0

If $a \mid c, b \mid c, \gcd (a,b)=1$ then $ab \mid c.$

I understand that given problem is true. however im struggling with writing to prove.

I let A=2 , B= 3 , C= 6

2 l 6= 3

3 I 6=2

3*2 l 6=1

I have shown my work to prove that the theorem is true however I can't write in words.

how can you write it in words to prove it?? I think proving the theorem or proof is the hardest part of the course

Martin Sleziak
  • 53,687
amie
  • 3
  • Use LaTeX for mathematics in this site. Directions in the FAQ section – DonAntonio Jun 01 '13 at 12:53
  • In words: If both $a,b$ divide $c$ and share no common factors, their product also divides $c$. – Pedro Jun 01 '13 at 12:53
  • is there any way to prove it by saying like " assume that a,b, c E Z such that a divides b , b divides c, ab divides C is gcd (a,b) is 1??? – amie Jun 01 '13 at 13:14
  • @amie I cannot think of a simpler proof than what I wrote. – Pedro Jun 01 '13 at 13:33
  • @amie If $\ a\mid b\ $ then $\ 1 = (a,b) = a\ $ so $\ ab = b\mid c\ $ by hypothesis. That trivial case is not equivalent to what you seek to prove. – Key Ideas Jun 01 '13 at 13:42

2 Answers2

1

Suppose that $a\mid c,b\mid c$. We have $a\mid \dfrac{c}{b}b$. Since $(a,b)=1$, this means $a\mid \dfrac c b$, which gives $ab\mid c$. Note we used Euclid's Lemma.

The other option is using Bezout: since $(a,b)=1$; there exist $s,t\in\Bbb Z$such that $$at+sb=1\tag 1$$

Write then $c=aj$, $c=bk$. Then Multiplying by $c$ in $(1)$ gives

$$act+scb=c\tag 2$$

And replacing $c$ by $aj$ and $bk$ appropriately in $(2)$, we get

$$a(bk)t+s(aj)b=c$$

which gives

$$(ab)(kt+sj)=c$$ which is saying $ab\mid c$

ADD (A version of) Euclid's Lemma says

Suppose $a\mid bc$ and $(a,b)=1$. Then $a\mid c$.

That is, if a number $a$ divides a product $bc$, and is coprime to the factor $b$, then it must divide the other.

P Since $(a,b)=1$, $ab=\operatorname{lcm}(a,b)$. But since $a\mid bc$ and $b\mid bc$; it follows $bc$ is a common multiple of $a$ and $b$. It follows the least common multiple of them, $ab$, divides them. That is $ab\mid bc$. But this means $a\mid c$.

Pedro
  • 122,002
  • Thank you. i figured it out now! <3 Peter muah – amie Jun 01 '13 at 16:03
  • @amie =) If this answer was helpful, you may upvote it or even accept it using the $\checkmark$. – Pedro Jun 01 '13 at 16:04
  • 1
    Can you explain what Euclid's Lemma is in your words?? i have note on Euclid's Lemma but it doesnt help me at all cause i dont undersand what prof wrote. I do understand what Bezout is though. – amie Jun 01 '13 at 16:13
1

We have $$\left(a|c\iff \exists k\in\mathbb{Z}:\quad c=ak\right)\quad\text{and}\quad (b|c\iff \exists k'\in\mathbb{Z}:\quad c=bk')$$ and by Bezout identity $$\gcd(a,b)=1\iff \exists (p,q)\in\mathbb{Z}^2:\quad pa+qb=1\tag{1}$$ then we multiply $(1)$ by $c$ we find $$pac+qbc=pk'ab+qkab=ab(pk'+qk)=c$$ hence $$ab|c$$

  • Could you explain all of that in words? I know it may takes some time and writing but i would really apprecite it if you can write it out for me so tha i can comprehend and not be confused. All in All thank you so much for your efforts and time – amie Jun 01 '13 at 13:46
  • ok but what you have not understood exactly? –  Jun 01 '13 at 13:50
  • i can read what you wrote, ((a|c⟺∃k∈Z:c=ak)and(b|c⟺∃k ′ ∈Z:c=bk ′ ) but i dont understand what it means. idk what it's saying. i was told that i should assume first that there is x or a,b,c E Z or R. and then come up with new variables and then solve to make a=c or a=b or ab=c to prove it. – amie Jun 01 '13 at 13:58
  • The first sentence means: $a$ divide $c$ if and only if $c$ is a multiple of $a$ so we write $c=ak$ for some $k$ and the same thing for $b|c$. The second sentence is the use of Bezout identity and you can see wikipedia for more explanation of this identity and finally we combine the two result to find that $ab(pk'+qk)=c$ which means that $c$ is a multiple of $ab$ or equivalently $ab$ divide $c$. –  Jun 01 '13 at 14:06