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Math Help - Problem involving rational number

  1. #1
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    Problem involving rational number[Solved]

    I can't seem to figure out a problem in one of my assignments. The question is:

    Suppose a, b, c, and d are integers and a does not equal c. Suppose also that x is a real number that satisfies the equation
    (ax+b)/(cx+d) = 1.

    Must x be rational? If so, express x as a ratio of two integers.



    So does that mean b can equal d? Confused on how to start it.
    Last edited by hellfire127; October 19th 2010 at 06:38 PM.
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  2. #2
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    Quote Originally Posted by hellfire127 View Post
    I can't seem to figure out a problem in one of my assignments. The question is:

    Suppose a, b, c, and d are integers and a does not equal c. Suppose also that x is a real number that satisfies the equation
    (ax+b)/(cx+d) = 1.

    Must x be rational? If so, express x as a ratio of two integers.



    So does that mean b can equal d? Confused on how to start it.
    Solve for x..

    ax + b = cx + d

    ax - cx = d - b

    x (a - c) = d - b

    x = (d - b) / (a - c)

    which is rational..
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  3. #3
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    wow, you made it look easy. thank you.
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    Instead of making a new thread, i have one more for the night which involves proofs. I can set it out but can't seem to get to the conclusion.

    I need to proof this by contradiction:
    If r>0 is irrational, then square root of r is irrational.

    Assume r>0, r is rational, and sqrt of r is irrational.

    Then sqrt of r is rational.
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    Quote Originally Posted by hellfire127 View Post
    Instead of making a new thread, i have one more for the night which involves proofs. I can set it out but can't seem to get to the conclusion.

    I need to proof this by contradiction:
    If r>0 is irrational, then square root of r is irrational.

    Assume r>0, r is rational, and sqrt of r is irrational.

    Then sqrt of r is rational.
    Proof by contradiction starts by assuming that the premise is true and the conclusion is false. You have done the opposite.

    Assume r>0 is irrational and sqrt(r) is rational.

    Consider (sqrt(r))^2. By closure of rational numbers, this number is rational. But it is equal to r, which is irrational. Contradiction.
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    All n=integer, there does not exist a k=integer such that n^2 - 2=4k. To proof this by contradiction I would assume that n=integer, k does not equal an integer, and n^2 - 2 does not equal 4k. What would be the best way to start it?
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  7. #7
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    All n=integer, there does not exist a k=integer such that n^2 - 2=4k.
    This should be written either in symbolic form: \forall n:\text{integer}\,\neg\exists k:\text{integer}.\,n^2-2=4k or in words: "For all integers n, there does not exist an integer k such that n^2 - 2=4k". In the first case, the notation "n : integer", as far as I know, comes from programming languages where it means "n is a variable of type integer". It is also possible to write n\in\mathbb{Z} where \mathbb{Z} is the standard notation for the set of integers. Equality should not be used to write this.

    To prove this statement by contradiction, you assume that n is an integer and then assume the negation of what you have to prove, i.e., that there does exist an integer k such that n^2 - 2=4k. One way to continue is to note that in this case n^2 is even and then to use an auxiliary fact that if a square of an integer x is even (this happens only when x itself is even), then x^2 is divisible by 4. From here, a contradiction can be obtained.

    Forum rules say, "9. Start a new thread for a new question. Don't tag a new question onto an existing thread. Otherwise the thread can become confusing and difficult to follow." I guess, another reason for this is that unless a new thread is opened, people don't know that there is a new question; they think that whoever answered your first question is better equipped to answer any follow-ups. Also, we love when a whole thread can be marked as [SOLVED]
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    Quote Originally Posted by emakarov View Post
    This should be written either in symbolic form: \forall n:\text{integer}\,\neg\exists k:\text{integer}.\,n^2-2=4k or in words: "For all integers n, there does not exist an integer k such that n^2 - 2=4k". In the first case, the notation "n : integer", as far as I know, comes from programming languages where it means "n is a variable of type integer". It is also possible to write n\in\mathbb{Z} where \mathbb{Z} is the standard notation for the set of integers. Equality should not be used to write this.

    To prove this statement by contradiction, you assume that n is an integer and then assume the negation of what you have to prove, i.e., that there does exist an integer k such that n^2 - 2=4k. One way to continue is to note that in this case n^2 is even and then to use an auxiliary fact that if a square of an integer x is even (this happens only when x itself is even), then x^2 is divisible by 4. From here, a contradiction can be obtained.

    Forum rules say, "9. Start a new thread for a new question. Don't tag a new question onto an existing thread. Otherwise the thread can become confusing and difficult to follow." I guess, another reason for this is that unless a new thread is opened, people don't know that there is a new question; they think that whoever answered your first question is better equipped to answer any follow-ups. Also, we love when a whole thread can be marked as [SOLVED]
    Ok, thanks for the rules, I'll keep that in mind. I see my mistake in the negation of the second part, but I still don't see how you can just state that n^2 is even. Should't have taken discrete math lol, I blame my adviser. Cal is so much easier.
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  9. #9
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    If n^2 - 2=4k, then n^2 = 2(2k + 1), so n^2 is even. This implies that n is even. Indeed, if n were odd, then n = 2m + 1 for some integer m. Then n^2=2(2m^2+2m) + 1, which is odd. Since n is even, n = 2i for some integer i. Therefore, n^2=4i^2, i.e., n^2 is divisible by 4. On the other hand, the first equation of this post implies that n^2 = 4k+2, i.e., n^2 produces the remainder 2 when divided by 4. A contradiction. Therefore, n^2 - 2=4k is false.
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