<p>Wikipedia is the best</p>
<p>i know why thats false ... you cant divide by a-b ... its mathematically incorrect because that would be dividing by a 0 which u cant do since a = b</p>
<p>too bad you can't cancel (a-b) because you can't divide by zero</p>
<p>:-P</p>
<p>Ok, there have been atleast 4 posts on this. For that proof, you have to divide by 0. The universe would implode before you get to 2=1.</p>
<p>Btw, OP, maybe you should consider that Algebra course ;)</p>
<p>:)</p>
<p>Try this then.</p>
<p>1-1+1-1+1-1+1-1...=0 (infinity)</p>
<p>Now if I put in brackets...</p>
<p>(1-1)+(1-1)+(1-1)...=0 (infinity again)</p>
<p>You all agree? <em>nod head</em> Another way to put in brackets...</p>
<p>1(-1+1)(-1+1)(-1+1)...=1 (infinty)</p>
<p>So therefore, 0=1. Now that I've proved 0=1, then we can assume that 1=2, 2=3, 3=4 etc.</p>
<p>(We can substitute 1 with 2, and we'll end up with 0=2)</p>
<p>Any ideas how to prove it wrong?</p>
<p>You can't just put brackets wherever you please. It has to agree with the originial equation; i.e, you can't turn addition into multiplication.</p>
<p>Based on your bracket assumptions 3+4 would be the same as 3(4) which we all know is not true.</p>
<p>ok, let me see if i get this...</p>
<p>1(-1+1)(-1+1)(-1+1)...=1?</p>
<p>Um...
-1+1 = 0, so 1(0)(0)(0) = 0</p>
<p>how exactly does this prove that 1=0? because it seems to me that the answer is 0=0</p>
<p>"1-1+1-1+1-1+1-1...=0 (infinity)"
This is incorrect, since the series (-1)^n does not converge.</p>
<p>"(1-1)+(1-1)+(1-1)...=0 (infinity again)"
This is correct.</p>
<p>"1+(-1+1)+(-1+1)+(-1+1)+...=1 (infinity)"
I assume you had a typo, but this is also correct.</p>
<p>The two statements don't follow from each other, so there is no contradiction here. When you move the brackets:
(1-1)+(1-1)+...+(1-1) = 0 becomes
1+(-1+1)+(-1+1)+...+(-1) = 0.
You can't just lop off the -1 at the end.</p>
<p>cmon you guys think it out b4 u post!</p>
<p>However you do it, you have two of the same sign at each end...</p>
<p>:)</p>
<p>The main problem is that infinite sums are only meaningful if they converge (or diverge to infinity). The sum 1+1/2+1/4+1/8+... can be said to equal 2 because it converges to that value (to be precise, the sums, 1, 3/2, 7/4, 15/8, etc. get arbitrarily close to 2). The "sum" 1-1+1-1+..., however, oscillates between 1 and 0 with each term, so it does not converge and is therefore a meaningless expression that cannot equal anything.</p>
<p>in my mind 2 does equal 0, but thats only because i havent been getting much sleep b/c im cramming for finals.</p>
<p>Sorry, I typed it wrong.</p>
<p>It's suppose to be...
1+(-1+1)+(-1+1)+(-1+1)...=0</p>
<p>"The two statements don't follow from each other, so there is no contradiction here. When you move the brackets:
(1-1)+(1-1)+...+(1-1) = 0 becomes
1+(-1+1)+(-1+1)+...+(-1) = 0.
You can't just lop off the -1 at the end."</p>
<p>Yes I can, because remember, it's infinity. It goes on, and on, and on, and on. It is still going to be the same equation.</p>
<p>um, there isnt really a polite way to say this, but</p>
<p>it seems to me that some of you havent really studied elementary algebra...</p>
<p>Serephia, your "theory" does not work because, like feuler said, your series does not converge to anything; it oscillates like the sin function. I don't know if you've taken Calc II yet, but you should learn about infinite series in that class! :D</p>
<p>Haha... finally someone figured that out. My brother was the one that gave the thing to me. He said that the equations are true, and it can work that way, but then the 0=1, 1=2, 2=3 thing is where it's wrong. Don't ask me about it, I don't have a solution nor have I taken ANY Calc classes yet.</p>
<p>Yes, as many who have studied series have pointed out already, 1-1+1-1+1-1+1... equals BOTH 1 and 0. It's a divergent series, and therefore has no single finite solution.</p>
<p>-20 = -20
16 - 36 = 25 - 45
4^2 - 9<em>4 = 5^2 - 9</em>5
4^2 - 9<em>4 + 81/4 = 5^2 - 9</em>5 + 81/4
(4 - 9/2)^2 = (5 - 9/2)^2
4 - 9/2 = 5 - 9/2
4 = 5</p>
<p>Ok enough. Who can solve this for me -</p>
<p>Sum the series: 1/1² + 1/2² + 1/3² ... + 1/n²</p>