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Irrationality of this number. Proof.

  1. Apr 30, 2012 #1
    1. The problem statement, all variables and given/known data

    Prove that if x is a real number then 2x2-3=0



    2. Relevant equations



    3. The attempt at a solution

    Well I think this proof is OK but it seems very to the point, please tell me if there is any fallacy in my reasoning.

    2x2-3=0

    2x2=3

    but 2x2 is even, while 3 is odd

    Hence x must be irrational for all real x.

    ??
     
  2. jcsd
  3. Apr 30, 2012 #2

    SammyS

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    All you proved is that x in not an integer.

    What is the problem as stated for you to solve?
     
  4. Apr 30, 2012 #3

    Hurkyl

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    There are many things wrong. :frown:

    This statement is false. x=0 is a counterexample: it satisfies the hypothesis, but not the conclusion.

    For this bit, I can't tell if there's any problem with your reasoning, because you don't say what you're reasoning is -- you just wrote two equations.

    The notion of "even" and "odd" applies to integers, but it doesn't apply to real numbers. If you are using x to denote a real number, then 2x2 is also 'merely' a real number, and so it doesn't make sense to talk about it being even.

    This is another false statement. x=0 is another counterexample. It is real, however it is not irrational. It also has nothing to do with the statement you said you were trying to prove.



    I hate to say it, but I think what you need to do is to focus on trying to understand logic, grammar, and the meaning of what you read/write, and also on explicitly writing out what you are thinking and doing.


    Would you please state exactly the question you were asked? I'm guessing it's something similar to
    If x is a real number satisfying the equation 2x2 - 3 = 0, then x is irrational.​
    Once you've stated the question, it may be useful to do one or all of the following:
    • Explain why your statement in the opening post means the same thing or doesn't mean the same thing as the question you were asked.
    • Explain why my statement means the same thing or doesn't mean the same thing as the question you were asked.
    • Explain why my statement means the same thing or doesn't mean the same thing as your statement in the opening post.
     
  5. Apr 30, 2012 #4
    Ok I miss read the question and I thought it was for x an integer...

    is this better.

    Suppose that

    2x^2-3=0

    adding 3 two both sides of the equation

    2x^2=3

    dividing both sides by 2

    x^2=3/2

    taking the square root of both sides

    x=√(3/2)

    Now suppose that x is not irrational, ie suppose

    x=p/q for some integers p/q

    so √(3/2)=p/q

    so √3/√2 =p/q

    but √3 and √2 are both irrational.

    Contradiction therefore x must be irrational.
     
  6. Apr 30, 2012 #5

    Office_Shredder

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    The ratio of two irrational numbers can be rational, for example the square root of eight divided by the square root of two.

    Do you know how to prove that the square root of two is irrational? A similar argument can be used here
     
  7. Apr 30, 2012 #6
    Yeah I think it should be along the lines...

    Assume that √2 is not irrational ie

    √2=a/b (by definition of rationality).

    so 2=a^2/b^2

    so..

    2b^2=a^2

    Now a^2 is divisible by 2 and clearly so is a.

    so let a=2k

    so 2b^2=(2k)^2

    2b^2=4k^2

    b^2=2k^2

    so b^2 is also divisible by 2.

    Contradiction Since any rational number is a quotient or relative primes.

    Therefore √2 is rational.
     
  8. Apr 30, 2012 #7

    HallsofIvy

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    You last statement should be "[itex]\sqrt{2}[/itex] is NOT rational"!

    Now apply those same ideas to [itex]\sqrt{3/2}[/itex].
     
  9. Apr 30, 2012 #8
    oops, Ok thanks Halls of Ivy!!
     
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