Need help with direct proof

Question

So I'm learning about direct proofs, and the first example shown is giving me a headache because I can't figure out how did the professor came up with the end solution. Here's what we need need to prove:

and the actual proof:

So, the third and the forth lines are really what is confusing to me, how he got there? Can anybody help with what is going on here ?

By this, I request you to completely confirm that you have understood the first two lines, namely that if $a \ | \ b$ and $a \ | \ c$, then $b = ad_1$ and $c = ad_2$ for some integers $d_1,d_2$. If you have understood thus far, then simply subtracting the second equation from the first gives $b- c = ad_1 - ad_2 = a(d_1-d_2)$. Now do you get the third line at least? — Sarvesh Ravichandran Iyer, Jul 19 '17 at 18:40
Subtract the two equations in the 2nd line, then factor out $,a,$ from the RHS, which yields $, b - c = a d_1! - a d_2 = a(d_1-d_2), $ so, by definition, $,a\mid b-c,,$ since $,d_1 - d_2,$ is an integer (since integers are closed under subtraction). — Bill Dubuque, Jul 19 '17 at 18:45

score 2 · Answer 1 · answered Jul 19 '17 at 18:41

2

You are given that $a\mid b$. By definition of "$\mid$", this means that there exists an integer $d_1$ such that $b=d_1a$.

You are given that $a\mid c$. By definition of "$\mid$", this means that there exists an integer $d_2$ such that $c=d_2a$.

You want to make a statement about $b-c$. From the above, we know that $b-c=d_1a-d_2a=(d_1-d_2)a$.

As $d_1-d_2$ is an integer, this fact shows that $a\mid b-c$.

answered Jul 19 '17 at 18:41

Hagen von Eitzen

374,180

Wow, finally something that I get, except the last thing, that if d1-d2 is an integer, that somehow proves a∣ b-c. I don't see why ? – Zed Jul 19 '17 at 18:47
Your goal (worth stating after the first two lines of this answer) is to find an integer $d_3$ such that $b-c = d_3a$. The last line shows you can do that by letting $d_3 = d_1 - d_2$. – Ethan Bolker Jul 19 '17 at 18:51
$a|b-c$ means there is an integer, $d$ so that $da = b-c$. Let $d= d_1-d_2$. That is an integer, and $da = (d_1-d_2)a = d_1a - d_2a = b- c$. So there DOES* exist an integer $d$ so that $d*a = b-c$. Everything that is required to conclude $a|b-c$ is satisfied and demonstrated. – fleablood Jul 19 '17 at 18:52

score 0 · Answer 2 · answered Jul 19 '17 at 18:45

We assume $a | b$ and $a | c$.

This means (by definition) that there are $d_1, d_2 \in \mathbb{Z}$ such that $b = d_1 a$ and $c = d_2 a$.

Now, $b-c = d_1 a - d_2 a$ (just subsitute the previous equations)
so $b-c = (d_1 - d_2) a$ (just expand the braces, we have a common factor $a$).
So we can write $(b-c)$ as "some number" ($d_1 - d_2$) times $a$, so again by definition $a | (b-c)$.

score 0 · Accepted Answer · answered Jul 19 '17 at 18:48

0

The definition of $n|m$ is that $n,m$ are both integers and there is an integer $d$ so that $m = n*d$.

Is that okay?

So if $a|b$ and $a|c$ then there exist integers $d_1$ and $d_2$ so that: $b = d_1*a$ and $c = d_2*a$.

Is that okay?

So $b-c = d_1*a - d_2*a = (d_1-d_2)*a$

Is that okay?

$d_1-d_2$ is an integer that exists. And $(d_1 - d_2)*a = (b-c)$?

Is that correct?

So there is an integer, namely $d_1-d_2$ so that, $(d_1-d_2)*a = (b-c)$. So, by definition, $a|(b-c)$?

Is that okay?

If not, where did I lose you?

Would it help if I had said:

$b - c = d_1*a - d_2*a = (d_1-d_2)*a$.

$d_1-d_2$ is an integer. Let's call it $d = d_2-d_2$.

So $b-c = (d_1-d_2)*a = d*a$. So there exists an integer, $d$, so that $d*a = (b-c)$.

That is the definition of $a|(b-c)$.

Is that okay?

answered Jul 19 '17 at 18:48

fleablood

124,253

Thanks, this extra explanation is what I really needed to understand. – Zed Jul 19 '17 at 18:50
"is an integer that exists" is a strange thing to say. – Bill Dubuque Jul 19 '17 at 18:51
Yes, it is a very strange thing to say. But I think the OP's confusion of doing a proof is a disconnect between "translating" concepts, to rules, to mechanical manipulations of expressions. The RULEZ say $a|b-c$ if there exists an integer so that $da = b-c$. And the OP knows* that $(d_1-d_2)a=b-c$. So why doesn't the OP conclude $a|b-c$? I believe it is a barrier in the language of proofs that the OP doesn't get that $d_1 - d_2$ is* the integer. So we need to show an integer exists that does it. And $d_1 - d_2$ is an integer. And it does it. And it .... exists. – fleablood Jul 19 '17 at 18:58

score 0 · Answer 4 · answered Jul 19 '17 at 18:51

So, we know that if $a|b$ and $c|d$, then there are integers $d_1,d_2$ so that $b=ad_1$ and $c=ad_2.$

To get the step on the third line we now substitute our expressions for $b$ and $c$. If we want to, we can look at the third step as $$(b-c)=[(ad_1)-(ad_2)]=a(d_1+d_2).$$

Now, when we look at the fourth step, we realize that since $d_1$ and $d_2$ are integers, $d_1+d_2$ is also an integer.

Looking back at the definition of $n|m,$ we see that the definition is $n|m$ if and only if there is an integer $k$ so that $m=kn.$ Here, in this instance, since $(d_1+d_2),\, a,$ and $(b-c)$ are all integers and $a(d_1+d_2)=(b-c)$, by the definition of $n|m$, we get that $a|(b-c)$.

The proof is saying that, just with much more brevity, and leaving some of the explanatory details out.

Need help with direct proof

4 Answers4