Direct Proofs Involving Divisibility

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In this video, we write direct proofs of four different statements that all involve divisibility of integers. I hope you find it helpful!

Timestamps:

0:00 - Intro
0:50 - Definition
3:30 - Proof 1
6:17 - Proof 2
9:41 - Proof 3
12:48 - Proof 4

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So glad that you're back! I'm taking Calc I this semester and you've been a ton of help with my precalc and college algebra classes.

Hcs
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I was wondering if this is another valid way to write the proof:

Prove the transitive property:
If a|b and b|c, then a|c for all positive integers a, b, c.

Let’s analyze our first given.
We know that a|b.
What does it mean for a to divide b?
Let l be any positive integer.
lℤ+.
This means that a times some integer l equals b.
Algebraically, we can say:
al=b
Manipulating this expression to solve for a, we find that:
a=b/l

Let’s analyze our second given:
We know that b|c.
What does it mean for b to divide c?
Let m be any positive integer.
mℤ+.
This means that b times some integer m equals c.
Algebraically, we can say:
bm=c

What are we trying to prove?
We are trying to prove that a|c.
What does it mean for a to divide c?
Let n be any positive integer.
nℤ+.
This means that a times some integer n equals c.
Algebraically, we can say:
an=c

We are trying to prove that a|c.
We are trying to prove that an=c where n is some integer.
This means we are trying to demonstrate that n=c/a is an integer.

We have n=c/a.
Recall that c=bm and a=b/l.
Let’s replace c with bm.
Let’s replace a with (b/l).
n=(bm)/(b/l)
We multiply bm by the reciprocal of (b/l).
n=bm*(l/b)
The bs here cancel, leaving us with:
n=ml

Recall that m is a positive integer.
Recall that l is a positive integer.
Positive integers are closed under multiplication.
This means that the product of two positive integers yields a positive integer.
Thus, n must be a positive integer.
Since we have confirmed that n is a positive integer, we have confirm that a indeed divides c.

Can you let me know whether or not this proof is valid?
Thank you :)

keldonchase