Is it possible to prove $a+b'=b+a', c+d'=d+c' implies ac+bd+a'd'+b'c' = ad+bc+a'c'+b'd'$ without using...












1












$begingroup$


Suppose that we are restricted to addition $(+)$ and multiplication $(cdot)$, and are NOT allowed to use subtraction $(-)$ and division $(div)$. Moreover, addition $(+)$ and multiplication $(cdot)$ are defined only on $Bbb N$. For convenience, I write $ab$ instead of $acdot b$.




The reason that I am not allowed to appeal to subtraction $(-)$ is that I am defining multiplication over $Bbb Z$ from ONLY multiplication and addition defined on $Bbb N$. We are allowed to use all properties about addition and multiplication defined on $Bbb N$.




I would like to prove below theorem:




begin{equation}
left.begin{aligned}
a,b,c,d,a',b',c',d' in Bbb N\
a+b'=b+a'\
c+d'=d+c'
end{aligned}right} implies ac+bd+a'd'+b'c' = ad+bc+a'c'+b'd'
end{equation}




While it is easy to come up with the desired equation by using subtraction $(-)$ defined on $Bbb Z$, I have struggled but to no avail in not using subtraction $(-)$ defined on $Bbb Z$.



Please shed me some light. Thank you for your help!



I added my attempt in case subtraction defined on $Bbb Z$ is allowed:




$a+b'=b+a' iff a-b=a'-b'$ and $c+d'=d+c' iff c-d = c'-d'$. Multiply two equalities, we get $(a-b)(c-d) = (a'-b')(c'-d')$ or equivalently $ac-ad-bc+bd = a'c'-a'd'-b'c'+b'd'$ or equivalently $ac+bd+a'd'+b'c' = ad+bc+a'c'+b'd'$.







algebra-precalculus







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$endgroup$








  • 2




    $begingroup$
    Well do you have the cancellation property?
    $endgroup$
    – Kenny Lau
    Dec 31 '18 at 10:11










  • $begingroup$
    Do you have any reasonable argument for not using -?
    $endgroup$
    – Martín Vacas Vignolo
    Dec 31 '18 at 10:12










  • $begingroup$
    @KennyLau, We are allowed to use all properties about addition and multiplication.
    $endgroup$
    – Akira
    Dec 31 '18 at 10:13






  • 3




    $begingroup$
    Then you can convert your proof using subtraction to a proof using cancellation.
    $endgroup$
    – Kenny Lau
    Dec 31 '18 at 10:13










  • $begingroup$
    @MartínVacasVignolo, I am defining multiplication over $Bbb Z$ from multiplication and addition over $Bbb N$.
    $endgroup$
    – Akira
    Dec 31 '18 at 10:15


















1












$begingroup$


Suppose that we are restricted to addition $(+)$ and multiplication $(cdot)$, and are NOT allowed to use subtraction $(-)$ and division $(div)$. Moreover, addition $(+)$ and multiplication $(cdot)$ are defined only on $Bbb N$. For convenience, I write $ab$ instead of $acdot b$.




The reason that I am not allowed to appeal to subtraction $(-)$ is that I am defining multiplication over $Bbb Z$ from ONLY multiplication and addition defined on $Bbb N$. We are allowed to use all properties about addition and multiplication defined on $Bbb N$.




I would like to prove below theorem:




begin{equation}
left.begin{aligned}
a,b,c,d,a',b',c',d' in Bbb N\
a+b'=b+a'\
c+d'=d+c'
end{aligned}right} implies ac+bd+a'd'+b'c' = ad+bc+a'c'+b'd'
end{equation}




While it is easy to come up with the desired equation by using subtraction $(-)$ defined on $Bbb Z$, I have struggled but to no avail in not using subtraction $(-)$ defined on $Bbb Z$.



Please shed me some light. Thank you for your help!



