I did so in one of my proofs.

(1/10 + 1/100 + 1/1000 +...) / 10 = (1/100 + 1/1000 + 1/10000+...)


1) In mathematical theory (which I believe is what were are dealing with), simplest terms are always important. You don't say 2A=2(pi)r^2. You don't say .77e=.77mc^2. You don't say a^2 + b^2 + 10=c^2 +10. You can, but you don't. Simplest terms are always used in mathematical theory. --SeedRankLou

I don't see any reason why you have to simplify everything. I've never heard this. Could you explain why this is true?

I'm fairly certain that the definition is specifying a ratio of two integers that is not also an integer itself, or the definition could also go: any real number that cannot be expressed as an integer (making fractions irrational number).

The definition of rational number is a ratio of two integers, where the denominator isn't 0. That's the only restriction. 1/1 is a rational number. 10/2 is a rational number. It doesn't matter if the rational number can be written in another form as an integer. Just a different way of representing the same value.

The list of rational numbers is like this (pretty sure):

...

... -1/2  -1/1  1/1  1/2  1/3  1/4  1/5 ...

... -2/2  -2/1  2/1  2/2  2/3  2/4  2/5 ...

... -3/2  -3/1  3/1  3/2  3/3  3/4   3/5 ...

...

If you want a set where no element is repeated, you can go through and eliminate all the repeats. (2/1, 3/1 for example)

1/1 = 2/1 = 3/1 = 1 = 1.0 = 9/9 = .9, all just different ways of representing the same rational number.

Is .999 repeating a rational number? Well, a number is rational if it can be written as A/B (A over B): .3 = 3/10 and .55555..... = 5/9, so these are both rational numbers. Now look at .99999999..... which is equal to 9/9 = 1. We have just written down 1 and .9999999 in the form A/B where A and B are both 9, so 1 and .9999999 are both rational numbers. In fact all repeating decimals like .575757575757... , all integers like 46, and all finite decimals like .472 are rational. -- http://mathforum.org/dr.math/faq/faq.integers.html

[Flying Mullet]

No, you can divide an infinite series by a number, or multiply it by a number, etc. I did so in one of my proofs.

(1/10 + 1/100 + 1/1000 +...) / 10 = (1/100 + 1/1000 + 1/10000+...)

True, but I meant that we aren't going to come up wih X = 1 after the division at that point, not that it's impossible to divide. Plus it was divided by 9, and when there are remainders it makes it a lot more difficult, then when they are divided by powers of 10.

[crono_logical]

Ok, I'll do this out in full too, and I might do it two ways again I'll do it the way you're having a problem with first

Method 1)

Start with 1.9 = 1.9.

You can split the integer and non-integer part one side as follows:

1.9 = 1 + 0.9

Now we need to do something about 0.9. Let 0.9 = X. X = 0.9 at this point in time only, and nothing else. We can rewrite that equation as

1.9 = 1 + X

This is Equation A.

Now, moving back to our definition of X, we have

X = 0.9

Multiply both sides by 10:

10X = 9.9.

Subtract X from both sides:

10X - X = 9.9 - X

Simplify the LHS:

9X = 9.9 - X

Since X = 0.9 at this point in time only, and nothing else, we substitute this value in the RHS:

9X = 9.9 - 0.9

Simplify the RHS:

9X = 9
X = 1

Now we're allowed to use X = 1 because we've shown it. We can stick this back in Equation A and get

1.9 = 1 + 1
1.9 = 2 ∎


Method 2)

We've got 1.9. Let's call this Y. Y = 1.9. Y is nothing else at this point in time. Y is only 1.9 at the moment.

Y = 1.9

Multiply both sides by 10:

10Y = 19.9

since when multiply by the base we're working in (10), you just shift the (decimal) point.

Subtract Y form both sides:

10Y - Y = 19.9 - Y
9Y = 19.9 - Y

Now Y = 1.9. Y is nothing else at this point in time. Y is only 1.9 at the moment. So we can only substitute this value in the RHS.

9Y = 19.9 - 1.9

Simplify the RHS:

9Y = 18

Divide both sides by 9:

Y = 2

Since Y = 1.9 though, then

2 = 1.9 ∎

[crono_logical]

Fine, how's this?
The first says that we haven't proved that 0.9 = 1

So 0.9 = X

10X = 9.9
10X - 0.9 = 9.9 - 0.9
10X - 0.9 = 9
10X/10 - 0.9/10 = 9/10
X - 0.09 = .9
X = .9 + 0.09
X = .99

That just proved that 0.9 is equal to itself, not that 0.9 is equal to 1.

I should hope you can prove anything equals itself Proving something equals itself doesn't rule out the possibility it equals something else at the same time, or has an alternate way of being represented. If you proved 0.5 = 0.5, are you saying 0.5 cannot equal 1/2?

[Flying Mullet]

Okay, I see how you're solving it, but here's what's bothering me:

Now, moving back to our definition of X, we have

X = 0.9

Multiply both sides by 10:

10X = 9.9.

Subtract X from both sides:

10X - X = 9.9 - X

You see, here, because X = 0.9, I can use 0.9 instead of X, so I get
10X - 0.9 = 9.9 - 0.9

And that throws this equation out of whack.

