Proving 1 = 0
BulletHead
Join Date: 2004-07-22 Member: 30049Members
Join Date: 2004-07-22 Member: 30049Members
in Discussions
<div class="IPBDescription">How to do it</div> The topic says it all- supposedly, 1 = 0, but this cannot be proven... nor disproven... and thus if 1 = 0, then all those people that wanna destroy the earth / all of humanity have succeeded for, if there is one earth, it means there are no earths...
DISCUSS!
DISCUSS!
Comments
actually the .999_ = 1 is an actual mathematical fact...
<a href='http://mathforum.org/library/drmath/view/53339.html' target='_blank'>http://mathforum.org/library/drmath/view/53339.html</a>
<a href='http://www.newton.dep.anl.gov/newton/askasci/1995/math/MATH070.HTM' target='_blank'>http://www.newton.dep.anl.gov/newton/askas...ath/MATH070.HTM</a>
actually the .999_ = 1 is an actual mathematical fact...
<a href='http://mathforum.org/library/drmath/view/53339.html' target='_blank'>http://mathforum.org/library/drmath/view/53339.html</a>
<a href='http://www.newton.dep.anl.gov/newton/askasci/1995/math/MATH070.HTM' target='_blank'>http://www.newton.dep.anl.gov/newton/askas...ath/MATH070.HTM</a> <!--QuoteEnd--> </td></tr></table><div class='postcolor'> <!--QuoteEEnd-->
As much as I have seen such answers, I have also seen answers that show how different the two can be. But that is hardly the subject at hand.
assume 1 = 0
if 1 = 0 then (1+1) = (0+0)
there yah go.
proof my fallicy (or whatever it is called)
oh, and that proof that 1 = 0 that floats around is faulty as it has you dividing by 0 at some point (a mathmatical falicy right there)
oh, yah
1 does = 2 for segnificantly large values of 1 <!--emo&:)--><img src='http://www.unknownworlds.com/forums/html/emoticons/smile-fix.gif' border='0' style='vertical-align:middle' alt='smile-fix.gif' /><!--endemo-->
Also, .9_ = 1 in most cases, however there are certain cases where it does not (basicaly in some fields of math it matters, and some it don't)
You're funny. I can't wait to see what thread you'll start next.
*edit* btw thansal your "disproof" of 1=0 doesn't actually disprove anything.
1 is the multiplicative identity. therefore
x * 1 = x.
0 is the additive identity. therefore
x + 0 = x.
Now, if 1 = 0, then necessarily
x + 1 = x, and x * 1 = x.
x + 1 = x * 1
Now, if 1 = 0, then x * 1 = x * 0 = 0.
However, obviously x + 1 != 0. Therefore, contradiction.
I was also ussing a specific instance (that is all youy need to disprove sometihng afik)
It has been about 4 years from any math class I have taken <!--emo&:p--><img src='http://www.unknownworlds.com/forums/html/emoticons/tounge.gif' border='0' style='vertical-align:middle' alt='tounge.gif' /><!--endemo-->
the principle I was thinking of was something about one of the rules of addition.. can't remember the name atm.
basicaly going on the fact that if 1 = 0 then 1+1, 0+0, 1+0, and 0+1 should all yield the same result.
If 1 = 0, then 0 + 0, 1 + 1, 1 + 0, 0 + 1 all mean the same thing
1 is just a visual representation of an abstract concept, call it by any other name and it'll still be 1. So saying 1 = 0 is almost the same thing as saying 0 = 0.
Educate me to this, and the whole theory <!--emo&:D--><img src='http://www.unknownworlds.com/forums/html/emoticons/biggrin-fix.gif' border='0' style='vertical-align:middle' alt='biggrin-fix.gif' /><!--endemo-->
In that sense, binary is not so much 0, 1, as it is (state A) and (state B).
In that sense, binary is not so much 0, 1, as it is (state A) and (state B). <!--QuoteEnd--></td></tr></table><div class='postcolor'><!--QuoteEEnd-->
Using that analogy State A = 1 and State B = 0. Two different values, because they are both different, like 1 and 0.
I am stupid when it comes to stuff like this, but putting the states in to an equasion just seems to come back to the same question for me.
They are different numbers, therefore they cannot equal one another.
No amount of calculations will ever disprove this.
same with 0.99 recurring = 1
0.9999... = 0.9999...
0.9999... != 1
Even a monkey could understand that
[EDIT:] What thaldarin says just sums it up.
If 0=1, then why doesn't 2=1, and 2=4, and 230587 = 23894761394876
<b>Because they are defferent numbers</b>, hence the word <b>different</b> in that sentence
<a href='http://www.newton.dep.anl.gov/newton/askasci/1995/math/MATH070.HTM' target='_blank'>http://www.newton.dep.anl.gov/newton/askas...ath/MATH070.HTM</a> <!--QuoteEnd--></td></tr></table><div class='postcolor'><!--QuoteEEnd-->
For the record, these proofs are bogus. Operations with repeating decimals will never be exact, and therefore you can't use them in proofs.
