Proof verification: Prove that the product of $2$ real negative numbers is positive.












1












$begingroup$



Can someone please verify any mistakes in my proof of this theorem. Is this a correct proof of the corollary of $(-1)times (-1) = 1$ that the product of two negative numbers is positive?




Theorem: The product of two real negative numbers is positive



Proof: Let $x,yinmathbb{R^+}$. We have



begin{equation}
begin{alignedat}{2}
(-1) times0=0quad
Rightarrowquad &&
(-1) times (-xy+xy) &= 0
\
Rightarrowquad &&
(-1)times((-1)times xy+xy)&= 0
\
Rightarrowquad &&
(-1)times(-1)times xy+(-1)times xy &= 0 \ Rightarrowquad && (-1)times x times (-1) times y +(-1)times xy&=0 \
Rightarrowquad &&(-x)times (-y)- xy&=0 \
Rightarrowquad &&(-x)times (-y)&=xy.
end{alignedat}
end{equation}






proof-verification proof-writing







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$endgroup$

















    1












    $begingroup$



    Can someone please verify any mistakes in my proof of this theorem. Is this a correct proof of the corollary of $(-1)times (-1) = 1$ that the product of two negative numbers is positive?




    Theorem: The product of two real negative numbers is positive



    Proof: Let $x,yinmathbb{R^+}$. We have



    begin{equation}
    begin{alignedat}{2}
    (-1) times0=0quad
    Rightarrowquad &&
    (-1) times (-xy+xy) &= 0
    \
    Rightarrowquad &&
    (-1)times((-1)times xy+xy)&= 0
    \
    Rightarrowquad &&
    (-1)times(-1)times xy+(-1)times xy &= 0 \ Rightarrowquad && (-1)times x times (-1) times y +(-1)times xy&=0 \
    Rightarrowquad &&(-x)times (-y)- xy&=0 \
    Rightarrowquad &&(-x)times (-y)&=xy.
    end{alignedat}
    end{equation}






    proof-verification proof-writing







    share|cite|improve this question











    $endgroup$















      1












      1








      1





      $begingroup$



      Can someone please verify any mistakes in my proof of this theorem. Is this a correct proof of the corollary of $(-1)times (-1) = 1$ that the product of two negative numbers is positive?




      Theorem: The product of two real negative numbers is positive



      Proof: Let $x,yinmathbb{R^+}$. We have



      begin{equation}
      begin{alignedat}{2}
      (-1) times0=0quad
      Rightarrowquad &&
      (-1) times (-xy+xy) &= 0
      \
      Rightarrowquad &&
      (-1)times((-1)times xy+xy)&= 0
      \
      Rightarrowquad &&
      (-1)times(-1)times xy+(-1)times xy &= 0 \ Rightarrowquad && (-1)times x times (-1) times y +(-1)times xy&=0 \
      Rightarrowquad &&(-x)times (-y)- xy&=0 \
      Rightarrowquad &&(-x)times (-y)&=xy.
      end{alignedat}
      end{equation}






      proof-verification proof-writing







      share|cite|improve this question











      $endgroup$





      Can someone please verify any mistakes in my proof of this theorem. Is this a correct proof of the corollary of $(-1)times (-1) = 1$ that the product of two negative numbers is positive?




      Theorem: The product of two real negative numbers is positive



      Proof: Let $x,yinmathbb{R^+}$. We have



      begin{equation}
      begin{alignedat}{2}
      (-1) times0=0quad
      Rightarrowquad &&
      (-1) times (-xy+xy) &= 0
      \
      Rightarrowquad &&
      (-1)times((-1)times xy+xy)&= 0
      \
      Rightarrowquad &&
      (-1)times(-1)times xy+(-1)times xy &= 0 \ Rightarrowquad && (-1)times x times (-1) times y +(-1)times xy&=0 \
      Rightarrowquad &&(-x)times (-y)- xy&=0 \
      Rightarrowquad &&(-x)times (-y)&=xy.
      end{alignedat}
      end{equation}






      proof-verification proof-writing




      proof-verification proof-writing






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      edited Jan 8 at 20:44









      Maria Mazur

      50k1361125




      50k1361125










      asked Jan 8 at 19:31







      user503154





























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          $begingroup$

          Yes it is correct, but perhaps this would be faster. Take any two negative number $x$ and $y$. Then $x=-a$ and $y=-b$ where $a,b$ are positive. Now we have:
          begin{eqnarray} xy &=& (-a)(-b)\ &=& (-1)a(-1)b\
          &=& (-1)(-1)ab \ &=& 1cdot ab \&=& ab>0end{eqnarray}






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            1 Answer
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            active

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            $begingroup$

            Yes it is correct, but perhaps this would be faster. Take any two negative number $x$ and $y$. Then $x=-a$ and $y=-b$ where $a,b$ are positive. Now we have:
            begin{eqnarray} xy &=& (-a)(-b)\ &=& (-1)a(-1)b\
            &=& (-1)(-1)ab \ &=& 1cdot ab \&=& ab>0end{eqnarray}






            share|cite|improve this answer









            $endgroup$


















              2












              $begingroup$

              Yes it is correct, but perhaps this would be faster. Take any two negative number $x$ and $y$. Then $x=-a$ and $y=-b$ where $a,b$ are positive. Now we have:
              begin{eqnarray} xy &=& (-a)(-b)\ &=& (-1)a(-1)b\
              &=& (-1)(-1)ab \ &=& 1cdot ab \&=& ab>0end{eqnarray}






              share|cite|improve this answer









              $endgroup$
















                2












                2








                2





                $begingroup$

                Yes it is correct, but perhaps this would be faster. Take any two negative number $x$ and $y$. Then $x=-a$ and $y=-b$ where $a,b$ are positive. Now we have:
                begin{eqnarray} xy &=& (-a)(-b)\ &=& (-1)a(-1)b\
                &=& (-1)(-1)ab \ &=& 1cdot ab \&=& ab>0end{eqnarray}






                share|cite|improve this answer









                $endgroup$



                Yes it is correct, but perhaps this would be faster. Take any two negative number $x$ and $y$. Then $x=-a$ and $y=-b$ where $a,b$ are positive. Now we have:
                begin{eqnarray} xy &=& (-a)(-b)\ &=& (-1)a(-1)b\
                &=& (-1)(-1)ab \ &=& 1cdot ab \&=& ab>0end{eqnarray}







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered Jan 8 at 19:36









                Maria MazurMaria Mazur

                50k1361125




                50k1361125






























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