<p>x = y<br>
x² = xy<br>
x² - y² = xy - y²<br>
(x + y)(x - y) = y(x - y)<br>
(x + y) = y<br>
y + y = y<br>
2y = y<br>
2 = 1 </p>
<p>…Discuss?</p>
<p><a href=“x%20+%20y”>quote</a>(x - y) = y(x - y)
[/quote]
</p>
<p>x = y => x-y=0</p>
<p>Division by 0 is undefined</p>
<p>You get 7,000 internets, good job.</p>
<p>How about, e^(i*x) yields a sinusoidal wave.</p>
<p>Discuss this much more interesting phenomenon.</p>
<p>2 = 1.999999999…
=> 2-1 = (1.99999999…) - 1
=> 1 = .9999999…</p>
<p>Ahahahahahaa, fizix2, best post in the thread by far.</p>
<p>
</p>
<p>You said “is” not “=”</p>
<p>for the “x=y” problem</p>
<p>once you get to this step:
(x+y)(x-y)=y(x-y)
divide by (x-y), then you get an indeterminant form, so then you do L’hopital’s rule
lim (as x->y) (x+y)(x-y)/(x-y)= lim(x->y) ((x-y)+(x+y))/1=2y</p>
<p>crap, that didn’t unless I did the product rule wrong.</p>
<p>Prove 1=0.</p>
<p>btw is .99999=1 because anything thats .xxxxx is x/9, so .99999 is 9/9=1.</p>
<p>
</p>
<p>This is like saying 3 does not equal 6/2 because they don’t look the same. One value can be represented in many different ways.</p>
<p>And ThisCouldBeHeavn is right; “is” is not the same thing is “equals.”</p>
<p>
</p>
<p>No, he’s right. 0.999999 is not 1. 0.999999… is 1, however.</p>
<p>^oh yea the … haha.</p>
<p>This thread is the nerdiest piece of crap on CC lol. I still say that 0.9999999999… doesn’t equal 1. It’s super duper close and the sig figs match up, but they just aren’t the same</p>
<p>Yeah, just go on denying mathematical truth. That’s the spirit.</p>
<p>Funny to see someone named proletariat say that.</p>
<p>by mathematical definition it does. but my logic it doesnt.</p>
<p>
Ask an Emory math professor</p>