If you found a fraction that equaled 9.99-repeating, it wouldn't work.
<!--QuoteBegin--></div><table border='0' align='center' width='95%' cellpadding='3' cellspacing='1'><tr><td><b>QUOTE</b> </td></tr><tr><td id='QUOTE'><!--QuoteEBegin-->*edit* btw thansal your "disproof" of 1=0 doesn't actually disprove anything.
1 is the multiplicative identity. therefore
x * 1 = x.
0 is the additive identity. therefore
x + 0 = x.
Now, if 1 = 0, then necessarily
x + 1 = x, and x * 1 = x.
x + 1 = x * 1
Now, if 1 = 0, then x * 1 = x * 0 = 0.<!--QuoteEnd--></td></tr></table><div class='postcolor'><!--QuoteEEnd-->
I smell circular logic, and it smells like burnt toast. If it's not that, then I have absolutely no clue what you're trying to do.
<!--QuoteBegin--></div><table border='0' align='center' width='95%' cellpadding='3' cellspacing='1'><tr><td><b>QUOTE</b> </td></tr><tr><td id='QUOTE'><!--QuoteEBegin-->Operations with repeating decimals will never be exact, and therefore you can't use them in proofs.<!--QuoteEnd--></td></tr></table><div class='postcolor'><!--QuoteEEnd-->
so you're saying 0.00_ != 0?
X = 1
X+0 = 1
If X is one and you are adding nothing to it then quite obviously it is 1. Although <b>I know</b> 1=0 can be proved. So far I am finding the logic of it too easy to disprove.
my theory reads
1 = 0 -> x = 0 for x {- R, not true, therefore 1 != 0
<!--QuoteBegin--></div><table border='0' align='center' width='95%' cellpadding='3' cellspacing='1'><tr><td><b>QUOTE</b> </td></tr><tr><td id='QUOTE'><!--QuoteEBegin-->Operations with repeating decimals will never be exact, and therefore you can't use them in proofs.<!--QuoteEnd--></td></tr></table><div class='postcolor'><!--QuoteEEnd-->
so you're saying 0.00_ != 0? <!--QuoteEnd--> </td></tr></table><div class='postcolor'> <!--QuoteEEnd-->
0.000... is not a recurring decimal, as adding on another zero doesn't change the value of the number
<!--QuoteBegin--></div><table border='0' align='center' width='95%' cellpadding='3' cellspacing='1'><tr><td><b>QUOTE</b> </td></tr><tr><td id='QUOTE'><!--QuoteEBegin-->Operations with repeating decimals will never be exact, and therefore you can't use them in proofs.<!--QuoteEnd--></td></tr></table><div class='postcolor'><!--QuoteEEnd-->
so you're saying 0.00_ != 0? <!--QuoteEnd--></td></tr></table><div class='postcolor'><!--QuoteEEnd-->
0.000... is not a recurring decimal, as adding on another zero doesn't change the value of the number <!--QuoteEnd--></td></tr></table><div class='postcolor'><!--QuoteEEnd-->
Anything which is .00000 etc. is classed as an integer, a whole number. Not recurring.
*Spelling.
By definition, two numbers are the same IFF there exist no other numbers between them. I.E. 1 < x < 1 means x = 1, because there is no number between 1 and 1 except for 1.
So...
if .999_ goes on for infinity down the line, give me a number between .999_ and 1 that isn't one. BUT STRANGE THINGS HAPPEN AT INFINITY!!! Of course they do, that's why we use such an abstract concept of amounts of digits so big we can't understand them.
Just think of this: Euclide, the Father of Modern Geometry , expressed that he could not imagine a number of anything less than zero. He said, it's pointless to conceive of -1 * an integer.
Now, while we mostly agree that imaginary numbers are quite useful, how many of use can say we can understand what sqrt(-1) is? Maybe 1500 years from now it'll be obvious!
What I'm tryin to say is that it's whacked to think you can "prove" anything in any case. Math pros and Logic buffs throw these things around like it's some kinda power of God they weild. It's PROVEN that you can't do this, or that this always works. And it's true, as far as we're concerned today, but whose to say we have the whole picture anyway?
Anything which is .00000 etc. is classed as an integer, a whole number. Not recurring.
*Spelling. <!--QuoteEnd--> </td></tr></table><div class='postcolor'> <!--QuoteEEnd-->
just challenging your definition of "never".
