For example, Say we have the number 10.
[1+9 = 10] , 1*9 = 9
[2+8 = 10] , 2*8 = 16
[3+7 = 10] , 3*7 = 21
[4+6 = 10] , 4*6 = 24
[5+5 = 10] , 5*5 = 25
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.
.
10+0 = 10
I want to know how to show that two different numbers
that their sum that adds up to 10. say (3,7) and (4,6) their products will
always be different. How can I show this in a general case? I trying to get
better with proofs
If $a + b = M$ and $a' + b' = M$
The average of $a$ and $b$ and the average of $a'$ and $b'$ is $\frac M2$ and we can rewrite $a = \frac M2 \pm d$ and $\frac M2 \mp d$ and $a' = \frac M2 \pm e$ and $b' = \frac M2\mp e$ for some $d,e\ge 0$.
If we let $\frac M2 = m$ we can restate the problem as.
For any pair of numbers $m +d$ and $m-d$ where $d \ge 0$ and another pair $m+e$ and $m-e$ where $e \ge 0$ but $d\ne e$ so that $(m+d)+ (m-d) = (m +e) + (m-e)=2m$. We want to prove that
$(m+d)(m-d) \ne (m+e)(m-e)$ if $d \ne e$ (and both $e,d \ge 0$).
$(m+d)(m-d) = m^2 -md +md -d^2 = m^2 -d^2$.
And $(m+e)(m-e) = m^2 - e^2$. So $ab = (m+d)(m-d) = (m+e)(m-e) = a'b'$ if and only if $d^2 = e^2$ but if we assume $e,d \ge 0$ then that means $d=e$.
So for example:
$3 + 7 = 10$ and $4+6 = 10$ but
$3*7 = (5-2)(5+2) = 5^2 - 2^2$ while $4*6 = (5-1)(5+1) = 5^2-1^2$. These are not equal because $2\ne 1$.