Proof clarification - If $ab = 0$ then $a = 0$ or $b =0$












1












$begingroup$


I came across a proof for the following theorem in Apostol Calculus 1. My question is regarding (1) in the proof, why is this part necessary? I don't see why you can't begin with (2)



Theorem 1.11



If $ab = 0$ then $a = 0$ or $b=0$



Proof



Let $a, b in mathbb{R}$ with $ab =0$



Then, if $a neq 0$ we know there exists $a^{-1} in mathbb{R}$ such that $a * a^{-1} = 1$



Thus,
$$begin{align}
ab = 0 &implies a^{-1}(ab) = a^{-1} cdot 0 = 0tag{1}label{1} \
end{align}$$
But,



$$begin{align}
a^{-1}(ab) = 0 &implies (a^{-1}a)b = 0tag{2}\
&implies 1cdot b = 0tag{3}\
&implies b = 0 tag{4}
end{align}$$






real-analysis proof-verification proof-writing real-numbers







share|cite|improve this question











$endgroup$












  • $begingroup$
    Line (2) uses the associativity of multiplication of the real numbers, that is, for any real numbers $a, b, c$ then $a(bc) = (ab)c$. The reason you can't start with (2) is because when you multiply $a^{-1}$ to both sides of $ab = 0$ then you are implicitly multiplying $a^{-1}$ to $ab$ and so you must use associativity at some point.
    $endgroup$
    – symchdmath
    Jan 12 at 11:49












  • $begingroup$
    ah ok, is it because line two doesn't show $a^{-1}(ab) = a^{-1} cdot 0 implies a^{-1}(ab) = 0$ (using $0 cdot a = a cdot 0 = 0$ which was proved as theorem 1.6 in the chapter)
    $endgroup$
    – Jake Kirsch
    Jan 12 at 11:58










  • $begingroup$
    and then associativity to get $(a^{-1} a)b = 0$
    $endgroup$
    – Jake Kirsch
    Jan 12 at 11:59










  • $begingroup$
    Yep, you can see it as combining it with transitivity of equality in one line
    $endgroup$
    – symchdmath
    Jan 12 at 12:19
















1












$begingroup$


I came across a proof for the following theorem in Apostol Calculus 1. My question is regarding (1) in the proof, why is this part necessary? I don't see why you can't begin with (2)



Theorem 1.11



If $ab = 0$ then $a = 0$ or $b=0$



Proof



Let $a, b in mathbb{R}$ with $ab =0$



Then, if $a neq 0$ we know there exists $a^{-1} in mathbb{R}$ such that $a * a^{-1} = 1$



Thus,
$$begin{align}
ab = 0 &implies a^{-1}(ab) = a^{-1} cdot 0 = 0tag{1}label{1} \
end{align}$$
But,



$$begin{align}
a^{-1}(ab) = 0 &implies (a^{-1}a)b = 0tag{2}\
&implies 1cdot b = 0tag{3}\
&implies b = 0 tag{4}
end{align}$$






real-analysis proof-verification proof-writing real-numbers







share|cite|improve this question











$endgroup$












  • $begingroup$
    Line (2) uses the associativity of multiplication of the real numbers, that is, for any real numbers $a, b, c$ then $a(bc) = (ab)c$. The reason you can't start with (2) is because when you multiply $a^{-1}$ to both sides of $ab = 0$ then you are implicitly multiplying $a^{-1}$ to $ab$ and so you must use associativity at some point.
    $endgroup$
    – symchdmath
    Jan 12 at 11:49












  • $begingroup$
    ah ok, is it because line two doesn't show $a^{-1}(ab) = a^{-1} cdot 0 implies a^{-1}(ab) = 0$ (using $0 cdot a = a cdot 0 = 0$ which was proved as theorem 1.6 in the chapter)
    $endgroup$
    – Jake Kirsch
    Jan 12 at 11:58










  • $begingroup$
    and then associativity to get $(a^{-1} a)b = 0$
    $endgroup$
    – Jake Kirsch
    Jan 12 at 11:59










  • $begingroup$
    Yep, you can see it as combining it with transitivity of equality in one line
    $endgroup$
    – symchdmath
    Jan 12 at 12:19














1












1








1





$begingroup$


I came across a proof for the following theorem in Apostol Calculus 1. My question is regarding (1) in the proof, why is this part necessary? I don't see why you can't begin with (2)