Method 2)

We've got 1.9. Let's call this Y. Y = 1.9. Y is nothing else at this point in time. Y is only 1.9 at the moment.

Y = 1.9

Multiply both sides by 10:

10Y = 19.9

since when multiply by the base we're working in (10), you just shift the (decimal) point.

Subtract Y form both sides:

10Y - Y = 19.9 - Y
9Y = 19.9 - Y

And again, the same thing here, I can insert 1.9 for Y, so I get
10Y - 1.9 - 10.9 - 1.9

[Flying Mullet]

I should hope you can prove anything equals itself Proving something equals itself doesn't rule out the possibility it equals something else at the same time, or has an alternate way of being represented. If you proved 0.5 = 0.5, are you saying 0.5 cannot equal 1/2?

That's my point, this proves that it equals itself, not that it equals 1.

EDIT:Nevermind, I remember that you have to use all X's or all it's value, and not a mix of the two.

Man, this has been a good thread. It's been helping to get the cobwebs out.

And here's another good thread debating this: http://www.experts-exchange.com/Misc..._20828798.html
I especially like reading some of the arguments against .999... = 1, there's convincing stuff on both sides.

[crono_logical]

The other thing you're forgetting when you want to say let's do

10X - 0.9 = 9.9 - 0.9

instead, is that as valid a step it is, it's not the only valid step you can do there. Which can make some other proofs, especially those involving trig functions and powers of such, rather nasty, since if you do the wrong step (which is still valid) at a certain place, you can make the equation more complicated or end up going round in circles instead (and proving something equals itself ), instead of actually doing the valid step you need to get to the result you're trying to show

[Flying Mullet]

Yeah, that's it, it's the equation's fault.

[Polaris]

Kill me, please!!!

ANything is better than Math!!!!!!

[SeeDRankLou]

1) In mathematical theory (which I believe is what were are dealing with), simplest terms are always important. You don't say 2A=2(pi)r^2. You don't say .77e=.77mc^2. You don't say a^2 + b^2 + 10=c^2 +10. You can, but you don't. Simplest terms are always used in mathematical theory. --SeedRankLou

I don't see any reason why you have to simplify everything. I've never heard this. Could you explain why this is true?

I'm fairly certain that the definition is specifying a ratio of two integers that is not also an integer itself, or the definition could also go: any real number that cannot be expressed as an integer (making fractions irrational number).

The definition of rational number is a ratio of two integers, where the denominator isn't 0. That's the only restriction. 1/1 is a rational number. 10/2 is a rational number. It doesn't matter if the rational number can be written in another form as an integer. Just a different way of representing the same value.

It's just math etiquette. Similar to writing etiquette. When you are just writing, you can write however you wish. But if you are writing a formal paper, there are rules you follow, like don't end a sentence with a preposition, use commas appropriately, ect. ect. When dealing with formal math (i.e. theory) there are rules you follow, and one of them is that all terms and equations must end as their simplest form. There are good reasons for this. One is irrelevence. In theory, extra terms are irrelevent. You are only dealing with what is at hand. Is .999....=1, not is .999....=1/1. For example, if you have an equation and you are solving for y and you get 10y=10x+20, in practicality that's fine, but in theory that's not fine. There is a factor 10 in that equation that in no way affect the value of y, and therefore the 10 is irrelevant and shouldn't be there. And to say 10y/10=(10x+20)/10 is also has irrelevent terms for the same reason. If people are going to be overly practical, then people could write something like 1(y+1-1)/1=1(x+2+1-1)/1. Yes, this is correct, but all of those 1s are irrelevent, they do nothing to the equation. Another reason is duplicity. Lets say I have the number 12. If I write 12/1, do I not still have 12? If I write 15-3, do I not still have 12? If I write (square root of 4)*(24/4), do I not still have 12? If I write log (base 2) 4096, do I not still have 12? If I write 24*(cos 45)^2, do I not still have 12? Numbers can be written in hundreds of differents fashions, but all of the above can be simplified to 12. Is that to say the number 12 is an integer, an arithmetic expression, a fraction, a compound fraction, a logarithm, and a trigonometric function, or is 12 just an integer? 12 can be expressed as all of the above things, but all of the above things can be expressed as 12. So which is right? Technically they're all right, but in mathematical theory, 12 is right. When numbers are in their simplest terms, they always lack duplicity, so the number is only what it is and nothing more, making a number concretely and inarguable what it is. If a number isn't in simplest terms, it can be something else, but if it is in simplest terms it is only what it is. That is probably the main reason for simplest terms. That and making equations as short as possible (i.e. laziness).

And your definition isn't entirely correct. A rational number is a number capable of being expressed as an integer or a ratio of two integers, excluding zero as the denominator. That distinction is made for a reason. A ratio of two integers is not itself suppose to represent an integer, or you would simply write integer and not the above phrase. Ratios in theory are in simplest terms, and if a 1 is in the denominator, then it is irrelevent.

[omnitarian]

Blast from the past

Your 0.000....1 doesn't work because of your contradiction of sticking something after the end of something with no end - the 0.999... has no contradiction in it's meaning.

Couldn't you consider .999... sticking a 9 after a series of infinite nines?

It comes down to which logic you use for infinity. And infinity isn't very logical to begin with. So I don't think there's a "real" answer here.