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Math Help - Please need help with direct proof

  1. #1
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    Please need help with direct proof

    Hi, I have a few problems which i can't solve

    1) need to prove (m+dk)mod d = m mod d (where m, d and k are integers and d does not equal to zero)

    2) if r is any rational number and s is any irrational, then r/s is irrational

    3) prove that there is at most one real number b with the property that br = r for all real numbers r.

    Thank you
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  2. #2
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    Hello, somestudent2!

    Prove: . m + dk \equiv m \pmod{d}
    where m,\,d,\,k are integers and d \neq 0.

    Definition: .  a \equiv b \pmod {d} \; \Longleftrightarrow \;a - b\:=\:dn

    In baby-talk: a and b are congruent (mod d) if their difference is a multiple of d.


    Since (m + dk) - m \:=\:dk, a multiple of d,
    . . then: . m + dk \equiv m \pmod{d}




    2) If r is any rational number and s is any irrational, then \frac{r}{s} is irrational.

    Proof by contradiction: .Assume \frac{r}{s} is rational.

    Then: . \frac{r}{s}\,=\,\frac{a}{b} .for some integers a and b. [1]

    Since r is rational: . r \,=\,\frac{c}{d} for some integers c and d.

    Then equation [1] becomes: . \frac{\frac{c}{d}}{s} \,=\,\frac{a}{b}\quad\Rightarrow\quad s \,=\,\frac{bc}{ad}


    Since bc and ad are integers (integers are closed under multiplication),
    . . s is a rational number. .But we are told that s is irrational.

    We have reached a contradiction, hence our assumption is incorrect.

    . . Therefore: . \frac{r}{s} is irrational.

    [Oh-oh! .You said direct proof, didn't you? . . . Never mind.]

    Last edited by Soroban; November 10th 2006 at 04:54 AM.
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  3. #3
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by somestudent2 View Post
    3) prove that there is at most one real number b with the property that br = r for all real numbers r.
    Say that some other real number, e, also has the property that er = r for all real numbers r. Then
    br = r = er

    br - er = 0

    (b - e)r = 0

    Thus b = e or r = 0. But r is any real number, so r is not 0 in general. Thus b = e.

    -Dan
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  4. #4
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    Quote Originally Posted by topsquark View Post
    But r is any real number, so r is not 0 in general.
    But zero is a real number, you cannot conclude it is non-zero. So his statement is wrong as you have shown.
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  5. #5
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    Thanks a lot for the quick response guys. I actually managed to solve 2nd and 3rd problem this morning today, and was glad to find out that my solution for 2nd matched with Soroban's solution(although I used square root of 2 to represent irrational number)

    For the last problem a sol. through contradiction came to my mind
    First we negate the statement, so we have
    Suppose there are two distinct real numbers a1 and a2 such that
    1) a1*r = r 2) a2*r = r

    then we have a1*a2= a2 (if r=a2)
    and a2*a1=a1 (if r=a1)
    this emplies

    a2=a1*a2=a2*a1= a1 (what means that a2=a1), but we assumed that a1 and a2 are two distinct real numbers, so we have a contradiction and thus our assumption was wrong.

    please tell me if this proof works.

    2)Is there a way to prove m+dk mod d = m mod d (without the use of the definition of mod, I don't think I can use it just yet)

    I was trying to get it going

    by representing m = dq +r and m+dk = dp +r
    then m=d(p-k) + r
    how do I prove that rs are the same in both cases?

    Thank you
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