Theorem 1.11



If $ab = 0$ then $a = 0$ or $b=0$



Proof



Let $a, b in mathbb{R}$ with $ab =0$



Then, if $a neq 0$ we know there exists $a^{-1} in mathbb{R}$ such that $a * a^{-1} = 1$



Thus,
$$begin{align}
ab = 0 &implies a^{-1}(ab) = a^{-1} cdot 0 = 0tag{1}label{1} \
end{align}$$
But,



$$begin{align}
a^{-1}(ab) = 0 &implies (a^{-1}a)b = 0tag{2}\
&implies 1cdot b = 0tag{3}\
&implies b = 0 tag{4}
end{align}$$






real-analysis proof-verification proof-writing real-numbers







share|cite|improve this question











$endgroup$




I came across a proof for the following theorem in Apostol Calculus 1. My question is regarding (1) in the proof, why is this part necessary? I don't see why you can't begin with (2)



Theorem 1.11



If $ab = 0$ then $a = 0$ or $b=0$



Proof



Let $a, b in mathbb{R}$ with $ab =0$



Then, if $a neq 0$ we know there exists $a^{-1} in mathbb{R}$ such that $a * a^{-1} = 1$



Thus,
$$begin{align}
ab = 0 &implies a^{-1}(ab) = a^{-1} cdot 0 = 0tag{1}label{1} \
end{align}$$
But,



$$begin{align}
a^{-1}(ab) = 0 &implies (a^{-1}a)b = 0tag{2}\
&implies 1cdot b = 0tag{3}\
&implies b = 0 tag{4}
end{align}$$






real-analysis proof-verification proof-writing real-numbers




real-analysis proof-verification proof-writing real-numbers






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Jan 12 at 11:54







Jake Kirsch

















asked Jan 12 at 11:45









Jake KirschJake Kirsch

687




687












  • $begingroup$
    Line (2) uses the associativity of multiplication of the real numbers, that is, for any real numbers $a, b, c$ then $a(bc) = (ab)c$. The reason you can't start with (2) is because when you multiply $a^{-1}$ to both sides of $ab = 0$ then you are implicitly multiplying $a^{-1}$ to $ab$ and so you must use associativity at some point.
    $endgroup$
    – symchdmath
    Jan 12 at 11:49












  • $begingroup$
    ah ok, is it because line two doesn't show $a^{-1}(ab) = a^{-1} cdot 0 implies a^{-1}(ab) = 0$ (using $0 cdot a = a cdot 0 = 0$ which was proved as theorem 1.6 in the chapter)
    $endgroup$
    – Jake Kirsch
    Jan 12 at 11:58










  • $begingroup$
    and then associativity to get $(a^{-1} a)b = 0$
    $endgroup$
    – Jake Kirsch
    Jan 12 at 11:59










  • $begingroup$
    Yep, you can see it as combining it with transitivity of equality in one line
    $endgroup$
    – symchdmath
    Jan 12 at 12:19


















  • $begingroup$
    Line (2) uses the associativity of multiplication of the real numbers, that is, for any real numbers $a, b, c$ then $a(bc) = (ab)c$. The reason you can't start with (2) is because when you multiply $a^{-1}$ to both sides of $ab = 0$ then you are implicitly multiplying $a^{-1}$ to $ab$ and so you must use associativity at some point.
    $endgroup$
    – symchdmath
    Jan 12 at 11:49












  • $begingroup$
    ah ok, is it because line two doesn't show $a^{-1}(ab) = a^{-1} cdot 0 implies a^{-1}(ab) = 0$ (using $0 cdot a = a cdot 0 = 0$ which was proved as theorem 1.6 in the chapter)
    $endgroup$
    – Jake Kirsch
    Jan 12 at 11:58










  • $begingroup$
    and then associativity to get $(a^{-1} a)b = 0$
    $endgroup$
    – Jake Kirsch
    Jan 12 at 11:59










  • $begingroup$
    Yep, you can see it as combining it with transitivity of equality in one line
    $endgroup$
    – symchdmath
    Jan 12 at 12:19
















$begingroup$
Line (2) uses the associativity of multiplication of the real numbers, that is, for any real numbers $a, b, c$ then $a(bc) = (ab)c$. The reason you can't start with (2) is because when you multiply $a^{-1}$ to both sides of $ab = 0$ then you are implicitly multiplying $a^{-1}$ to $ab$ and so you must use associativity at some point.
$endgroup$
– symchdmath
Jan 12 at 11:49






