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anonymous
 4 years ago
Does this seem legit?
anonymous
 4 years ago
Does this seem legit?

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anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Prove that \(\sqrt{8}\) is irrational. Proof: Assume \(\sqrt{8}\) is rational. Then, for some integers \(p\) and \(q\) where \(q\neq0\) and \(p\) and \(q\) have no common factors, we have \(\sqrt{8}=p/q\), \(8=p^2/q^2\), \(p^2=8q^2\), \(p^2=2(4q^2)\). This implies that \(p^2\) is an even number, which in turn implies \(p\) is an even number. Therefore, for some integer \(r\), we can rewrite \(p=2r\). Substituting, we obtain \(r^2=2q^2\). Similarly, this implies \(r\) is an even number, and, for some integer \(s\), we can rewrite \(r=2s\). Substituting, we obtain \(2s^2=q^2\). This implies that \(q\) is an even number. Therefore, we have shown that both \(p\) and \(q\) are even numbers, and must then have the common factor \(2\). This is a contradiction of our hypothesis, hence \(\sqrt{8}\) must be an irrational number. \(\blacksquare\)\[\]

JamesJ
 4 years ago
Best ResponseYou've already chosen the best response.2Yes, it's a straight imitation of the proof \( \sqrt{2} \) is irrational. Which suggests another proof. If you know that \( \sqrt{2} \) is irrational, then \[ \sqrt{8} = 2\sqrt{2} \] must be irrational also.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0once you prove the product of irrational and rational is irrational

JamesJ
 4 years ago
Best ResponseYou've already chosen the best response.2...but I know prealg has done that already. Now, here's a small challenge: what is it about this proof that wouldn't work for \( \sqrt{4} \). Suppose, in other words, we didn't know this was equal to 2. What about imitating to proof above fails?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Hi prealgebra :) wanna chat

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0not to be a pain, but i always wonder why this is the preferred proof. if you want to prove that the square root of any natural number that is not a perfect square is irrational, the fundamental theorem of arithmetic gets it quickly. the "even/odd" bit seems unnecessarily forced, just because we have a word "even" for divisible by 2, and no such word for "divisible by 7" for example

JamesJ
 4 years ago
Best ResponseYou've already chosen the best response.2**correction: what is it about imitating the proof above that fails?

JamesJ
 4 years ago
Best ResponseYou've already chosen the best response.2I think it's preferred because it very elementary and doesn't require any other larger results. It's something you can explain to a HS school student.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0in my experience, the problem with it is that after a minute or two the "even/odd" bit gets all muddled up. i am not exactly sure, but i believe all the proofs use the method of infinite descent somewhere.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Well i am soo not intrested in this conversation anyomrre :p
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