Mixing
Prove that $(-a).b=-(a.b)$ , $(-a)^{-1}=-(a^{-1})$ and $(a.b)^{-1}=a^{-1}.b^{-1}$.
$begingroup$
Prove that
1)$forall a, b in mathbb{R}$, $(-a).b=-(a.b)$.
$(-a).b=(-a).b+0$
$=(-a).b+[a.b+(-(a.b))]$
$=[(-a).b+a.b]+(-(a.b))$
$=[((-a)+a)b]+(-(a.b))$
$=0.b+(-(a.b))$
$=0+(-(a.b))$
$=-(a.b)$
2) $forall ain mathbb{R}setminus lbrace 0rbrace, (-a)^{-1}=-(a^{-1})$.
$(-a)^{-1}=(-a)^{-1}.1$
$=(-a)^{-1}.[(-a).(-a^{-1})]$
$=[(-a)^{-1}.(-a)].(-a^{-1})$
$=1.(-a^{-1})$
$=-a^{-1}$
3) If $aneq 0$ and $b neq 0$, then $ab neq 0$ and $(a.b)^{-1}=a^{-1}.b^{-1}$.
$(a.b)^{-1}=(a.b)^{-1}.1$
$=(a.b)^{-1}.[a.b.b^{-1}.a^{-1}]$
$=(a.b)^{-1}.[(a.b).(b^{-1}.a^{-1})]$
$=[(a.b)^{-1}.(a.b)].(b^{-1}.a^{-1})$
$=1.(b^{-1}.a^{-1})$
$=b^{-1}.a^{-1}$
$=a^{-1}.b^{-1}$
Is that true please? Thanks.
I used these axioms:
real-analysis proof-verification
asked Jan 26 at 22:30
DimaDima
837516
$endgroup$
|
show 2 more comments
$begingroup$
Prove that
1)$forall a, b in mathbb{R}$, $(-a).b=-(a.b)$.
$(-a).b=(-a).b+0$
$=(-a).b+[a.b+(-(a.b))]$
$=[(-a).b+a.b]+(-(a.b))$
$=[((-a)+a)b]+(-(a.b))$
$=0.b+(-(a.b))$
$=0+(-(a.b))$
$=-(a.b)$
2) $forall ain mathbb{R}setminus lbrace 0rbrace, (-a)^{-1}=-(a^{-1})$.
$(-a)^{-1}=(-a)^{-1}.1$
$=(-a)^{-1}.[(-a).(-a^{-1})]$
$=[(-a)^{-1}.(-a)].(-a^{-1})$
$=1.(-a^{-1})$
$=-a^{-1}$
3) If $aneq 0$ and $b neq 0$, then $ab neq 0$ and $(a.b)^{-1}=a^{-1}.b^{-1}$.
$(a.b)^{-1}=(a.b)^{-1}.1$
$=(a.b)^{-1}.[a.b.b^{-1}.a^{-1}]$
$=(a.b)^{-1}.[(a.b).(b^{-1}.a^{-1})]$
$=[(a.b)^{-1}.(a.b)].(b^{-1}.a^{-1})$
$=1.(b^{-1}.a^{-1})$
$=b^{-1}.a^{-1}$
$=a^{-1}.b^{-1}$
Is that true please? Thanks.
I used these axioms:
real-analysis proof-verification
asked Jan 26 at 22:30
DimaDima
837516
$endgroup$
$begingroup$
For 1) you first have to prove that $0.b = 0$ which usually is not a given.
$endgroup$
– fleablood
Jan 26 at 22:36
$begingroup$
In 2) you have no reason to assume $(-a)(-a^{-1}) = 1$. Unless you are assuming $(-a)(-b) = ab$ and so $(-a)(-a^{-1}) = a.a^{-1} = 1$. If so you should point that out, it wasn't clear. ANd to you do need to prove that $(-a)(-b) = ab$. It's not a given.
$endgroup$
– fleablood
Jan 26 at 22:39
$begingroup$
3) Is $.$ commutative? Was that an axiom? it isn't true if $.$ isn't commutative. So perhaps before either of had even gotten started I should have asked? What exactly are you doing? Is $.$ supposed to be multiplication on the reals? The rationals? An arbitrary field? What class is this for? What are your axioms? Are you just assuming the field axioms, etc? Or is this high school algebra? Or elementary school arithmetic? Context does matter.
