But as you should have noticed by now, logic doesn't work on these people.Rabid Toilet said:I had the same reaction, but dammit, it's just so tempting to prove them wrong.
But as you should have noticed by now, logic doesn't work on these people.Rabid Toilet said:I had the same reaction, but dammit, it's just so tempting to prove them wrong.
You misunderstand how variables work. The variable has only one value at a time, but can have an infinite amount of values. So in this case, the solutions for x[sup]2[/sup] = x are, indeed, x = 0, 1.emeraldrafael said:Secondly 0^2 =/= 1. 2^0 = 1. But thats oka.
Would you kindly provide this 'very very slight' difference between 1 and 0.(9)? I'm very very much interested in seeing it.SomethingAmazing said:No.
.9999 is lesser than 1. Very very slightly lesser, but still lesser.
I know! That's what makes it even more tempting!Ravek said:But as you should have noticed by now, logic doesn't work on these people.Rabid Toilet said:I had the same reaction, but dammit, it's just so tempting to prove them wrong.
No matter how many threes you have, it still won't quite be a third. Sorry.BlacklightVirus said:1/3 > 1.333...? No... just no.PoisonUnagi said:But 1/3 is greater than 1.33333...BlacklightVirus said:There is an even simpler proof.havass said:I myself have my doubts about it, but I just can't find anything wrong in any step of the proof! Every step is perfectly logical.crudus said:Every math major I have talked to and showed that to has described that as "shady".
1/3 = 0.333...
1/3 + 1/3 + 1/3 = 3/3 = 1
But 0.333... + 0.333... + 0.333 = 0.999...
Hence 0.999 = 1
So you've failed this round.
acer840 said:That's because it's my equation that is in the first quote box.
The original equation from the first page.If x = 0.999999...
Then 10x = 9.9999...
Therefore, 10x - x = 9
Which implies 9x = 9
Thus, x = 1
x also = 0.99999...
In conclusion, I have just proven 1 = 0.9999...
If you have an infinite number of threes, it does indeed equal a third.PoisonUnagi said:No matter how many threes you have, it still won't quite be a third. Sorry.BlacklightVirus said:1/3 > 1.333...? No... just no.PoisonUnagi said:But 1/3 is greater than 1.33333...BlacklightVirus said:There is an even simpler proof.havass said:I myself have my doubts about it, but I just can't find anything wrong in any step of the proof! Every step is perfectly logical.crudus said:Every math major I have talked to and showed that to has described that as "shady".
1/3 = 0.333...
1/3 + 1/3 + 1/3 = 3/3 = 1
But 0.333... + 0.333... + 0.333 = 0.999...
Hence 0.999 = 1
So you've failed this round.
A tempting fairytale, but nonsense. If x is in S, then x = y implies y is also in S. 0.33(3) is also a rational number, because it's the exact same thing as 1/3. By the same reasoning, 1/3 is also a real number. Of course, all rational numbers are also real numbers, so that was obvious anyway.IMakeIce said:No one would ever dispute the fact that 1/3 = .3333 however, you're talking about two numbers living in totally different worlds. 1/3 lives in Q (rational numbers), .(3) lives in R (real numbers). These are two ways, in different number sets to express the same idea (after all, numbers are simply ideas). Likewise, we have 1 (which lives in N (natural numbers), or Z, Q, or R for that matter) and .(9) which lives in R (real numbers).
When you have an infinite amount of threes, it will be exactly a third. But I'm not sorry, that's just how math works.PoisonUnagi said:No matter how many threes you have, it still won't quite be a third. Sorry.
It's true for any finite number of nines, but not when the number of nines is infinite.Sprntr_Zomby said:Ok for simplicities sake and to end this arguement
x=.99999 (the amount of nines doesn't matter it is true for any number of nines, I'm just truncating at a smaller amount of nines.)
10x=9.9999 (this is the step that everyone gets wrong as multipling (.9*.1^n)! by 10 gets (9*.1^n)!, not (9*.1^(n+1))!. (9*.1^(n+1))! is 9.99999 when n+5)
thus 9.9999-.99999=8.99991=9x
8.99991/9=9x/9,8.99991/9=.99999
QED x=.99999.
Try this with a calculator for any amount of nines and you'll see that my math is right.
