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March 26th, 2014, 10:51 PM  #1 
Senior Member Joined: Oct 2013 From: Far far away Posts: 422 Thanks: 18  Prove 1+sqrt(3) is irrational
Question) Prove that 1 + sqrt3 is an irrational number. Show similarly that m + n(sqrt3) is an irrational number for all rationals m and n (n not 0) In the question, sqrt3 means square root of 3. I will use the same notation everywhere in the question My attempt: We will assume (correctly) that sqrt3 is irrational Let 1 + sqrt3 = a/b implies sqrt3 = (a/b)  1 implies sqrt3 = (ab)/b implies sqrt3 is rational but sqrt3 is irrational Therefore by contradiction, 1 + sqrt3 is irrational Second part of question: Let m + n(sqrt3) = a/b implies n(sqrt3) = (a/b)  m implies n(sqrt3) = (a  bm)/b implies sqrt3 = (abm)/bn implies sqrt3 is rational but sqrt3 is irrational Therefore by contradiction, m + n(sqrt3) is irrational Are my proofs acceptable? Is there a better way to do this because after looking at similar problems, this has become rather mechanical for me. Thanks 
March 26th, 2014, 11:51 PM  #2 
Senior Member Joined: Jun 2013 From: London, England Posts: 1,316 Thanks: 116  Re: Prove 1+sqrt(3) is irrational
Those proofs are fine.

March 28th, 2014, 05:39 PM  #3 
Senior Member Joined: Nov 2013 Posts: 247 Thanks: 2 
Ah but you can multiply any square root by another square root by just multiplying what is in the radicals. Sqrt 2 * Sqrt 50. Both of these are irrational but multiplying them gives 10, a rational number. Why? Because you multiply 2 * 50 = 100 and 100 is a perfect square.

March 28th, 2014, 09:52 PM  #4 
Math Team Joined: Dec 2013 From: Colombia Posts: 7,655 Thanks: 2633 Math Focus: Mainly analysis and algebra  I would prove this rather than assuming it. The best proof I know of is to suppose that m, n, p and q are integers, both and are in their lowest terms and Now, since p and q share no factors, every factor of q^2 must divide n. In other words But since m and n are coprime, and so . In other words, there is no rational number that is a perfect square unless both m and n are perfect squares. In particular, when n=1, an integer cannot be the square of a rational number, unless that number itself is an integer. And is not an integer because so . 

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