I found this in some other website, its basically the manipulation of a formula to finally give 1=0!. Here it is:
Let x = 1
Then, x^2=x
x^2-1 = x-1
Dividing both sides by (x-1), we get
x+1 = 1
x+1-1 = 0, so
Since x=1,
1=0
Any explanations?
I found this in some other website, its basically the manipulation of a formula to finally give 1=0!. Here it is:
Let x = 1
Then, x^2=x
x^2-1 = x-1
Dividing both sides by (x-1), we get
x+1 = 1
x+1-1 = 0, so
Since x=1,
1=0
Any explanations?
You are correct that $\displaystyle \displaystyle 1 = 0!$, but $\displaystyle \displaystyle 1 \neq 0$.
$\displaystyle \displaystyle \begin{align*} n! &= n(n - 1)! \\ \frac{n!}{n} &= (n - 1)! \end{align*}$
Now letting $\displaystyle \displaystyle n = 1 $ we have
$\displaystyle \displaystyle \begin{align*} (1-1)! &= \frac{1!}{1} \\ 0! &= \frac{1}{1} \\ 0! &= 1 \end{align*}$