A community for students.
Here's the question you clicked on:
 0 viewing
Bee_see
 one year ago
Prove that square root 2 + square root 2 is irrational. Your proof is allowed to use the fact that square root 2 is irrational.
Bee_see
 one year ago
Prove that square root 2 + square root 2 is irrational. Your proof is allowed to use the fact that square root 2 is irrational.

This Question is Closed

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I think this might be similar to what you're looking for http://www.homeschoolmath.net/teaching/proof_square_root_2_irrational.php It's not exact but you can follow the same process perhaps.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Then again, if you're allowed to say that the sqrt of two is already irrational, you might not have to do that. Any number plus an irrational number is going to be irrational anyway

Bee_see
 one year ago
Best ResponseYou've already chosen the best response.0I think I understand it with one square root of 2, but I don't know what to do if I have two...

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Well it says that you're allowed to say that the sqrt of 2 is irrational, so I don't think you're suppose to prove it. https://www.khanacademy.org/math/algebra/rationalandirrationalnumbers/rationalandirrationalexpressions/v/proofthatsumofrationalandirrationalisirrational This is what I think what you are supposed to prove ^^

jim_thompson5910
 one year ago
Best ResponseYou've already chosen the best response.1Hint: \[\large \sqrt{2}+\sqrt{2} = \frac{a}{b}\] \[\large 2\sqrt{2} = \frac{a}{b}\] \[\large (2\sqrt{2})^2 = \left(\frac{a}{b}\right)^2\] \[\large 8 = \frac{a^2}{b^2}\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Oh I completely missed the sqrt in the first term. My apologies :(

Bee_see
 one year ago
Best ResponseYou've already chosen the best response.0Right so then 8b^2=a^2...so a is even and then change a to 2k then 8=(2k)^2/b^2?

freckles
 one year ago
Best ResponseYou've already chosen the best response.1\[\sqrt{2}+\sqrt{2}=2 \sqrt{2} =\frac{a}{b} \\ \sqrt{2}=\frac{a}{2b} \text{ I think we could have stopped here }\]

freckles
 one year ago
Best ResponseYou've already chosen the best response.1it says we can use that sqrt(2) is irrational

freckles
 one year ago
Best ResponseYou've already chosen the best response.1but here we have expressed sqrt(2) as an integer over an integer

Bee_see
 one year ago
Best ResponseYou've already chosen the best response.0aren't I supposed to get rid of the 8 besides b^2? Because then I would get 1/2k^2 and that's not correct since the statement should say even number.....

freckles
 one year ago
Best ResponseYou've already chosen the best response.1you don't have to do all of that

jim_thompson5910
 one year ago
Best ResponseYou've already chosen the best response.1@freckles has a much better way to do it if you decide to go another route, you could say a = 4k (a is still even), so 8=a^2/b^2 8=(4k)^2/b^2 8b^2=16k^2 b^2 = 2k^2 which implies b is even. But if b is even, then a/b isn't fully reduced. So that's another contradiction

Loser66
 one year ago
Best ResponseYou've already chosen the best response.0We have a theorem : " If x is a solution of a polynomial, then either \( x\in\mathbb Z\) or \(x \notin \mathbb Q\)." Are you allowed to use it?

Loser66
 one year ago
Best ResponseYou've already chosen the best response.0It allows us to prove this kind of problem nicely.

Bee_see
 one year ago
Best ResponseYou've already chosen the best response.0when are you supposed to use the proof by contradiction?

Loser66
 one year ago
Best ResponseYou've already chosen the best response.0Ok, whatever, I contribute what I know. Consider \((x + 2\sqrt 2)(x2\sqrt2)=0\\x^2 8 =0 \) Hence either \(2\sqrt2 \in \mathbb Z~~or ~~2\sqrt2 \notin \mathbb Q\) Suppose \(2\sqrt2\in \mathbb Z\) we know that \(1< \sqrt 2<2\\2< 2\sqrt 2<4\) Hence if \(2\sqrt2 \in\mathbb Z\), then \(2\sqrt2 = 3\) \(\implies 8 =9\) contradiction. Hence \(2\sqrt2 \notin \mathbb Z\) therefore \(2\sqrt2 \notin \mathbb Q\)

Bee_see
 one year ago
Best ResponseYou've already chosen the best response.0The class hasn't gone over what you wrote..The problems just mentions "prove," so how do I know for sure which proof to use between contradiction, indirect proof, and direct proof?
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.