and now also assume that [math]a_{1} \neq b_{1}[/math]
This second statement is logically equivalent to [math]a_{1} < b_{1} \vee a_{1} > b_{1}[/math]
Suppose the left case is true [math]a_{1} < b_{1} \Longrightarrow a_{1} + a_{2}*10^{-1} +... < b_{1} + b_{2}*10^{-1} + ... \Longleftrightarrow a_{1}.a_{2}a_{3}... < b_{1}.b_{2} b_{3}... [/math]
This immediately contradicts our hypothesis that these two decimals are equal. If you consider the other case a similar contradiction arises.
If this general proof works (I see literally no mistake with it, can you?) then how come 1 = 0.9999... even though 0 < 1
It makes no sense. This proof is really killing me inside. Is mathematics fundamentally wrong?
Colton Walker
>a1.a2a3...=b1.b2b3... >and now also assume that >a1≠b1 p sure that's not possible senpai.
David Bell
>p sure that's not possible senpai.
That is the point. When we say that 1.0000... = 0.9999... is a true statement we are assuming the equality but also 0 is not equal to 1.
I simply generalized to see what happened.
This is a common practice in mathematics, to assume two things and then find a contradiction that arises from this choice, proving that these two statements are contradictory.
If these two statements are contradictory, as I have proven, then how come 1=0.999... works?
Kevin Davis
Oh, my bad. I didn't read past the first bit. Having read the whole post now, I thing the problem is with >a1
Mason Brown
That's the problem with your proof. You did a conditional proof so your result is entirely dependent on a!=b. Prove this statement is given and not an assumption to complete your proof.
Jaxson Smith
>assume that 0.999... ≠1.0 >set out to prove that 0.999...≠1.0 >LOOK GUISE IT WORKS
Elijah Parker
True not given.....
Benjamin Hernandez
But I assume that 0.999... = 1.00....
Look at the two assumptions I make. That the decimals are equal and that their leading digit is not equal.
This is a case satisfied by 1 and 0.999... Pay attention.
Jason Cook
>you haven't proved it for infinite series
This may be the thing but the problem I have is that there is no immediate way to see why this would not be the case for an infinite series.
If you see my proof, from the beginning i assume infinite decimals with the ... after the third term of the decimal.
It is not at all obvious why infinity would break down the logic of my statement and that is seriously disturbing. Anyone who did not have the previous knowledge of 1 being equal to 0.999... would look at my proof and say 'yup, sounds about right'.
Jack Lewis
>You did a conditional proof so your result is entirely dependent on a!=b
A conditional proof is true when the conditions are met.
1 = 0.999... first condition met 1 != 0 second condition met
Nathan Diaz
>It is not at all obvious why infinity would break down the logic of my statement Infinity breaks down the logic of a lot of stuff.
Grayson Myers
Infinite series fuck everything up. I can promise you the proof will fall apart if you try this.
Adam Wilson
that's not a contradiction, because strict inequalities don't hold under infinite summation
basically, open intervals are not closed sets
Jaxson Bailey
Your implication statement only holds if a1
Brody Brooks
>It is not at all obvious why infinity would break down the logic of my statement you should NEVER assume something works if you haven't proved it. assuming something is right unless it's "obviously wrong" is ridiculous
Jaxon Wood
>a1
Hunter Campbell
But 1 doesn't equal .999...
1=1
That's it you fucking retards
Luke Jackson
An infinite sum is really the limit of a sequence (namely the sequence of partial sums). It's pretty easy to convince yourself that if [math] a_n< b_n [/math] for every n, then it could be the case that [math] lim a_n=lim b_n [/math] (e.g. let a_n=1/n and b_n=2/n).
So, for sequences, strict inequalities don't hold when you pass to the limit, and a series is a special case of a sequence.
David Reed
this
Dylan Moore
There is a lot to point out in this beyond of retarded thread.
Andrew Bailey
> ---> a1+a2∗10−1+...
Carson Thompson
>[math]a_{1} < b_{1} \Longrightarrow a_{1} + a_{2}*10^{-1} +... < b_{1} + b_{2}*10^{-1} + ...[/math] As others have ever-so-gently pointed out, this part isn't true.
If you have numbers [math]a_i[/math] and [math]b_i[/math] for [math]i \in \{ 1, 2, 3, \ldots \}[/math], such that [math]0 \leq a_i \leq 9[/math] and [math]0 \leq b_i \leq 9[/math] for all [math]i[/math], and [math]a_1 < b_1[/math], then it does not follow that [eqn]\sum_{i=1}^{\infty} a_i \times 10^{1 - i} < \sum_{i=1}^{\infty} b_i \times 10^{1 - i}[/eqn] . Try proving it; you'll find that you can't, because it isn't actually true.
Gabriel Nguyen
If you have two sequences of numbers [math]a_n, b_n[/math] it is completely possible that: [math]a_n< b_n[/math] for all [math]n[/math] but the limits are still the same. And exactly that is your problem:
[math]a_{1} < b_{1} \Longrightarrow a_{1} + a_{2}*10^{-1} +... < b_{1} + b_{2}*10^{-1} + ... \Longleftrightarrow a_{1}.a_{2}a_{3}... < b_{1}.b_{2} b_{3}...[/math] is wrong it should be:
which leads to no contradiction and you have rather proved that they are the same because if you take the other case you have shown the inequalities from both sides.
Camden Robinson
All math is made up bullshit anyway. You can literally just use this,
0.999... = 1
And call it a day. No need to make anything else. You can also use,
Apple = 1
or,
OP is a Faggot = 1
All the same shit. That is how symbols and variables work.
Daniel Brooks
No, he's "controversial" (=faggot).
Luke Wood
maybe you want 1 but can not afford it 0.9 too little 0.99 too little 0.999999999 is ok lets use it
Fucking math poor people hobby
Austin Carter
Obviously 0.999... isn't 1, you don't need a prove for that. Just as 1/2 isn't 2/4.
But it's the same number.
sage obviously and holy fuck how did this get 25 replies?!?