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Math Help - prove sqrt(3) is irrational

  1. #1
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    prove sqrt(3) is irrational

    proof:
    Assume sqrt(3)= a/b where a,b are integers and b cannot equal 0. (sqrt(3)=(a/b)^2 turns 3=a^2/b^2 and 3b^2=a^2. We know that 3|a^2 iff 3|a. Using this theorem we know that 3(b^2)=a^2 must be a multiple of 3|a. 3|a then becomes 3r=a where r is some integer. Then squaring both side 3r=a becomes 9r^2=a^2. Since 3b^2=a^2 that means that 3b^2=9r^2 which simplifies into b/r=sqrt(3).
    Since b/r is a rational number this is a contradiction

    I think its right not sure though.
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  2. #2
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    Re: prove sqrt(3) is irrational

    Quote Originally Posted by bonfire09 View Post
    proof:
    Assume sqrt(3)= a/b where a,b are integers and b cannot equal 0. (sqrt(3)=(a/b)^2 turns 3=a^2/b^2 and 3b^2=a^2. We know that 3|a^2 iff 3|a. Using this theorem we know that 3(b^2)=a^2 must be a multiple of 3|a. 3|a then becomes 3r=a where r is some integer. Then squaring both side 3r=a becomes 9r^2=a^2. Since 3b^2=a^2 that means that 3b^2=9r^2 which simplifies into b/r=sqrt(3).
    Since b/r is a rational number this is a contradiction
    I will comment on your proof because I do not follow it.
    There is a standard proof.
    If the integer n is not a square then \sqrt n is irrational.
    Prove that by using the floor function.
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  3. #3
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    Re: prove sqrt(3) is irrational

    I havent learned the floor function yet. The book tells me to use this as a lemma: 3|a^2 if and only if 3|a. Since 3(b^2)=a^2 and 9r^2=a^2 from squaring 3|a I can just set them equal to each other and it becomes
    3b^2=9r^2? and then show that b/r=sqrt(3) and b/r is rational which is a contradiction.
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  4. #4
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    Re: prove sqrt(3) is irrational

    Quote Originally Posted by bonfire09 View Post
    I havent learned the floor function yet. The book tells me to use this as a lemma: 3|a^2 if and only if 3|a. Since 3(b^2)=a^2 and 9r^2=a^2 from squaring 3|a I can just set them equal to each other and it becomes
    3b^2=9r^2? and then show that b/r=sqrt(3) and b/r is rational which is a contradiction.
    Sorry to read this.
    You have an out-of-date textbook.
    May be someone else can help you.
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  5. #5
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    Re: prove sqrt(3) is irrational

    Hello, bonfire09!

    I'll give you the standard proof.


    \text{Prove that }\sqrt{3}\text{ is irrational.}

    Assume that \sqrt{3} is rational.

    We have: \sqrt{3} \,=\,\frac{a}{b}\;\text{ where }\:\begin{Bmatrix}(1) & a\text{ and }b\text{ are integers.} \\ (2) & \text{The denominator }b\text{ is not zero.} \\ (3) & \text{The fraction }\dfrac{a}{b}\text{ is reduced to lowest terms.}\end{Bmatrix}

    Square the equation: . 3 \:=\:\frac{a^2}{b^2} \quad\Rightarrow\quad 3b^2 \:=\:a^2 .[4]

    We see that a^2 is a multiple of 3; hence a is a multiple of 3.

    Let a = 3r

    Then [4] becomes: . 3b^2 \:=\:9r^2\quad\Rightarrow\quad b^2 \:=\:3r^2

    We see that b^2 is a multiple of 3; hence b is a multiple of 3.


    But if a is a multiple of 3 and b is a multiple of 3,
    . . then the fraction \frac{a}{b} is not reduced to lowest terms.


    We have arrived at a contradiction.
    Therefore, \sqrt{3} is not rational.
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  6. #6
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    Re: prove sqrt(3) is irrational

    Here is my new revised proof then
    Assume sqrt(3)= a/b where a,b are integers and b cannot equal 0. We can also assume that a/b has been reduced to lowest terms. Sqrt(3)=(a/b)^2 turns 3=a^2/b^2 and 3b^2=a^2. We know that 3|a^2 iff 3|a. Using this theorem we know that 3(b^2)=a^2 must be a multiple of 3|a. 3|a then becomes 3r=a where r is some integer. Then squaring both side 3r=a becomes 9r^2=a^2. Since 3b^2=a^2 that means that 3b^2=9r^2 which implies 3r^2=b^2 where r^2 is an integer. By the previous thereom stated we know that if 3|b^2 then 3|b. Since a and b both share a divisor of 3 that means it has not been reduced to its lowest terms. Hence by contradiction sqrt(3) is irrational.
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  7. #7
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    Re: prove sqrt(3) is irrational

    Looks correct except for the sentence "Using this theorem we know that 3(b^2)=a^2 must be a multiple of 3|a"
    Maybe modify it to be something like "From 3(b^2)=a^2, we know that 3|a^2, and so this theorem shows that 3|a."
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