## 0 = -1 proof

Posted: December 9, 2012 in Mathematics

I think some of you may have seen many 0=1 proofs already.  Today I am going to show you another one.  So let’s begin.

Theorem:  1=0.

Proof:

Just for fun let’s integrate $\tan x$,

$\displaystyle \int \tan x\;dx=\int \frac{\sin x}{\cos x}\;dx$

Next we let,

$\displaystyle u=\frac{1}{\cos x}, dv=\sin x\;dx$

$\displaystyle du=\frac{\sin x}{\cos^2 x} \;dx, v=-\cos x$

Integration by Parts then give us,

$\displaystyle \int\tan x\;dx=\frac{-\cos x}{\cos x}-\int(-\cos x)\frac{\sin x}{\cos^2 x}\;dx=-1+\int\tan x\;dx$

Subtract the integral from both sides, we have,

$0=-1$

add 1 onto both sides of the equation, and therefore 1=0. $\square$

Of course everyone knows zero can’t possible equal to one, but can you find where the error is? 😀

1. sigmundlok says:

”Integration by Parts” was proved as the following….first, [d(uv)/dx]=v[du/dx]+u[dv/dx]. Now, integrate both side and we done. So, int[u dv]= uv- int[v du]. But, we have a mistake here. int(d(uv))=uv+Constant =/= uv. Of course, we are always accustomed to add the constant when we finish the integration. So, this mistake is not easy to find it

2. mathgarage says:

You’re right but you shouldn’t post this so soon and let others play. 🙂

3. sigmundlok says:

ha~ agree~ i’m sorry about it.

4. malihahalawy says:

Reblogged this on malihahalawy.