Irrational number - Wikipedia, the free encyclopedia
I saw this the other day. Look under the "History" section. Hippasus' proof that an odd number is even. It's not exactly the same as a 1=2 "proof," but is at the very least relevant to what you are touching upon here.
It's worth mentioning that--if I've understood correctly--a major difference is that Hippasus's proof is valid. No fallacies in logic can be pointed out as in the examples here. I therefore speculate that it is conceivable that a valid 1=2 proof can exist, and that it would merely prove, as Hippasus' proof does, that irrational numbers exist.
My speculation is probably totally hare-brained though, as I am not a real mathematician.
Actually, in his proof he assumes (in contradiction), that there are no irrational numbers. He then reaches a cotradiction - that the number b, which he specified, must be both even and odd, however no such number exists. therefore, the claim that there are no irratioal numbers is false.
I'm surprised no one brought this one up, since it's been quite a number of years since I first heard of it.
Here is a 1=2 proof using differentiation:
Start with:
x^2 = x^2
x^2 is x*x, so we can represent one side by addition:
x + x + x + ... + x = x^2
where there are x x's.
Now, we differentiate both sides:
1 + 1 + 1 + ... + 1 = 2x
There are x 1's, so we can sum all the 1's to get
x = 2x
And dividing by x we get
1=2
let's look at the equation:
$\displaystyle \sum_{i=1}^kx = x^2$
when we solve for k, we get k = x.
now, let's differentiate that same equation:
$\displaystyle \sum_{i=1}^k 1 = 2x$.
when we solve for k, we get k = x/2.
evidently, x = x/2, so 2x = x, so x = 0.
therefore, the flaw in the proof is the very last step...we have divided by 0.
-1 = 1
0 = 1 + 1
0 = 2
so it will zero not 1 =2 .
word problem help
you made a mistake here bro:
$\displaystyle \cos x\ =\ \sqrt{\cos2x+\sin^2x}$ (taking square root)
there should absolute function over cosx. since sqrt(x^2)=abs(x). Its the most common mistake that students makes while learning Pre-calculus.But we soon realize the importance of absolute function when we study limits.
$\displaystyle \cos^2x-\sin^2x\ =\ \cos2x$ (identity)
$\displaystyle \cos^2x\ =\ \cos2x+\sin^2x$ (rearranging)
$\displaystyle |\cos x}|\ =\ \sqrt{\cos2x+\sin^2x}$ (taking square root)
And thus there is no contradiction when $\displaystyle x\ =\ \frac{3\pi}4$. This absolute value sign must be included if we are taking the positive square root on the right hand side of the equation.