$endgroup$
– fleablood
Jan 26 at 22:43
$begingroup$
Lest I seem too critical. Yes, your proofs and reasoning are perfectly correct and the way to go. But you must prove $0.b = 0$ for all $b$ and that $(-a)(-b) = ab$.
$endgroup$
– fleablood
Jan 26 at 22:46
$begingroup$
Thank you so much prof, I used the 9 axioms [in the question]. For $0.b=b$ my prof was prof if, I will prove that $(-a)(-b)=ab$,
$endgroup$
– Dima
Jan 26 at 22:50
|
show 2 more comments
$begingroup$
Prove that
1)$forall a, b in mathbb{R}$, $(-a).b=-(a.b)$.
$(-a).b=(-a).b+0$
$=(-a).b+[a.b+(-(a.b))]$
$=[(-a).b+a.b]+(-(a.b))$
$=[((-a)+a)b]+(-(a.b))$
$=0.b+(-(a.b))$
$=0+(-(a.b))$
$=-(a.b)$
2) $forall ain mathbb{R}setminus lbrace 0rbrace, (-a)^{-1}=-(a^{-1})$.
$(-a)^{-1}=(-a)^{-1}.1$
$=(-a)^{-1}.[(-a).(-a^{-1})]$
$=[(-a)^{-1}.(-a)].(-a^{-1})$
$=1.(-a^{-1})$
$=-a^{-1}$
3) If $aneq 0$ and $b neq 0$, then $ab neq 0$ and $(a.b)^{-1}=a^{-1}.b^{-1}$.
$(a.b)^{-1}=(a.b)^{-1}.1$
$=(a.b)^{-1}.[a.b.b^{-1}.a^{-1}]$
$=(a.b)^{-1}.[(a.b).(b^{-1}.a^{-1})]$
$=[(a.b)^{-1}.(a.b)].(b^{-1}.a^{-1})$
$=1.(b^{-1}.a^{-1})$
$=b^{-1}.a^{-1}$
$=a^{-1}.b^{-1}$
Is that true please? Thanks.
I used these axioms:
real-analysis proof-verification
asked Jan 26 at 22:30
DimaDima
837516
$endgroup$
Prove that
1)$forall a, b in mathbb{R}$, $(-a).b=-(a.b)$.
$(-a).b=(-a).b+0$
$=(-a).b+[a.b+(-(a.b))]$
$=[(-a).b+a.b]+(-(a.b))$
$=[((-a)+a)b]+(-(a.b))$
$=0.b+(-(a.b))$
$=0+(-(a.b))$
$=-(a.b)$
2) $forall ain mathbb{R}setminus lbrace 0rbrace, (-a)^{-1}=-(a^{-1})$.
$(-a)^{-1}=(-a)^{-1}.1$
$=(-a)^{-1}.[(-a).(-a^{-1})]$
$=[(-a)^{-1}.(-a)].(-a^{-1})$
$=1.(-a^{-1})$
$=-a^{-1}$
3) If $aneq 0$ and $b neq 0$, then $ab neq 0$ and $(a.b)^{-1}=a^{-1}.b^{-1}$.
$(a.b)^{-1}=(a.b)^{-1}.1$
$=(a.b)^{-1}.[a.b.b^{-1}.a^{-1}]$
$=(a.b)^{-1}.[(a.b).(b^{-1}.a^{-1})]$
$=[(a.b)^{-1}.(a.b)].(b^{-1}.a^{-1})$
$=1.(b^{-1}.a^{-1})$
$=b^{-1}.a^{-1}$
$=a^{-1}.b^{-1}$
Is that true please? Thanks.
I used these axioms:
real-analysis proof-verification
real-analysis proof-verification
asked Jan 26 at 22:30
DimaDima
837516
asked Jan 26 at 22:30
DimaDima
837516
edited Jan 26 at 22:48
Dima
asked Jan 26 at 22:30
DimaDima
837516
asked Jan 26 at 22:30
DimaDima
837516
asked Jan 26 at 22:30
DimaDima
837516
837516
$begingroup$
For 1) you first have to prove that $0.b = 0$ which usually is not a given.