I added my attempt in case subtraction defined on $Bbb Z$ is allowed:




$a+b'=b+a' iff a-b=a'-b'$ and $c+d'=d+c' iff c-d = c'-d'$. Multiply two equalities, we get $(a-b)(c-d) = (a'-b')(c'-d')$ or equivalently $ac-ad-bc+bd = a'c'-a'd'-b'c'+b'd'$ or equivalently $ac+bd+a'd'+b'c' = ad+bc+a'c'+b'd'$.







algebra-precalculus







share|cite|improve this question











$endgroup$








  • 2




    $begingroup$
    Well do you have the cancellation property?
    $endgroup$
    – Kenny Lau
    Dec 31 '18 at 10:11










  • $begingroup$
    Do you have any reasonable argument for not using -?
    $endgroup$
    – Martín Vacas Vignolo
    Dec 31 '18 at 10:12










  • $begingroup$
    @KennyLau, We are allowed to use all properties about addition and multiplication.
    $endgroup$
    – Akira
    Dec 31 '18 at 10:13






  • 3




    $begingroup$
    Then you can convert your proof using subtraction to a proof using cancellation.
    $endgroup$
    – Kenny Lau
    Dec 31 '18 at 10:13










  • $begingroup$
    @MartínVacasVignolo, I am defining multiplication over $Bbb Z$ from multiplication and addition over $Bbb N$.
    $endgroup$
    – Akira
    Dec 31 '18 at 10:15
















1












1








1





$begingroup$


Suppose that we are restricted to addition $(+)$ and multiplication $(cdot)$, and are NOT allowed to use subtraction $(-)$ and division $(div)$. Moreover, addition $(+)$ and multiplication $(cdot)$ are defined only on $Bbb N$. For convenience, I write $ab$ instead of $acdot b$.




The reason that I am not allowed to appeal to subtraction $(-)$ is that I am defining multiplication over $Bbb Z$ from ONLY multiplication and addition defined on $Bbb N$. We are allowed to use all properties about addition and multiplication defined on $Bbb N$.




I would like to prove below theorem:




begin{equation}
left.begin{aligned}
a,b,c,d,a',b',c',d' in Bbb N\
a+b'=b+a'\
c+d'=d+c'
end{aligned}right} implies ac+bd+a'd'+b'c' = ad+bc+a'c'+b'd'
end{equation}




While it is easy to come up with the desired equation by using subtraction $(-)$ defined on $Bbb Z$, I have struggled but to no avail in not using subtraction $(-)$ defined on $Bbb Z$.



Please shed me some light. Thank you for your help!



I added my attempt in case subtraction defined on $Bbb Z$ is allowed:




$a+b'=b+a' iff a-b=a'-b'$ and $c+d'=d+c' iff c-d = c'-d'$. Multiply two equalities, we get $(a-b)(c-d) = (a'-b')(c'-d')$ or equivalently $ac-ad-bc+bd = a'c'-a'd'-b'c'+b'd'$ or equivalently $ac+bd+a'd'+b'c' = ad+bc+a'c'+b'd'$.







algebra-precalculus







share|cite|improve this question











$endgroup$




Suppose that we are restricted to addition $(+)$ and multiplication $(cdot)$, and are NOT allowed to use subtraction $(-)$ and division $(div)$. Moreover, addition $(+)$ and multiplication $(cdot)$ are defined only on $Bbb N$. For convenience, I write $ab$ instead of $acdot b$.




The reason that I am not allowed to appeal to subtraction $(-)$ is that I am defining multiplication over $Bbb Z$ from ONLY multiplication and addition defined on $Bbb N$. We are allowed to use all properties about addition and multiplication defined on $Bbb N$.




I would like to prove below theorem:




begin{equation}
left.begin{aligned}
a,b,c,d,a',b',c',d' in Bbb N\
a+b'=b+a'\
c+d'=d+c'
end{aligned}right} implies ac+bd+a'd'+b'c' = ad+bc+a'c'+b'd'
end{equation}




While it is easy to come up with the desired equation by using subtraction $(-)$ defined on $Bbb Z$, I have struggled but to no avail in not using subtraction $(-)$ defined on $Bbb Z$.



Please shed me some light. Thank you for your help!