Anyway, care to point out the circular logic in my proof?
e^i(pi) is imaginary, a zero in the realm of real numbers
e^i(pi) = 0
0 - 1 = 0
-1 = 0
-1 * -1 = -1 * 0
1 = 0
voila <!--emo&:D--><img src='http://www.unknownworlds.com/forums/html/emoticons/biggrin-fix.gif' border='0' style='vertical-align:middle' alt='biggrin-fix.gif' /><!--endemo-->
Incidentally, this also proves that -1 = 1 ^_^
My bad, I read your proof totally wrong at first. Nothing to see here, folks...
<!--QuoteBegin--></div><table border='0' align='center' width='95%' cellpadding='3' cellspacing='1'><tr><td><b>QUOTE</b> </td></tr><tr><td id='QUOTE'><!--QuoteEBegin-->e^i(pi) is imaginary, a zero in the realm of real numbers<!--QuoteEnd--></td></tr></table><div class='postcolor'><!--QuoteEEnd-->
Man, now we're getting silly (I hope). It's time for a phase to off-topic.
e^(i*x) = cos x + i * sin x for x {- R. It's got a real component, which is how you're getting e^(i*pi) = -1 in the first place.
Anyway, 0.99_ = 1 by definition, so i don't see what there is to argue about.
As a challenge, you give me any false statement, and I will prove it if you allow me to assume that 1=0 <!--emo&:)--><img src='http://www.unknownworlds.com/forums/html/emoticons/smile-fix.gif' border='0' style='vertical-align:middle' alt='smile-fix.gif' /><!--endemo-->.
e^i(pi) is <b>imaginary</b>, a zero in the realm of real numbers
e^i(pi) = 0
0 - 1 = 0
-1 = 0
-1 * -1 = -1 * 0
1 = 0
voila <!--emo&:D--><img src='http://www.unknownworlds.com/forums/html/emoticons/biggrin-fix.gif' border='0' style='vertical-align:middle' alt='biggrin-fix.gif' /><!--endemo-->
Incidentally, this also proves that -1 = 1 ^_^ <!--QuoteEnd--> </td></tr></table><div class='postcolor'> <!--QuoteEEnd-->
if e^i(pi) is imaginary, then how can it = 0?
If a number is imaginary, it does not exist, and there is no way to write it down on paper, so therefore it cannot equal 0.
e^i(pi) is <b>imaginary</b>, a zero in the realm of real numbers
e^i(pi) = 0
0 - 1 = 0
-1 = 0
-1 * -1 = -1 * 0
1 = 0
voila <!--emo&:D--><img src='http://www.unknownworlds.com/forums/html/emoticons/biggrin-fix.gif' border='0' style='vertical-align:middle' alt='biggrin-fix.gif' /><!--endemo-->
Incidentally, this also proves that -1 = 1 ^_^ <!--QuoteEnd--></td></tr></table><div class='postcolor'><!--QuoteEEnd-->
if e^i(pi) is imaginary, then how can it = 0?
If a number is imaginary, it does not exist, and there is no way to write it down on paper, so therefore it cannot equal 0. <!--QuoteEnd--></td></tr></table><div class='postcolor'><!--QuoteEEnd-->
GHAAAAH EYES BLLLEEEEDING MAKE IT STOOOOP!! AGIEOSGH!!!
please, I beg of you, don't let this stuff spread. Look these things up. Learn a little bit about them.
Here, I'll help you <a href='http://en.wikipedia.org/wiki/Imaginary_numbers' target='_blank'>Imaginary numbers</a>
for example:
<!--c1--></div><table border='0' align='center' width='95%' cellpadding='3' cellspacing='1'><tr><td><b>CODE</b> </td></tr><tr><td id='CODE'><!--ec1-->
x = 0 . 9 9 9 9 ... 9
0 1 2 3 4 ... n
x^2 = 0 . 9 9 9 ... 8 0 0 0 ... 1
0 1 2 3 ... n n+1 .... 2n
for all n {- Z+ (positive integers for you guys not familiar with notation)
<!--c2--></td></tr></table><div class='postcolor'><!--ec2-->
now let's see:
if 0.999_ = 1, then
0.999_^2 = 1^2.
1^2 = 1.
0.999_^2 > 0.999_8.... != 0.999_, therefore 0.999_ != 1
this proof only works if you disregard the fact that inductive reasoning from finite multiplication fails to carry over to the infinite, and also that any number after the _ doesn't, by definition, matter.
also,
i^2 = -1, i^4 = -1 * -1 = 1.
what's sqrt (i^5)?
sqrt (i^5) = sqrt (i^4 * i) = sqrt (i * 1) = sqrt (i)
but,
sqrt (i^5) = sqrt (i^2 * i^2 * i ) = i^2 * sqrt (i) = - sqrt (i).
Therefore, i = -i -> 1 = -1.
1 + 1 = 2 = 1 + -1 = 0 -> 0 = 2 -> 0/2 = 2/2 -> 0 = 1.
Something to think about.
*edit* left out some critical stuff.