My Math Forum Prove 1+sqrt(3) is irrational

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 March 26th, 2014, 10:51 PM #1 Senior Member     Joined: Oct 2013 From: Far far away Posts: 429 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 = (a-b)/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 = (a-bm)/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
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Quote:
 Originally Posted by shunya We will assume (correctly) that sqrt3 is irrational
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 $\frac{m}{n}$ and $\frac{p}{q}$ are in their lowest terms and
\begin{align*}\left(\frac{p}{q}\right)^2 &= \frac{m}{n} \\
np^2 &= mq^2\end{align*}

Now, since p and q share no factors, every factor of q^2 must divide n. In other words
\begin{align*}n &= \lambda q^2 \\
\lambda q^2 p^2 &= mq^2 \\
\lambda p^2 &= m \\
\end{align*}

But since m and n are coprime, $\lambda= 1$ and so $m= p^2, n=q^2$.
In other words, there is no rational number $\frac{m}{n}$ 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 $\sqrt{3}$ is not an integer because $1^2 \lt 3 \lt 2^2$ so $1 \lt \sqrt{3} \lt 2$.

 Tags irrational, prove, sqrt3

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# Prove that square root of plus another number is irrational

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