# Thread: Good Day! just need Guidance on my process.

1. ## Good Day! just need Guidance on my process.

If a|b+c then a|b and b|c

Suppose a|(b+c) = t for some t element of Z such that t = (k+p) for some k and p elements of Z

b+c = a(t)
b+c = a (k+p) by substitution
b+c = ak + ap by distributive property

then b = ak and c = ap
this implies a|b = k and a|c =p

thus if a|b and b|c , then a|(b+c)

please correct me if im mistaken... thnaks

2. ## Re: Good Day! just need Guidance on my process.

If a|b+c then a|b and b|c
did you mean ...

If a|(b+c) then a|b and b|c ?

are you proving this statement or its converse?

"If a|(b+c) then a|b and b|c." is false ... consider a = 2, b = 3, and c = 5

The converse "If a|b and b|c, then a|(b+c)" is true.

3. ## Re: Good Day! just need Guidance on my process.

Originally Posted by rcs
If a|b+c then a|b and b|c

Suppose a|(b+c) = t for some t element of Z such that t = (k+p) for some k and p elements of Z

b+c = a(t)
b+c = a (k+p) by substitution
b+c = ak + ap by distributive property

then b = ak and c = ap THIS DOES NOT FOLLOW
this implies a|b = k and a|c =p

thus if a|b and b|c , then a|(b+c)

please correct me if im mistaken... thnaks
$a= 2,\ b = 13,\ c = 19,\ k = 11,\ and\ p = 5.$

$b + c = 13 + 19 = 32 \implies \dfrac{b + c}{a} = \dfrac{32}{2} = 16 = 11 + 5 = k + p.$

$ak = 2 * 11 = 22.$

$ap = 2 * 5 = 10.$

$\therefore ak + ap = 22 + 10 = 32 = b + c.$

$But\ 13 \ne 2 * 11 = 22\ and\ 19 \ne 2 * 5 = 10.$

4. ## Re: Good Day! just need Guidance on my process.

Your basic problem is that you are trying to prove something that isn't true! 7 divides 4+ 3 but does not divide either 4 or 3.