Simple mathematical problem (?)

[LuxFerum]

Note: there are three dots at both ends.

And then the rest of your proof falls apart.

Ok just for you then

            ...888888888888888888888...  This is infinity
            ...777777777777777777777...  This is infinity too


When we subtry infinite from infinte the result is zero
sooo 


           ...888888888888888888888...  
         - ...777777777777777777777...  
           ----------------------------
           ...111111111111111111111...

Wich proof that zero is infinity.

 

[patoco12]

Lux,

Let's make this real easy to avoid any hand waving or playing with infinity or summations or limits.

Just give us two integers.

Since .9~ is a rational number, then you should be able to give us two integers A and B such that A/B = .9~ and A != B.

There is no need for any limits, summations, series, infinities, or anything. Just give us those two integers so we can end this thread.

I have a feeling, though, that this request will again be ignored.

 

[LuxFerum]

Lux,

Let's make this real easy to avoid any hand waving or playing with infinity or summations or limits.

Just give us two integers.

Since .9~ is a rational number, then you should be able to give us two integers A and B such that A/B = .9~ and A != B.

There is no need for any limits, summations, series, infinities, or anything. Just give us those two integers so we can end this thread.

I have a feeling, though, that this request will again be ignored.

I already did.

xouper: Find a real number N such that N is less than 1 and N is greater than 0.999...

LuxFerum: If you say that
0,999.. = sum 9*10(-n), n=1 to infinity

That number is

1-sum 99*10(-2n), n=1 to infinity

 

[xouper]

LuxFerum: no I claim that N tends to zero

How can a number "tend" to a value. As 69dodge already pointed out, a number does not "tend" towards a thing, it is a thing. We don't say that 3+1 tends to 4. We say 3+1 is another way of representing the value 4. When I ask for a number N that is less than 1 but greater than 0.999... I am asking for a number, not something that "tends" to a number. Give me a specific value of N.

and I agree that 0,999.... tends to 1.

That's not good enough. 0.999... is a real number, right? And any real number has a specific value, right? Real numbers do not "tend" towards a value, they have a value.

having a infinity nimbers of digits shows that you cant work with that digits the same way you do with non infinity string of numbers.

OK, smartass, prove it. This I'd like to see. And I'm going to hold you to strict formal and rigorous requirements of mathematic proof. I'd really like to know why you think you are right and why you think mathematicians have all lost their marbles.

 

[patoco12]

I already did.

Please remind me where you gave two INTEGERS A and B where A/B = .9~. I must have missed it.

 

[LuxFerum]

Lux, one more nit-pick of you big proof, I ignored it in the previous post since I was rebutting you conclusion about finding something that tends to infinity faster.

You multiplied different fractions of a whole by different numbers. Lets take two different sequences that equal 1 and subtract them.

0.8 + 0.2 = 1
0.7 + 0.3 = 1

You multiplied the n-th terms in your two series, (0.9, 0.09, ...) and (0.5, 0.25, ...) by 10ⁿ. If I do this in my two series I get

80 + 200 =280
70 + 300 = 370
(80 -70) + (200 - 300) = -90

By your reasoning I have somehow that my first to series are not equal. You abused the communative property of multiplication. In your case in series 1 (0.9, 0.09, ...) you ended up multiplying 90% of the sum by 10, 9% by 100 and so on. In the second case you multiplied 50% of the sum by 10, 25% by 100 and so on. So you should not be surprised when you result is gibberish.

Walt

humm now I'm geting.

 

[LuxFerum]

Please remind me where you gave two INTEGERS A and B where A/B = .9~. I must have missed it.

why that?

 

[Walter Wayne]

80-70=10
200-300=-100
10-100=-90
That is still correct.

But when we change sum 9.10^(-n) n=1... infinity buy 1
and sum 2^(-n) n=1... infinity
whe got this

Sum [(1 -1)*10^n] n=1.. infinity

And that is zero.

the result is different from infinity wich whe know that is the correct answer.

sum 9.10(-n) n=1... infinity tends to 1. and is not 1.

First: No, we did not get Sum[(1 -1)*10^n] n=1.. infinity. You got that. However, my example has only two terms, each of the sums equal 1, if I subtract the two sums term by term I get zero. But when I multiply the terms by 10^n I don't get 0.

You have erroneously applied math laws
S₁ = sum [t_1i] over some range
S₂ = sum [t_2i] over same range

sum[(t_1i -t_2i)] over said range =S₁ - S₂

sum[(t_1i10ⁱ -t_2i)10ⁱ] over said range != sum[(S₁ - S₂) 10ⁱ] <-- arrow to emphasize does not equal sign

So your math is wrong, this is waht my series of two terms demonstrate.

Second I have already shown that 0 * infinity can take on a variety of values, so why are you still using this proof to show that 0.999...) != 0.5 + 0.25 + ...

Walt

Edited to add: Sorry lux, I started this reply before your post two above this, and didn't notice it in time.

