## Prove me wrong

For students of class 11-12 (age 16+)
tanvirab
Posts: 446
Joined: Tue Dec 07, 2010 2:08 am

### Re: Prove me wrong

Actually it is undefined. Division is defined as multiplication by inverse, and there is no multiplicative inverse of zero. So division by zero is undefined.

Abdul Muntakim Rafi
Posts: 173
Joined: Tue Mar 29, 2011 10:07 pm

### Re: Prove me wrong

I agree with Masum bhai.
$a/b=c$
$a=b c$
Now if you take a and b to be 0 then
$0=0*c$
It is satisfied for any value of c. So we can't determine the value of c here. So it is undetermined.

Again,
$a = b c$
$R= 0* c$
Now if take a to be a any number(except 0) and b to be 0. you can't find any value of c. Its horrible. We can't define c using any of our logic. So its undefined.
Man himself is the master of his fate...

tanvirab
Posts: 446
Joined: Tue Dec 07, 2010 2:08 am

### Re: Prove me wrong

Abdul Muntakim Rafi wrote:I agree with Masum bhai.
$a/b=c$
$a=b c$
Now if you take a and b to be 0 then
Then your first line $a/b$ is undefined (not undetermined). The word undetermined has a very specific meaning; it means something is definable, but cannot determined. You cannot define something else like $a/b$ where $b \neq 0$ and then take $b = 0$.

Abdul Muntakim Rafi
Posts: 173
Joined: Tue Mar 29, 2011 10:07 pm

### Re: Prove me wrong

Bhaiya, we can't define $x/0$ where x is not equal to 0.
but we can define $0/0$
The above process proves that... Yet we can't determine the value... so $indeterminate$ ...
Maybe the convention is to use the term $indeterminate$... However, I meant $indeterminate$ by $undetermined$....
Man himself is the master of his fate...

tanvirab
Posts: 446
Joined: Tue Dec 07, 2010 2:08 am

### Re: Prove me wrong

Abdul Muntakim Rafi wrote:Bhaiya, we can't define $x/0$ where x is not equal to 0.
but we can define $0/0$
The above process proves that...
The above process is wrong.

tanvirab
Posts: 446
Joined: Tue Dec 07, 2010 2:08 am

### Re: Prove me wrong

Lets start from definitions.

Multiplicative inverse: The multiplicative inverse of a number $x$ is a number $y$ such that $xy=1$.
Division: The division of a number $x$ by a number $y$ is $xy^{-1}$ where $y^{-1}$ is the multiplicative inverse of $y$.

So you see dividing by $0$ is not defined because it has no multiplicative inverse.

tanvirab
Posts: 446
Joined: Tue Dec 07, 2010 2:08 am

### Re: Prove me wrong

We usually write the multiplicative inverse of $x$ as $1/x$.

You process is wrong because you first write $a/b$ which actually means $a \times \frac{1}{b}$. Now you cannot take $b=0$ because there is no such thing as $1/0$.

Abdul Muntakim Rafi
Posts: 173
Joined: Tue Mar 29, 2011 10:07 pm

### Re: Prove me wrong

Why $0$ doesn't have any multiplicative inverse?
Man himself is the master of his fate...

tanvirab
Posts: 446
Joined: Tue Dec 07, 2010 2:08 am

### Re: Prove me wrong

Look at the definition. Is it possible for $0$ to have a multiplicative inverse?

Masum
Posts: 592
Joined: Tue Dec 07, 2010 1:12 pm
But I think definitions as usual makes not that much sense in real. If $\frac10$ was real, it must be a real number, say it is $x$. Then we could say, $x.0=1$
But that is impossible for any $x\in\mathbb R$. So, it never can be a real number. Hope it makes a better sense.
One one thing is neutral in the universe, that is $0$.