## Saturday, January 05, 2008

### Elementary Lemmas

Lemma 1:

If:

gcd(e,f)=1
e divides k
f divides k

Then:

ef divides k

Proof:

(1) Assume that ef does not divide k.

(2) Then, there must exist a prime p and a power c such that pc divides ef but pc does not divide k.

(3) Since gcd(e,f)=1, it follows that pc must entirely divide e or pc must entirely divide f.

(4) Let's assume that pc divides e. [We can make a parallel argument if pc divides f.

(5) But then we have a contradiction since e divides k implies that pc divides k.

(6) So we reject our assumption in step #1.

QED

Lemma 2:

If a ≥ b and b ≥ a, then a = b

Proof:

(1) a ≥ b

(2) So it follows that a = b or a is greater than b.

(3) But a is not greater than b since b ≥ a.

(4) So then it follows that a = b.

QED

Corollary 2.1:

If a,b are positive and a divides b and b divides a, then a = b

Proof:

(1) If a divides b, then b ≥ a.

(2) If b divides a, then a ≥ b

(3) Since a ≥ b and b ≥ a, we can use Lemma 2 above to conclude that a= b.

QED

Lemma 3:

if gcd(e,f)=1 and e divides af

then:

e divides a

Proof:

(1) Assume that e does not divide a.

(2) Then, there exists a prime p such that pc divides e but does not divide a.

(3) Since pc divides e, it follows that pc must divide af.

(4) Using Euclid's Lemma (see Lemma 2, here), we know that if p divides af, then it divides a or it divides f. But it cannot divide f since gcd(e,f)=1 so it divides a.

(5) But then pc must divide a since no p can divide f. This contradicts step #2.

(6) Therefore, we reject our assumption in step #1.

QED

Anonymous said...

Corollary 2.1:

If a divides b and b divides a, then a = b

I think this should instead be "If a divides b and b divides a, then |a| = |b|" because if you let a = -3 and b = 3, then -3|3 , 3|-3 but 3 doesn't equal -3. However, |3| = |-3|.

Larry Freeman said...

Thanks very much for your comment.

You are correct. I've updated the lemma to say if a,b are positive integers.

Cheers,

-Larry