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  1. #1
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    prove irrational

    Prove that sqrt(1 + sqrt(3)) is irrational!
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    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by mandy123 View Post
    Prove that sqrt(1 + sqrt(3)) is irrational!
    you can do a proof by contradiction

    assume it is rational. then we can write $\displaystyle \sqrt{1 + \sqrt {3}} = \frac ab$ for $\displaystyle a,b \in \mathbb{Z}$, $\displaystyle b \ne 0$.

    we square both sides to obtain: $\displaystyle 1 + \sqrt{3} = \frac {a^2}{b^2}$

    so $\displaystyle \sqrt{3} = \frac {a^2}{b^2} - 1$

    now the right side of the equation is a rational number (by our assumption). thus we must have that $\displaystyle \sqrt{3}$ is rational. therefore, we can prove the claim by showing that this is false.

    now proving $\displaystyle \sqrt{3}$ is irrational is also done by contradiction, and usually done the same way $\displaystyle \sqrt{2}$ is proven to be irrational (see here)

    but there is a shorter way.

    consider the equation $\displaystyle x^2 - 3 = 0$

    by the rational roots theorem, if rational roots exist, they will be one of $\displaystyle \pm 1, \mbox{ or } \pm 3$. but none of these solve the equation. however, $\displaystyle \sqrt{3}$ solves the equation and is therefore not a rational number.

    so seeing that $\displaystyle \sqrt{3}$ is not rational, we have our contradiction
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  3. #3
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    Quote Originally Posted by Jhevon View Post
    [snip]

    now proving $\displaystyle \sqrt{3}$ is irrational is also done by contradiction, and usually done the same way $\displaystyle \sqrt{2}$ is proven to be irrational (see here)

    [snip]
    I'll just toss in my 2 cents worth here and say that proving $\displaystyle \sqrt{3}$ is irrational is subtlely more difficulty than proving $\displaystyle \sqrt{2}$ is irrational ......

    If you're assuming the usual thing, $\displaystyle \sqrt{3} = \frac{m}{n}$ where m,n are relatively prime integers, then at some stage you must prove that if m^2 is divisible by 3, then m is also divisible by 3.

    This is slightly more difficult than proving that if m^2 is divisible by 2 then m is divisible by 2 .... (but can nevertheless be easily done by using the fact that every positive integer has a unique prime factorisation).
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    Quote Originally Posted by mr fantastic View Post
    I'll just toss in my 2 cents worth here and say that proving $\displaystyle \sqrt{3}$ is irrational is subtlely more difficulty than proving $\displaystyle \sqrt{2}$ is irrational ......

    If you're assuming the usual thing, $\displaystyle \sqrt{3} = \frac{m}{n}$ where m,n are relatively prime integers, then at some stage you must prove that if m^2 is divisible by 3, then m is also divisible by 3.

    This is slightly more difficult than proving that if m^2 is divisible by 2 then m is divisible by 2 .... (but can nevertheless be easily done by using the fact that every positive integer has a unique prime factorisation).
    I think this may be overkill, they may already know that the square root of
    any integer that is not a perfect square is irrational (or even n-th root of
    an integer not an n-th power).

    In this form of question I woukd assume they know that $\displaystyle \sqrt{3}$ is irrational.

    RonL
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    Quote Originally Posted by mr fantastic View Post
    I'll just toss in my 2 cents worth here and say that proving $\displaystyle \sqrt{3}$ is irrational is subtlely more difficulty than proving $\displaystyle \sqrt{2}$ is irrational ......

    If you're assuming the usual thing, $\displaystyle \sqrt{3} = \frac{m}{n}$ where m,n are relatively prime integers, then at some stage you must prove that if m^2 is divisible by 3, then m is also divisible by 3.

    This is slightly more difficult than proving that if m^2 is divisible by 2 then m is divisible by 2 .... (but can nevertheless be easily done by using the fact that every positive integer has a unique prime factorisation).
    It is extremely simple.

    Definition: Given a positive integer $\displaystyle a$ and a prime $\displaystyle p$ we define $\displaystyle \mbox{ord}_p a = n$ ($\displaystyle n\geq 0$)as the number of times that $\displaystyle p$ divides $\displaystyle a$, i.e. $\displaystyle p^n | a$ but $\displaystyle p^{n+1} \not | a$. (Note, this definition is a good definition because eventually powers of $\displaystyle p$ will exceeded $\displaystyle a$ and therefore will not divide it any further thus such an $\displaystyle n$ has to exist).

    Lemma 1: Given two positive integers $\displaystyle a \mbox{ and }b$ with $\displaystyle p$ prime we have that $\displaystyle \mbox{ord}_p (ab) = \mbox{ord}_p a + \mbox{ord}_p b$.

    Proof: If $\displaystyle \mbox{ord}_p a= n$ then $\displaystyle p^n|a$ but $\displaystyle p^n\not | a$. If $\displaystyle \mbox{ord}_p b = m$ then $\displaystyle p^m | b$ but $\displaystyle p^m \not | b$. This means that $\displaystyle p^{n+m}| ab$ but if $\displaystyle p^{n+m+1}|ab$ then $\displaystyle p^{n+1}$ (or $\displaystyle p^{m+1})$ divides $\displaystyle a$ (or $\displaystyle b$ respectively) which is a contradiction.

    Lemma 2: Given a positive integer $\displaystyle a$ the integer is a power of $\displaystyle n$ if and only if $\displaystyle \mbox{ord}_p a$ is divisible by $\displaystyle n$ for all primes $\displaystyle p$.

    Proof: If $\displaystyle a=b^n$ then $\displaystyle \mbox{ord}_p a = \mbox{ord}_p b^n = n\mbox{ord}_p b$. Conversely, if $\displaystyle \mbox{ord}_p a$ is divisible by $\displaystyle n$ then by unique factorization we know that $\displaystyle a= \prod_{p \text{ prime}} p^{\text{ord}_p a}$ and thereofore each exponent is prime and divisible by $\displaystyle n$ hence $\displaystyle a$ is a power of $\displaystyle n$.

    Theorem: If $\displaystyle a\geq 1$ is not a power of $\displaystyle n$ then $\displaystyle \sqrt[n]{a}$ is irrational.

    Proof: Proof by contrapositive. Suppose that $\displaystyle \sqrt[n]{a} = r/s$ for positive integers $\displaystyle r,s$. Then, $\displaystyle (r/s)^n = a $ thus $\displaystyle r^n = as^n$. Now for any prime $\displaystyle p$ we have $\displaystyle \mbox{ord}_p r^n = \mbox{ord}_p as^n$ thus $\displaystyle n \mbox{ord}_p r = n\mbox{ord}_p s + \mbox{ord}_p a$. This shows us that $\displaystyle \mbox{ord}_p a$ is divisible by $\displaystyle n$ for any prime $\displaystyle p$. Thus, by Lemma 2 we have that $\displaystyle a$ is a power of $\displaystyle n$.
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