A community for students.
Here's the question you clicked on:
 0 viewing
anonymous
 3 years ago
Can anyone prove the following?
If each of a and b is in ℤ+ and a I b then a <= b
anonymous
 3 years ago
Can anyone prove the following? If each of a and b is in ℤ+ and a I b then a <= b

This Question is Closed

KingGeorge
 3 years ago
Best ResponseYou've already chosen the best response.0What does it mean that ab?

KingGeorge
 3 years ago
Best ResponseYou've already chosen the best response.0I.e., what's the definition of ab?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0means that there is an integer c such that b=a*c =====>>>> a  b

KingGeorge
 3 years ago
Best ResponseYou've already chosen the best response.0Precisely, and both a,b are in \(\mathbb{Z}^+\). Now, assume that \(a>b\), and work towards a contradiction.

KingGeorge
 3 years ago
Best ResponseYou've already chosen the best response.0I would then break it up into two cases, and show why these are both impossible. (also, why is \(c\neq 0\)?) Case 1: \(c<0\) Case 2: \(c>0\)

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0contradiction when a>b?

KingGeorge
 3 years ago
Best ResponseYou've already chosen the best response.0Right. So assume a>b, and then divide into the two cases where c<0 and c>0. Reach a contradiction with both cases, so a<=b.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0i'm sorry, i've never been good with contradictions. So like if c>0, then a would HAVE to be < b since (b/c) would result in a fraction?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0therefore making a smaller than b?

KingGeorge
 3 years ago
Best ResponseYou've already chosen the best response.0Let's suppose \(c>0\). Then, \(c\ge 1\) since \(c\in\mathbb{Z}\). Hence, \[ac=b\ge a\]Since we assumed \(b<a\), this is a contradiction, so \(a\le b\). Now you just need to work out \(c\le 0\). That case should be easier.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0ohhh ok. Gotcha. Sorry for the bother
Ask your own question
Sign UpFind more explanations on OpenStudy
Your question is ready. Sign up for free to start getting answers.
spraguer
(Moderator)
5
→ View Detailed Profile
is replying to Can someone tell me what button the professor is hitting...
23
 Teamwork 19 Teammate
 Problem Solving 19 Hero
 Engagement 19 Mad Hatter
 You have blocked this person.
 ✔ You're a fan Checking fan status...
Thanks for being so helpful in mathematics. If you are getting quality help, make sure you spread the word about OpenStudy.