A community for students. Sign up today!
Here's the question you clicked on:
 0 viewing

This Question is Closed

Boblovesmath
 2 years ago
Best ResponseYou've already chosen the best response.0Sorry I meant it is an irrational number. So it is false.

UnkleRhaukus
 2 years ago
Best ResponseYou've already chosen the best response.1prove by contradiction , assuming the square root of two is rational , it can be written as a ratio of natural numbers that have no common factors p,q say \[√2=p/q\] \[2=p^2/q^2\]\[2q^2=p^2\] this implies \(p^2\) is even if \(p^2\) is even then \(p\) is even if \(p \) is even it can be written as two times another natural number \(r\) \[p=2r\] \[2q^2=p^2=(2r)^2=4r^2\]\[q^2=2r^2\] which similarly implies that \(q\) is even With \(p,q\) both even the natural numbers necessarily have a common factor of two , which contradicts the statement of rationality, \[\sqrt2\not\in\text{rational numbers}\] \[\square\]

Boblovesmath
 one year ago
Best ResponseYou've already chosen the best response.0So in order to be rational the none of them should be even or at least one of them should be even and the other not even in order not to have a common factor of 2? Can you please explain this situation to me if you may?

UnkleRhaukus
 one year ago
Best ResponseYou've already chosen the best response.1if the numer is rational there must be a way to express it as a ratio of (relatively prime) natural numbers, (ie no common factors)
Ask your own question
Ask a QuestionFind 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.