I bolded the problem for you. You can't do any of this with a calculator, because a calculator can only deal with finite numbers of decimals. We're talking about the nines going on for infinity, in which case there is no infinity - 1, because that is still infinity, which can't be represented with a calculator.Sprntr_Zomby said:Ok for simplicities sake and to end this arguement
x=.99999 (the amount of nines doesn't matter it is true for any number of nines, I'm just truncating at a smaller amount of nines.)
10x=9.9999 (this is the step that everyone gets wrong as multipling (.9*.1^n)! by 10 gets (9*.1^n)!, not (9*.1^(n+1))!. (9*.1^(n+1))! is 9.99999 when n+5)
thus 9.9999-.99999=8.99991=9x
8.99991/9=9x/9,8.99991/9=.99999
QED x=.99999.
Try this with a calculator for any amount of nines and you'll see that my math is right.
I think you're trolling. 1/3 will always be less than 1.(3). That's a mathematical fact.PoisonUnagi said:No matter how many threes you have, it still won't quite be a third. Sorry.BlacklightVirus said:1/3 > 1.333...? No... just no.PoisonUnagi said:But 1/3 is greater than 1.33333...BlacklightVirus said:There is an even simpler proof.havass said:I myself have my doubts about it, but I just can't find anything wrong in any step of the proof! Every step is perfectly logical.crudus said:Every math major I have talked to and showed that to has described that as "shady".
1/3 = 0.333...
1/3 + 1/3 + 1/3 = 3/3 = 1
But 0.333... + 0.333... + 0.333 = 0.999...
Hence 0.999 = 1
So you've failed this round.
I think he meant to say 0.33..., not 1.33...BlacklightVirus said:I think you're trolling. 1/3 will always be less than 1.(3). That's a mathematical fact.PoisonUnagi said:No matter how many threes you have, it still won't quite be a third. Sorry.BlacklightVirus said:1/3 > 1.333...? No... just no.PoisonUnagi said:But 1/3 is greater than 1.33333...BlacklightVirus said:There is an even simpler proof.havass said:I myself have my doubts about it, but I just can't find anything wrong in any step of the proof! Every step is perfectly logical.crudus said:Every math major I have talked to and showed that to has described that as "shady".
1/3 = 0.333...
1/3 + 1/3 + 1/3 = 3/3 = 1
But 0.333... + 0.333... + 0.333 = 0.999...
Hence 0.999 = 1
So you've failed this round.
1.333... = 4/3 > 1/3 = 0.333...PoisonUnagi said:No matter how many threes you have, it still won't quite be a third. Sorry.BlacklightVirus said:1/3 > 1.333...? No... just no.PoisonUnagi said:But 1/3 is greater than 1.33333...BlacklightVirus said:There is an even simpler proof.havass said:I myself have my doubts about it, but I just can't find anything wrong in any step of the proof! Every step is perfectly logical.crudus said:Every math major I have talked to and showed that to has described that as "shady".
1/3 = 0.333...
1/3 + 1/3 + 1/3 = 3/3 = 1
But 0.333... + 0.333... + 0.333 = 0.999...
Hence 0.999 = 1
So you've failed this round.
No it isn't.SomethingAmazing said:Well then your own damn numbers are confusing you. Because .9999... is still less than 1.BlacklightVirus said:This is the problem which people not educated in math always find. As I have already said the reals do not permit numbers infinitesimally small.
no, the problem is that you said 1/3 > 1.333...PoisonUnagi said:No matter how many threes you have, it still won't quite be a third. Sorry.BlacklightVirus said:1/3 > 1.333...? No... just no.PoisonUnagi said:But 1/3 is greater than 1.33333...BlacklightVirus said:There is an even simpler proof.havass said:I myself have my doubts about it, but I just can't find anything wrong in any step of the proof! Every step is perfectly logical.crudus said:Every math major I have talked to and showed that to has described that as "shady".
1/3 = 0.333...
1/3 + 1/3 + 1/3 = 3/3 = 1
But 0.333... + 0.333... + 0.333 = 0.999...
Hence 0.999 = 1
So you've failed this round.
YOU FOOL! we're humans logic does not apply to us!Rabid Toilet said:I know! That's what makes it even more tempting!