 

[xouper]

LuxFerum: Ok just for you then

            ...888888888888888888888...  This is infinity
            ...777777777777777777777...  This is infinity too


When we subtry infinite from infinte the result is zero
sooo 


           ...888888888888888888888...  
         - ...777777777777777777777...  
           ----------------------------
           ...111111111111111111111...

Wich proof that zero is infinity.

Please pay attention. Those numbers are not real numbers. Let me repeat that in case you didn't get it the first time. Those are not real numbers. Your analysis has no relevance to real numbers.

 

[LuxFerum]

Please pay attention. Those numbers are not real numbers. Let me repeat that in case you didn't get it the first time. Those are not real numbers. Your analysis has no relevance to real numbers.

then 0,999.. are not a real number too.

 

[patoco12]

why that?

Well, .9~ is a rational number. Every rational number can be written as A/B such that A and B are integers. If you claim that .9~ != 1, then you should be able to give two integers, A and B, where A/B = .9~ but does not equal 1.





Now as for .9~ being rational...

The union of rationals and irrationals is the real numbers.
.9~ is NOT irrational because it has a pattern.
Therefore, it is rational. Easy, huh?


Good luck finding the integers!

 

[LuxFerum]

First: No, we did not get Sum[(1 -1)*10^n] n=1.. infinity. You got that. However, my example has only two terms, each of the sums equal 1, if I subtract the two sums term by term I get zero. But when I multiply the terms by 10^n I don't get 0.

You have erroneously applied math laws
S₁ = sum [t_1i] over some range
S₂ = sum [t_2i] over same range

sum[(t_1i -t_2i)] over said range =S₁ - S₂

sum[(t_1i10ⁱ -t_2i)10ⁱ] over said range != sum[(S₁ - S₂) 10ⁱ] <-- arrow to emphasize does not equal sign

So your math is wrong, this is waht my series of two terms demonstrate.

Second I have already shown that 0 * infinity can take on a variety of values, so why are you still using this proof to show that 0.999...) != 0.5 + 0.25 + ...

Walt

Edited to add: Sorry lux, I started this reply before your post two above this, and didn't notice it in time.

no problem, thank you.

 

[69dodge]

Originally posted by LuxFerum
the limit is 1 then.

The limit of the sequence 0.9, 0.99, 0.999, . . . is 1.

The question is, does "0.999. . ." represent the sequence, which tends to 1, or does it represent the limit of the sequence, which is 1?

My understanding is that "0.999 . . ." represents the limit of the sequence, not the sequence itself.

A sequence is an infinitely long list of numbers. We can talk about what limit it tends to, if any; we can talk about how fast it approaches its limit; we can compare its rate of approach with that of other sequences; etc.

The limit of a sequence is a number. A single number. It doesn't tend to anything, quickly or slowly. It's just a number.

 

[xouper]

LuxFerum: then 0,999.. are not a real number too.

You are abusing my patience.

Any real number has a finite value.

0.999... has a finite value and is a real number.

...999... does not have a finite value and thus is not a real number.

 

[LuxFerum]

How can a number "tend" to a value. As 69dodge already pointed out, a number does not "tend" towards a thing, it is a thing. We don't say that 3+1 tends to 4. We say 3+1 is another way of representing the value 4. When I ask for a number N that is less than 1 but greater than 0.999... I am asking for a number, not something that "tends" to a number. Give me a specific value of N.

I really dont know what you are talking about

0,999.... is sum 9.10^(-n) n=1... infinity or not?

 

[ceptimus]

I really dont know what you are talking about

0,999.... is sum 9.10^(-n) n=1... infinity or not?

Yes, but it's not 'going there' it already 'is there'.

2+2=4

2+2 is 4, it's not 'going towards' 4

 

[LuxFerum]

The limit of the sequence 0.9, 0.99, 0.999, . . . is 1.

ok

The question is, does "0.999. . ." represent the sequence, which tends to 1, or does it represent the limit of the sequence, which is 1?

good

My understanding is that "0.999 . . ." represents the limit of the sequence, not the sequence itself.

If 0,99.. represents the limit of the sequence, then it "is" 1.

A sequence is an infinitely long list of numbers. We can talk about what limit it tends to, if any; we can talk about how fast it approaches its limit; we can compare its rate of approach with that of other sequences; etc.

The limit of a sequence is a number. A single number. It doesn't tend to anything, quickly or slowly. It's just a number.

right

 

[xouper]

69dodge: My understanding is that "0.999 . . ." represents the limit of the sequence, not the sequence itself.

Exactly. I tried to make this point pages ago (in color, even), but is has apperently fallen on deaf ears.

LuxFerum: 0,999.... is sum 9.10^(-n) n=1... infinity or not?

Yes, they both represent the same value, and that value is 1. They are just two different ways of expressing the value 1.

 

[xouper]

LuxFerum: If 0,99.. represents the limit of the sequence, then it "is" 1.

Yes, that's what it represents. We are finally in agreement.

I've been trying to say that all along, but thanks to 69dodge for finding a way to say it that LuxFerum can agree with.

 

[LuxFerum]

cool

If "0.999 . . ." represents the limit of the sequence
then we agree that 0,999... "is" 1.

Thanks for your time