1. not understanding

have to do a couple problems for homework and im not understanding exactly what is going on and why its the answer

27. Let m, n, and d be integers. Show that if d \ m and d \ n, then d\ (m - n).

my professor gave us the answer to this but i do not understand why it works
answer: m = dq1 n = dq2 m - n = dq1 - dq2 = d(q1 - q2) therefore d \ (m - n)

these are the other problems i have to do

28. Let m, n, and d be integers. Show that if d \ m, then d \ mn.

31. Let a, b, and c be integers. Show that if a \ b and b \ c, then a \ c.

33. Give an example of consecutive primes p1 = 2, p2 ..., pn
where
p1p2... pn + 1
is not prime.

any help would be much appreciated

2. Originally Posted by pooshipple
have to do a couple problems for homework and im not understanding exactly what is going on and why its the answer

27. Let m, n, and d be integers. Show that if d \ m and d \ n, then d\ (m - n).
If $\displaystyle d\mid m,d\mid n$ then $\displaystyle m=dz,n=dz'$ for some $\displaystyle z,z'\in\mathbb{Z}$. Clearly then $\displaystyle m-n=dz-dz'=d\left(z-z'\right)$ and since $\displaystyle z,z'\in\mathbb{Z}\implies z-z'\in\mathbb{Z}$ the conclusion follows.
my professor gave us the answer to this but i do not understand why it works
answer: m = dq1 n = dq2 m - n = dq1 - dq2 = d(q1 - q2) therefore d \ (m - n)
Sorry, I just noticed this. What it means to say that $\displaystyle d\mid n$ is that $\displaystyle d$ divides $\displaystyle n$ or that $\displaystyle \frac{n}{d}\in\mathbb{Z}$. Does that make a little more sense? So if $\displaystyle m-n=d\left(z-z'\right)$ then $\displaystyle \frac{m-n}{d}=\frac{d\left(z-z'\right)}{d}=z-z'\in\mathbb{Z}$

these are the other problems i have to do

28. Let m, n, and d be integers. Show that if d \ m, then d \ mn.
$\displaystyle d\mid m\implies m=dz$ so then $\displaystyle mn=dzn$ or $\displaystyle \frac{mn}{d}=zn$ and since the integers are closed under multiplication the conclusion follows.

31. Let a, b, and c be integers. Show that if a \ b and b \ c, then a \ c.
These all can be done the same. $\displaystyle a\mid b\implies b=za$ and $\displaystyle b\mid c\implies c=bz'$. So then $\displaystyle c=bz'=\left(az\right)z'$ and the conclusion follows.

33. Give an example of consecutive primes p1 = 2, p2 ..., pn
where
p1p2... pn + 1
is not prime.
What does "consecutive" mean? That $\displaystyle p_2=p_1+1$ or that is merely the next prime in the list of primes? I'm assuming the latter. What do you think?