I added my attempt in case subtraction defined on $Bbb Z$ is allowed:




$a+b'=b+a' iff a-b=a'-b'$ and $c+d'=d+c' iff c-d = c'-d'$. Multiply two equalities, we get $(a-b)(c-d) = (a'-b')(c'-d')$ or equivalently $ac-ad-bc+bd = a'c'-a'd'-b'c'+b'd'$ or equivalently $ac+bd+a'd'+b'c' = ad+bc+a'c'+b'd'$.







algebra-precalculus




algebra-precalculus






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 31 '18 at 11:45







Akira

















asked Dec 31 '18 at 10:08









AkiraAkira

21415




21415








  • 2




    $begingroup$
    Well do you have the cancellation property?
    $endgroup$
    – Kenny Lau
    Dec 31 '18 at 10:11










  • $begingroup$
    Do you have any reasonable argument for not using -?
    $endgroup$
    – Martín Vacas Vignolo
    Dec 31 '18 at 10:12










  • $begingroup$
    @KennyLau, We are allowed to use all properties about addition and multiplication.
    $endgroup$
    – Akira
    Dec 31 '18 at 10:13






  • 3




    $begingroup$
    Then you can convert your proof using subtraction to a proof using cancellation.
    $endgroup$
    – Kenny Lau
    Dec 31 '18 at 10:13










  • $begingroup$
    @MartínVacasVignolo, I am defining multiplication over $Bbb Z$ from multiplication and addition over $Bbb N$.
    $endgroup$
    – Akira
    Dec 31 '18 at 10:15
















  • 2




    $begingroup$
    Well do you have the cancellation property?
    $endgroup$
    – Kenny Lau
    Dec 31 '18 at 10:11










  • $begingroup$
    Do you have any reasonable argument for not using -?
    $endgroup$
    – Martín Vacas Vignolo
    Dec 31 '18 at 10:12










  • $begingroup$
    @KennyLau, We are allowed to use all properties about addition and multiplication.
    $endgroup$
    – Akira
    Dec 31 '18 at 10:13






  • 3




    $begingroup$
    Then you can convert your proof using subtraction to a proof using cancellation.
    $endgroup$
    – Kenny Lau
    Dec 31 '18 at 10:13










  • $begingroup$
    @MartínVacasVignolo, I am defining multiplication over $Bbb Z$ from multiplication and addition over $Bbb N$.
    $endgroup$
    – Akira
    Dec 31 '18 at 10:15










2




2




$begingroup$
Well do you have the cancellation property?
$endgroup$
– Kenny Lau
Dec 31 '18 at 10:11




$begingroup$
Well do you have the cancellation property?
$endgroup$
– Kenny Lau
Dec 31 '18 at 10:11












$begingroup$
Do you have any reasonable argument for not using -?
$endgroup$
– Martín Vacas Vignolo
Dec 31 '18 at 10:12




$begingroup$
Do you have any reasonable argument for not using -?
$endgroup$
– Martín Vacas Vignolo
Dec 31 '18 at 10:12












$begingroup$
@KennyLau, We are allowed to use all properties about addition and multiplication.
$endgroup$
– Akira
Dec 31 '18 at 10:13




$begingroup$
@KennyLau, We are allowed to use all properties about addition and multiplication.
$endgroup$
– Akira
Dec 31 '18 at 10:13




3




3




$begingroup$
Then you can convert your proof using subtraction to a proof using cancellation.
$endgroup$
– Kenny Lau
Dec 31 '18 at 10:13




$begingroup$
Then you can convert your proof using subtraction to a proof using cancellation.
$endgroup$
– Kenny Lau
Dec 31 '18 at 10:13












$begingroup$
@MartínVacasVignolo, I am defining multiplication over $Bbb Z$ from multiplication and addition over $Bbb N$.
$endgroup$
– Akira
Dec 31 '18 at 10:15






$begingroup$
@MartínVacasVignolo, I am defining multiplication over $Bbb Z$ from multiplication and addition over $Bbb N$.
$endgroup$
– Akira
Dec 31 '18 at 10:15












1 Answer
1






active

oldest

votes


















1












$begingroup$

I have figured out the proof and posted it here:




WLOG, we assume $a ge b$ and $c ge d$. Then $b' le a'$ and $d' le c'$.