$begingroup$
Line (2) uses the associativity of multiplication of the real numbers, that is, for any real numbers $a, b, c$ then $a(bc) = (ab)c$. The reason you can't start with (2) is because when you multiply $a^{-1}$ to both sides of $ab = 0$ then you are implicitly multiplying $a^{-1}$ to $ab$ and so you must use associativity at some point.
$endgroup$
– symchdmath
Jan 12 at 11:49














$begingroup$
ah ok, is it because line two doesn't show $a^{-1}(ab) = a^{-1} cdot 0 implies a^{-1}(ab) = 0$ (using $0 cdot a = a cdot 0 = 0$ which was proved as theorem 1.6 in the chapter)
$endgroup$
– Jake Kirsch
Jan 12 at 11:58




$begingroup$
ah ok, is it because line two doesn't show $a^{-1}(ab) = a^{-1} cdot 0 implies a^{-1}(ab) = 0$ (using $0 cdot a = a cdot 0 = 0$ which was proved as theorem 1.6 in the chapter)
$endgroup$
– Jake Kirsch
Jan 12 at 11:58












$begingroup$
and then associativity to get $(a^{-1} a)b = 0$
$endgroup$
– Jake Kirsch
Jan 12 at 11:59




$begingroup$
and then associativity to get $(a^{-1} a)b = 0$
$endgroup$
– Jake Kirsch
Jan 12 at 11:59












$begingroup$
Yep, you can see it as combining it with transitivity of equality in one line
$endgroup$
– symchdmath
Jan 12 at 12:19




$begingroup$
Yep, you can see it as combining it with transitivity of equality in one line
$endgroup$
– symchdmath
Jan 12 at 12:19










1 Answer
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$begingroup$

Looks a bit weird. If $ane 0$, then $a$ is invertible and so
$$b = 1b = (a^{-1}a)b = a^{-1}(ab) = a^{-1}0 = 0.$$
In the last step, $b0=0$, I used the fact that $0$ is absorbing.



In the general case, if you have a ring $R$ and a unit $ain R$, then the above proof shows that units are not zero divisors.






share|cite|improve this answer











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    1 Answer
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    1 Answer
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    $begingroup$

    Looks a bit weird. If $ane 0$, then $a$ is invertible and so
    $$b = 1b = (a^{-1}a)b = a^{-1}(ab) = a^{-1}0 = 0.$$
    In the last step, $b0=0$, I used the fact that $0$ is absorbing.



    In the general case, if you have a ring $R$ and a unit $ain R$, then the above proof shows that units are not zero divisors.






    share|cite|improve this answer











    $endgroup$


















      0












      $begingroup$

      Looks a bit weird. If $ane 0$, then $a$ is invertible and so
      $$b = 1b = (a^{-1}a)b = a^{-1}(ab) = a^{-1}0 = 0.$$
      In the last step, $b0=0$, I used the fact that $0$ is absorbing.



      In the general case, if you have a ring $R$ and a unit $ain R$, then the above proof shows that units are not zero divisors.






      share|cite|improve this answer











      $endgroup$
















        0












        0








        0





        $begingroup$

        Looks a bit weird. If $ane 0$, then $a$ is invertible and so
        $$b = 1b = (a^{-1}a)b = a^{-1}(ab) = a^{-1}0 = 0.$$
        In the last step, $b0=0$, I used the fact that $0$ is absorbing.



        In the general case, if you have a ring $R$ and a unit $ain R$, then the above proof shows that units are not zero divisors.






        share|cite|improve this answer











        $endgroup$



        Looks a bit weird. If $ane 0$, then $a$ is invertible and so
        $$b = 1b = (a^{-1}a)b = a^{-1}(ab) = a^{-1}0 = 0.$$
        In the last step, $b0=0$, I used the fact that $0$ is absorbing.



        In the general case, if you have a ring $R$ and a unit $ain R$, then the above proof shows that units are not zero divisors.







        share|cite|improve this answer














        share|cite|improve this answer



        share|cite|improve this answer








        edited Jan 12 at 12:24

























        answered Jan 12 at 11:49









        WuestenfuxWuestenfux

        4,9921513




        4,9921513






























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