Results 1 to 4 of 4

Math Help - Proof by Contradiction

  1. #1
    Junior Member
    Joined
    Sep 2010
    Posts
    48

    Proof by Contradiction

    I need help with this proof. We have to prove the following statement by contradiction:

    "Show that the product of a non-zero rational number and an irrational number is irrational."

    This is what I have so far, based off of examples we did in class:

    We proceed by way of contradiction. Assume that the product of a non-zero rational number and an irrational number is rational. Let p and r be rational numbers, and q be an irrational number. By definition, there exist integers a and b, b not equal to 0, and integers c and d, d not equal to 0, for which p = (a/b) and r = (c/d). So, pq = r, which implies (a/b)q = (c/d). Without loss of generality, we assume (a/b) and (c/d) are in lowest terms.

    I don't know where to go from here. If anyone could help, I would greatly appreciate it.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor undefined's Avatar
    Joined
    Mar 2010
    From
    Chicago
    Posts
    2,340
    Awards
    1
    Quote Originally Posted by Lprdgecko View Post
    I need help with this proof. We have to prove the following statement by contradiction:

    "Show that the product of a non-zero rational number and an irrational number is irrational."

    This is what I have so far, based off of examples we did in class:

    We proceed by way of contradiction. Assume that the product of a non-zero rational number and an irrational number is rational. Let p and r be rational numbers, and q be an irrational number. By definition, there exist integers a and b, b not equal to 0, and integers c and d, d not equal to 0, for which p = (a/b) and r = (c/d). So, pq = r, which implies (a/b)q = (c/d). Without loss of generality, we assume (a/b) and (c/d) are in lowest terms.

    I don't know where to go from here. If anyone could help, I would greatly appreciate it.
    Let p be a nonzero rational and q be irrational and suppose pq = r is rational. Then r/p = q is irrational. But \mathbb{Q} is closed under multiplication. Contradiction.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Super Member

    Joined
    May 2006
    From
    Lexington, MA (USA)
    Posts
    11,746
    Thanks
    648
    Hello, Lprdgecko!

    I think you're there . . .


    \text{Show that the product of a non-zero rational number}
    \text{and an irrational number is irrational.}

    Suppose that: . \text{(rational)} \times \text{(irrational)} \;=\;\text{(rational)}

    \text{That is: }\;p \cdot q \:=\:r\quad\text{ where: }\;\begin {Bmatrix} p &=& \text{rational} \\ q &=& \text{irrational} \\ r &=& \text{rational}\end{Bmatrix}


    \text{Then }p = \dfrac{a}{b},\;\text{ where }a\text{ and }b\text{ are integers and }b \ne 0.

    \text{And }r = \dfrac{c}{d},\;\text{ where }c\text{ and }d\text{ are integers and }d \ne 0.


    \text{Hence, we have: }\;\dfrac{a}{b}\cdot q \:=\:\dfrac{c}{d} \quad\Rightarrow\quad q \;=\;\dfrac{bc}{ad}

    \text{Since }bc\text{ and }ad\text{ are integers, }\dfrac{bc}{ad}\text{ is rational.}


    Therefore, \,q is rational . . . We have reached a contradiction.

    Follow Math Help Forum on Facebook and Google+

  4. #4
    Junior Member
    Joined
    Sep 2010
    Posts
    48
    Thank you both. I keep forgetting that I can go back to what I have stated in the proof and use that to complete the proof. Thanks again
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. [SOLVED] Proof by contradiction
    Posted in the Number Theory Forum
    Replies: 5
    Last Post: February 28th 2011, 10:06 AM
  2. Proof by Contradiction
    Posted in the Discrete Math Forum
    Replies: 5
    Last Post: September 28th 2010, 09:23 PM
  3. [SOLVED] direct proof and proof by contradiction
    Posted in the Number Theory Forum
    Replies: 2
    Last Post: February 27th 2010, 10:07 PM
  4. Proof by contradiction
    Posted in the Number Theory Forum
    Replies: 1
    Last Post: February 5th 2010, 04:17 PM
  5. Proof by contradiction
    Posted in the Discrete Math Forum
    Replies: 2
    Last Post: April 11th 2009, 04:12 PM

Search Tags


/mathhelpforum @mathhelpforum