It follows that there exist unique $overline b, overline {b'}, overline d, overline {d'}$ such that $a = b+overline b, a' = b' + overline {b'}, c = d + overline d, c' = d' + overline {d'}$.



Then $a+b'=b+a' iff (b+overline b) + b' = b + (b' + overline {b'}) iff overline b = overline {b'}$ and $c+d'=d+c' iff (d + overline d) + d' = d + (d' + overline {d'}) iff overline d = overline {d'}$.



Then $a = b+overline b, a' = b' + overline b, c = d + overline d, c' = d' + overline d$.



$LHS=ac+bd+a'd'+b'c' = (b+overline b) (d + overline d) +bd +(b' + overline b)d' + b'(d' + overline d) =bd+boverline d +overline b d+overline b overline d+bd+b'd'+overline bd'+b'd'+b'overline d=2(bd+b'd')+boverline d +overline b d+overline b overline d+overline bd'+b'overline d.$



$RHS=ad+bc+a'c'+b'd'=(b+overline b)d+b(d + overline d)+(b' + overline b)(d' + overline d)+b'd'=bd+overline b d+bd+boverline d +b'd'+b'overline d+overline b d'+overline boverline d+b'd'=2(bd+b'd')+boverline d +overline b d+overline b overline d+overline bd'+b'overline d.$



Thus $LHS=RHS$.






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    $begingroup$

    I have figured out the proof and posted it here:




    WLOG, we assume $a ge b$ and $c ge d$. Then $b' le a'$ and $d' le c'$.



    It follows that there exist unique $overline b, overline {b'}, overline d, overline {d'}$ such that $a = b+overline b, a' = b' + overline {b'}, c = d + overline d, c' = d' + overline {d'}$.



    Then $a+b'=b+a' iff (b+overline b) + b' = b + (b' + overline {b'}) iff overline b = overline {b'}$ and $c+d'=d+c' iff (d + overline d) + d' = d + (d' + overline {d'}) iff overline d = overline {d'}$.



    Then $a = b+overline b, a' = b' + overline b, c = d + overline d, c' = d' + overline d$.



    $LHS=ac+bd+a'd'+b'c' = (b+overline b) (d + overline d) +bd +(b' + overline b)d' + b'(d' + overline d) =bd+boverline d +overline b d+overline b overline d+bd+b'd'+overline bd'+b'd'+b'overline d=2(bd+b'd')+boverline d +overline b d+overline b overline d+overline bd'+b'overline d.$



    $RHS=ad+bc+a'c'+b'd'=(b+overline b)d+b(d + overline d)+(b' + overline b)(d' + overline d)+b'd'=bd+overline b d+bd+boverline d +b'd'+b'overline d+overline b d'+overline boverline d+b'd'=2(bd+b'd')+boverline d +overline b d+overline b overline d+overline bd'+b'overline d.$



    Thus $LHS=RHS$.






    share|cite|improve this answer









    $endgroup$


















      1












      $begingroup$

      I have figured out the proof and posted it here:




      WLOG, we assume $a ge b$ and $c ge d$. Then $b' le a'$ and $d' le c'$.



      It follows that there exist unique $overline b, overline {b'}, overline d, overline {d'}$ such that $a = b+overline b, a' = b' + overline {b'}, c = d + overline d, c' = d' + overline {d'}$.



      Then $a+b'=b+a' iff (b+overline b) + b' = b + (b' + overline {b'}) iff overline b = overline {b'}$ and $c+d'=d+c' iff (d + overline d) + d' = d + (d' + overline {d'}) iff overline d = overline {d'}$.



      Then $a = b+overline b, a' = b' + overline b, c = d + overline d, c' = d' + overline d$.



