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Math Help - [SOLVED] Proving Irrational Numbers (Easy?)

  1. #1
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    [SOLVED] Proving Irrational Numbers (Easy?)

    Hello All,

    I had a quick question about irrational numbers. We know that Sqrt[15] is irrational, so it can be proven that (Sqrt[3] + Sqrt[5]) is an irrational number as well. How is it proven?

    Out of curiosity, why is Sqrt[15] connected to Sqrt[3] + Sqrt[5] other than 15=3x5 ? What is the connection?
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  2. #2
    Super Member Failure's Avatar
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    Quote Originally Posted by Samson View Post
    Hello All,

    I had a quick question about irrational numbers. We know that Sqrt[15] is irrational, so it can be proven that (Sqrt[3] + Sqrt[5]) is an irrational number as well. How is it proven?
    If \sqrt{3}+\sqrt{5} were rational then \sqrt{3}-\sqrt{5} would also be rational (for consider that \sqrt{3}-\sqrt{5}=\frac{3-5}{\sqrt{3}+\sqrt{5}}).
    But from this it would follow that both \sqrt{3} and \sqrt{5} are rational, which is not the case.

    Out of curiosity, why is Sqrt[15] connected to Sqrt[3] + Sqrt[5] other than 15=3x5 ? What is the connection?
    Whether the square root of a natural number is rational or not is determined by wether or not its (essentially) unique prime factor decomposition consists only of even powers of prime numbers or not. In the former case its square root is a natural number, in the latter case its square root is irrational. So \sqrt{3^1\cdot 5^1}, \sqrt{3^2\cdot 5^1}, \sqrt{3^1\cdot 5^2},\sqrt{3^{101}\cdot 5^{78}} are all irrational, whereas \sqrt{3^2\cdot 5^2}, \sqrt{3^{102}\cdot 5^{78}} are natural numbers.
    Last edited by Failure; June 7th 2010 at 08:23 AM.
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    Quote Originally Posted by Failure View Post
    If \sqrt{3}+\sqrt{5} were rational then \sqrt{3}-\sqrt{5} would also be rational (for consider that \sqrt{3}-\sqrt{5}=\frac{3-5}{\sqrt{3}+\sqrt{5}}).
    But from this it would follow that both \sqrt{3} and \sqrt{5} are rational, which is not the case.


    Whether the square root of a natural number is rational or not is determined by wether or not its (essentially) unique prime factor decomposition consists only of even powers of prime numbers or not. In the former case its square root is a natural number, in the latter case its square root is irrational. So \sqrt{3^1\cdot 5^1}, \sqrt{3^2\cdot 5^2}, \sqrt{3^1\cdot 5^2},\sqrt{3^{101}\cdot 5^{78}} are all irrational, whereas \sqrt{3^2\cdot 5^2}, \sqrt{3^{102}\cdot 5^{78}} are natural numbers.
    I believe you stated that one of your solutions was both rational and irrational Sqrt[(3^2) * (5^2)] ... is that just a mistake or am I missing something?
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  4. #4
    Senior Member roninpro's Avatar
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    Note that (\sqrt{3}+\sqrt{5})^2=8+2\sqrt{15}, and consider the following fact: the square root of an irrational number is again irrational. Since the right hand side is irrational, we have \sqrt{3}+\sqrt{5}=\sqrt{8+2\sqrt{15}} is also.
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  5. #5
    Super Member Failure's Avatar
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    Quote Originally Posted by Samson View Post
    I believe you stated that one of your solutions was both rational and irrational Sqrt[(3^2) * (5^2)] ... is that just a mistake or am I missing something?
    Right, that's a typo that I am going to correct right away, thanks.
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    Thank you to both of you for your inputs!
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