$endgroup$
– fleablood
Jan 26 at 22:36
$begingroup$
In 2) you have no reason to assume $(-a)(-a^{-1}) = 1$. Unless you are assuming $(-a)(-b) = ab$ and so $(-a)(-a^{-1}) = a.a^{-1} = 1$. If so you should point that out, it wasn't clear. ANd to you do need to prove that $(-a)(-b) = ab$. It's not a given.
$endgroup$
– fleablood
Jan 26 at 22:39
$begingroup$
3) Is $.$ commutative? Was that an axiom? it isn't true if $.$ isn't commutative. So perhaps before either of had even gotten started I should have asked? What exactly are you doing? Is $.$ supposed to be multiplication on the reals? The rationals? An arbitrary field? What class is this for? What are your axioms? Are you just assuming the field axioms, etc? Or is this high school algebra? Or elementary school arithmetic? Context does matter.
$endgroup$
– fleablood
Jan 26 at 22:43
$begingroup$
Lest I seem too critical. Yes, your proofs and reasoning are perfectly correct and the way to go. But you must prove $0.b = 0$ for all $b$ and that $(-a)(-b) = ab$.
$endgroup$
– fleablood
Jan 26 at 22:46
$begingroup$
Thank you so much prof, I used the 9 axioms [in the question]. For $0.b=b$ my prof was prof if, I will prove that $(-a)(-b)=ab$,
$endgroup$
– Dima
Jan 26 at 22:50
|
show 2 more comments
$begingroup$
For 1) you first have to prove that $0.b = 0$ which usually is not a given.
$endgroup$
– fleablood
Jan 26 at 22:36
$begingroup$
In 2) you have no reason to assume $(-a)(-a^{-1}) = 1$. Unless you are assuming $(-a)(-b) = ab$ and so $(-a)(-a^{-1}) = a.a^{-1} = 1$. If so you should point that out, it wasn't clear. ANd to you do need to prove that $(-a)(-b) = ab$. It's not a given.
$endgroup$
– fleablood
Jan 26 at 22:39
$begingroup$
3) Is $.$ commutative? Was that an axiom? it isn't true if $.$ isn't commutative. So perhaps before either of had even gotten started I should have asked? What exactly are you doing? Is $.$ supposed to be multiplication on the reals? The rationals? An arbitrary field? What class is this for? What are your axioms? Are you just assuming the field axioms, etc? Or is this high school algebra? Or elementary school arithmetic? Context does matter.
$endgroup$
– fleablood
Jan 26 at 22:43
$begingroup$
Lest I seem too critical. Yes, your proofs and reasoning are perfectly correct and the way to go. But you must prove $0.b = 0$ for all $b$ and that $(-a)(-b) = ab$.
$endgroup$
– fleablood
Jan 26 at 22:46
$begingroup$
Thank you so much prof, I used the 9 axioms [in the question]. For $0.b=b$ my prof was prof if, I will prove that $(-a)(-b)=ab$,
$endgroup$
– Dima
Jan 26 at 22:50
$begingroup$
For 1) you first have to prove that $0.b = 0$ which usually is not a given.
$endgroup$
– fleablood
Jan 26 at 22:36
$begingroup$
For 1) you first have to prove that $0.b = 0$ which usually is not a given.
$endgroup$
– fleablood
Jan 26 at 22:36
$begingroup$
In 2) you have no reason to assume $(-a)(-a^{-1}) = 1$. Unless you are assuming $(-a)(-b) = ab$ and so $(-a)(-a^{-1}) = a.a^{-1} = 1$. If so you should point that out, it wasn't clear. ANd to you do need to prove that $(-a)(-b) = ab$. It's not a given.
$endgroup$
– fleablood
Jan 26 at 22:39
$begingroup$
In 2) you have no reason to assume $(-a)(-a^{-1}) = 1$. Unless you are assuming $(-a)(-b) = ab$ and so $(-a)(-a^{-1}) = a.a^{-1} = 1$. If so you should point that out, it wasn't clear. ANd to you do need to prove that $(-a)(-b) = ab$. It's not a given.
