# Is 1 equal to 0?

• Jun 20th 2011, 05:02 AM
IBstudent
Is 1 equal to 0?
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?
• Jun 20th 2011, 05:03 AM
Quacky
Re: Is 1 equal to 0?
Quote:

Originally Posted by IBstudent
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, x2=x
x2-1 = 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?

This is very old.
"x = 1"
"Dividing both sides by (x-1)"
• Jun 20th 2011, 05:05 AM
earboth
Re: Is 1 equal to 0?
Quote:

Originally Posted by IBstudent
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 <--- here you divide by zero which isn't allowed
x+1 = 1
x+1-1 = 0, so
Since x=1,
1=0

Any explanations?

...
• Jun 20th 2011, 05:07 AM
Prove It
Re: Is 1 equal to 0?
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*}
• Jun 20th 2011, 05:09 AM
Quacky
Re: Is 1 equal to 0?
Quote:

Originally Posted by Prove It
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*}

(Rofl)
• Jun 20th 2011, 05:19 AM
Soroban
Re: Is 1 equal to 0?

$\displaystyle \boxed{\begin{array}{c}\text{Theorem} \\[2mm]1 = 0\:\text{ for large values of }0.\end{array}}$