You can't do it because multiplication by \( 0 \) destroys information. That is, for \( z = x \times y \), \( y \neq 0 \), we can recover \( x \) via \( \frac {z} {y} \). However, when \( y = 0\), \( 0 = x \times y \) for all \( x \), so there's no way to recover \( x \) through a division by \( y \)--the information about \( x \) has been destroyed. Anthropomorphically speaking, the division operator "doesn't know" what \( x \) was--\( x \) could have been any number.

This isn't true of things like \( i \), since there's a reversal operation to recover what \( i \) "used to be"--namely, squaring it.

It's possible you could put together a system of math where division by zero is defined in cases where you know the value of \( x \), since then you know how to reverse \( x \times 0 \). That is, \( z = x \times 0 \), and \( \frac {z} {0} = x \). I leave that as an exercise for the reader. :)
–
Christopher Johnson
·
3 years, 12 months ago

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@Christopher Johnson
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That is a clear and charitable answer, Christopher. Thanks for doing this.
–
Silas Hundt
Staff
·
3 years, 12 months ago

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TopNewestYou can't do it because multiplication by \( 0 \) destroys information. That is, for \( z = x \times y \), \( y \neq 0 \), we can recover \( x \) via \( \frac {z} {y} \). However, when \( y = 0\), \( 0 = x \times y \) for all \( x \), so there's no way to recover \( x \) through a division by \( y \)--the information about \( x \) has been destroyed. Anthropomorphically speaking, the division operator "doesn't know" what \( x \) was--\( x \) could have been any number.

This isn't true of things like \( i \), since there's a reversal operation to recover what \( i \) "used to be"--namely, squaring it.

It's possible you could put together a system of math where division by zero is defined in cases where you know the value of \( x \), since then you know how to reverse \( x \times 0 \). That is, \( z = x \times 0 \), and \( \frac {z} {0} = x \). I leave that as an exercise for the reader. :) – Christopher Johnson · 3 years, 12 months ago

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– Silas Hundt Staff · 3 years, 12 months ago

That is a clear and charitable answer, Christopher. Thanks for doing this.Log in to reply

– Ajmal Ima · 3 years, 12 months ago

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Umm ... \(\dfrac{z}{0}\) is undefined (not a constant) for \(z \in \mathbb{C}\). – Jimmy Kariznov · 3 years, 12 months ago

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