Prove me wrong

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tanvirab
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Re: Prove me wrong

Unread post by tanvirab » Fri Dec 09, 2011 4:36 pm

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.

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Abdul Muntakim Rafi
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Re: Prove me wrong

Unread post by Abdul Muntakim Rafi » Fri Dec 09, 2011 7:13 pm

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.
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tanvirab
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Re: Prove me wrong

Unread post by tanvirab » Sat Dec 10, 2011 2:35 am

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 $b = 0$.

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Abdul Muntakim Rafi
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Re: Prove me wrong

Unread post by Abdul Muntakim Rafi » Sat Dec 10, 2011 11:05 am

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
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Re: Prove me wrong

Unread post by tanvirab » Sat Dec 10, 2011 11:28 am

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
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Re: Prove me wrong

Unread post by tanvirab » Sat Dec 10, 2011 11:33 am

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
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Re: Prove me wrong

Unread post by tanvirab » Sat Dec 10, 2011 11:36 am

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 $1/0$.

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Abdul Muntakim Rafi
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Re: Prove me wrong

Unread post by Abdul Muntakim Rafi » Sat Dec 10, 2011 12:11 pm

Why $0$ doesn't have any multiplicative inverse?
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tanvirab
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Re: Prove me wrong

Unread post by tanvirab » Sat Dec 10, 2011 12:14 pm

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

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Masum
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Re: Prove me wrong

Unread post by Masum » Sat Dec 10, 2011 10:35 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 $0$.

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