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Thread: more contrapositive

  1. #1
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    more contrapositive

    given: if the sum of two numbers is strictly less than 50 then at least one number is strictly less than 25.

    I am supposed to do this by contrapositive

    Is the contrapositive, "if two numbers are equal or great than 25, then their sum is 50 or greater" ??

    if so then

    llet $\displaystyle a,b \in \mathbb{Z}, s.t. a \ge 25;b \ge 25$

    then $\displaystyle a+b \ge 50$

    therefore a or b must be less than 25 in order for the sum to be less than 50.
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  2. #2
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    Re: more contrapositive

    that's it
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  3. #3
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    Re: more contrapositive

    More work is needed in the proof imho.

    If $\displaystyle \begin{align*} a \geq 25 \end{align*}$ then $\displaystyle \begin{align*} a = 25 + c \end{align*}$ where $\displaystyle \begin{align*} c \geq 0 \end{align*}$.

    If $\displaystyle \begin{align*} b \geq 25 \end{align*}$ then $\displaystyle \begin{align*} b = 25 + d \end{align*}$ where $\displaystyle \begin{align*} d \geq 0 \end{align*}$.

    So $\displaystyle \begin{align*} a + b = 25 + c + 25 + d = 50 + c + d \geq 50 \end{align*}$ as $\displaystyle \begin{align*} c,d \geq 0 \end{align*}$.
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  4. #4
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    Re: more contrapositive

    More work is needed in the proof imho.

    If $\displaystyle \begin{align*} a \geq 25 \end{align*}$ then $\displaystyle \begin{align*} a = 25 + c \end{align*}$ where $\displaystyle \begin{align*} c \geq 0 \end{align*}$.

    If $\displaystyle \begin{align*} b \geq 25 \end{align*}$ then $\displaystyle \begin{align*} b = 25 + d \end{align*}$ where $\displaystyle \begin{align*} d \geq 0 \end{align*}$.

    So $\displaystyle \begin{align*} a + b = 25 + c + 25 + d = 50 + c + d \geq 50 \end{align*}$ as $\displaystyle \begin{align*} c,d \geq 0 \end{align*}$.
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