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I noticing that some numbers when multiplied together give the same product as their sum. After some trial and error I came up with a simple rule for this.

$$n\times\frac{n}{n-1} = n+\frac{n}{n-1}$$

For example

$$7\times\frac{7}{6} = \frac{49}{6}$$ $$ 7+\frac{7}{6} = \frac{49}{6}$$

I was wondering a few things,

  1. Is there any significance to this "theorem"?
  2. If it has already been discovered is there a name for it?
  3. Is there a similar theorem for division and subtraction?
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  • $\begingroup$ Well... $ab=a+b\iff a(b-1)=b\iff a=\frac b{b-1}$ so it's just a matter of rearrangement. (note that neither term can be $1$ as $a=a+1$ is impossible). $\endgroup$ – lulu Apr 17 '18 at 0:26
  • $\begingroup$ $n + \frac {n}{n-1} = \frac {n(n-1)}{n-1} + \frac {n}{n-1} = \frac {n(n-1) + n}{n-1}= \frac {n^2 -n + n}{n-1} = \frac {n^2}{n-1} = \frac {n\times n}{n-1} = n \times \frac {n}{n-1}$. $\endgroup$ – fleablood Apr 17 '18 at 0:39
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This is a consequence of the following: $$ n + \frac{n}{n - 1} = \frac{n(n -1)}{n - 1} + \frac{n}{n - 1} = \frac{n^2 - n + n}{n - 1} = \frac{n^2}{n - 1} = n\times \frac{n }{n - 1}. $$

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  • $\begingroup$ Damn! You beat me to it! $\endgroup$ – Shaun Apr 17 '18 at 0:33
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We have $$\begin{align} n + \frac{n}{n - 1} &= \frac{n(n -1)}{n - 1} + \frac{n}{n - 1} \\ &= \frac{n^2 - n + n}{n - 1} \\ &= \frac{n^2}{n - 1} \\ &= n\times \frac{n }{n - 1}. \end{align} $$

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Notice $1 + \frac 1{n-1} = \frac {n-1}{n-1} + \frac 1{n-1} = \frac n{n-1}$

So $n\times \frac n{n-1} = n(1 + \frac 1{n-1}) = n + \frac n{n-1}$

Also note: If we were asked to solve $m*n = m + n$ then $m*n -m = n$ and $m(n-1) = n$.

Then if $n -1 = 0$ then $n = 1$ and $m*0 = 1$ which is impossible. So $n\ne 1$ and $n -1 \ne 0$ and so $m = \frac n{n-1}$.

So the only solutions are if $n \frac n{n-1}=n + \frac n{n-1}$ and we must solve for $n$.

If $n = 0$ then $n \frac n{n-1}=0$ and $n + \frac n{n-1} = 0$ so $n = 0$ is a solution.

If $n \ne 0$ then $n = \frac {n-1}n (n + \frac n{n-1}) = (n-1) + 1 = n$ so an number (integer or not) will be a solution.

So $nm = n + m \iff (n,m) = (0,0) $ or $m = \frac n{n-1}$ (and $m = \frac n{n-1} \iff n = \frac m{m-1}$).

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