      $LHS=ac+bd+a'd'+b'c' = (b+overline b) (d + overline d) +bd +(b' + overline b)d' + b'(d' + overline d) =bd+boverline d +overline b d+overline b overline d+bd+b'd'+overline bd'+b'd'+b'overline d=2(bd+b'd')+boverline d +overline b d+overline b overline d+overline bd'+b'overline d.$



      $RHS=ad+bc+a'c'+b'd'=(b+overline b)d+b(d + overline d)+(b' + overline b)(d' + overline d)+b'd'=bd+overline b d+bd+boverline d +b'd'+b'overline d+overline b d'+overline boverline d+b'd'=2(bd+b'd')+boverline d +overline b d+overline b overline d+overline bd'+b'overline d.$



      Thus $LHS=RHS$.






      share|cite|improve this answer









      $endgroup$
















        1












        1








        1





        $begingroup$

        I have figured out the proof and posted it here:




        WLOG, we assume $a ge b$ and $c ge d$. Then $b' le a'$ and $d' le c'$.



        It follows that there exist unique $overline b, overline {b'}, overline d, overline {d'}$ such that $a = b+overline b, a' = b' + overline {b'}, c = d + overline d, c' = d' + overline {d'}$.



        Then $a+b'=b+a' iff (b+overline b) + b' = b + (b' + overline {b'}) iff overline b = overline {b'}$ and $c+d'=d+c' iff (d + overline d) + d' = d + (d' + overline {d'}) iff overline d = overline {d'}$.



        Then $a = b+overline b, a' = b' + overline b, c = d + overline d, c' = d' + overline d$.



        $LHS=ac+bd+a'd'+b'c' = (b+overline b) (d + overline d) +bd +(b' + overline b)d' + b'(d' + overline d) =bd+boverline d +overline b d+overline b overline d+bd+b'd'+overline bd'+b'd'+b'overline d=2(bd+b'd')+boverline d +overline b d+overline b overline d+overline bd'+b'overline d.$



        $RHS=ad+bc+a'c'+b'd'=(b+overline b)d+b(d + overline d)+(b' + overline b)(d' + overline d)+b'd'=bd+overline b d+bd+boverline d +b'd'+b'overline d+overline b d'+overline boverline d+b'd'=2(bd+b'd')+boverline d +overline b d+overline b overline d+overline bd'+b'overline d.$



        Thus $LHS=RHS$.






        share|cite|improve this answer









        $endgroup$



        I have figured out the proof and posted it here:




        WLOG, we assume $a ge b$ and $c ge d$. Then $b' le a'$ and $d' le c'$.



        It follows that there exist unique $overline b, overline {b'}, overline d, overline {d'}$ such that $a = b+overline b, a' = b' + overline {b'}, c = d + overline d, c' = d' + overline {d'}$.



        Then $a+b'=b+a' iff (b+overline b) + b' = b + (b' + overline {b'}) iff overline b = overline {b'}$ and $c+d'=d+c' iff (d + overline d) + d' = d + (d' + overline {d'}) iff overline d = overline {d'}$.



        Then $a = b+overline b, a' = b' + overline b, c = d + overline d, c' = d' + overline d$.



        $LHS=ac+bd+a'd'+b'c' = (b+overline b) (d + overline d) +bd +(b' + overline b)d' + b'(d' + overline d) =bd+boverline d +overline b d+overline b overline d+bd+b'd'+overline bd'+b'd'+b'overline d=2(bd+b'd')+boverline d +overline b d+overline b overline d+overline bd'+b'overline d.$



        $RHS=ad+bc+a'c'+b'd'=(b+overline b)d+b(d + overline d)+(b' + overline b)(d' + overline d)+b'd'=bd+overline b d+bd+boverline d +b'd'+b'overline d+overline b d'+overline boverline d+b'd'=2(bd+b'd')+boverline d +overline b d+overline b overline d+overline bd'+b'overline d.$



        Thus $LHS=RHS$.







        share|cite|improve this answer












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        share|cite|improve this answer










        answered Jan 1 at 2:58









        AkiraAkira

        21415




        21415






























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