$endgroup$
– fleablood
Jan 26 at 22:39
$begingroup$
3) Is $.$ commutative? Was that an axiom? it isn't true if $.$ isn't commutative. So perhaps before either of had even gotten started I should have asked? What exactly are you doing? Is $.$ supposed to be multiplication on the reals? The rationals? An arbitrary field? What class is this for? What are your axioms? Are you just assuming the field axioms, etc? Or is this high school algebra? Or elementary school arithmetic? Context does matter.
$endgroup$
– fleablood
Jan 26 at 22:43
$begingroup$
3) Is $.$ commutative? Was that an axiom? it isn't true if $.$ isn't commutative. So perhaps before either of had even gotten started I should have asked? What exactly are you doing? Is $.$ supposed to be multiplication on the reals? The rationals? An arbitrary field? What class is this for? What are your axioms? Are you just assuming the field axioms, etc? Or is this high school algebra? Or elementary school arithmetic? Context does matter.
$endgroup$
– fleablood
Jan 26 at 22:43
$begingroup$
Lest I seem too critical. Yes, your proofs and reasoning are perfectly correct and the way to go. But you must prove $0.b = 0$ for all $b$ and that $(-a)(-b) = ab$.
$endgroup$
– fleablood
Jan 26 at 22:46
$begingroup$
Lest I seem too critical. Yes, your proofs and reasoning are perfectly correct and the way to go. But you must prove $0.b = 0$ for all $b$ and that $(-a)(-b) = ab$.
$endgroup$
– fleablood
Jan 26 at 22:46
$begingroup$
Thank you so much prof, I used the 9 axioms [in the question]. For $0.b=b$ my prof was prof if, I will prove that $(-a)(-b)=ab$,
$endgroup$
– Dima
Jan 26 at 22:50
$begingroup$
Thank you so much prof, I used the 9 axioms [in the question]. For $0.b=b$ my prof was prof if, I will prove that $(-a)(-b)=ab$,
$endgroup$
– Dima
Jan 26 at 22:50
|
show 2 more comments
2 Answers
2
active
oldest
votes
$begingroup$
Your proofs are mostly good.
But before you can prove 1) you must prove:
$0cdot b = 0$ for all $b$.
Pf: $0cdot b = (0 + 0)cdot b$ (as $0 = 0 + 0$)$
$= 0cdot b + 0cdot b$ (via distribution)
So $0 = 0cdot b + (-0cdot b) $ (as every number has an additive inverse)
$= (0cdot b + 0cdot b) + (-0cdot b) $ (by substitution)
$= 0cdot b + (0cdot b + (-0cdot b))$ (associativity)
$= 0cdot b + 0$ (definition of inverse)
$= 0cdot b$ (definition of 0)$
In your prove of 2) you substituted $1$ with $(-a)(-a^{-1})$ which I don't think is justified without proof.
But you have proven in 1) that $(-a)(-a^{-1}) = -(a(-a^{-1}))$ and by commutativity $-(a(-a^{-1})) = -((-a^{-1})a)$ and by 1 again $= -(-(a^{-1}a)) = -(-1)$.
But we still need to prove that $-(-a) = a$ which can be done as
$-(-a) + (-a) =0$
$-(-a) + (-a) + a = 0 + a$
$-(-a) + 0 = a$
$-(-a) = a$.
Your 3) is just fine
answered Jan 26 at 23:10
fleabloodfleablood
73k22789
$endgroup$
add a comment |
$begingroup$
Why not: $(-a)b + ab = (-a + a)b= 0cdot b = 0$ (assuming you know the latter, which si not an axiom but a consequence of them, hopefully shown before), so that $ab$ is the additive inverse of $(-a)b$ and as inverses are unique in any group we have $(-a)b = -(ab)$.
For the second $acdot a^{-1}=1$ by definition. Assuming you have shown $(-x)cdot(-y)=xy$ we write $(-a) cdot -(a^{-1})=1$ and so $-(a^{-1})$ is the multiplicative inverse of $-a$ so by unicity again $(-a)^{-1} = -(a^{-1})$.
For 3, $(ab)(a^{-1}b^{-1}) =1 $ by applying commutativity and the definition of inverse twice. So again by unicity of multiplicative inverses: $(ab)^{-1} = a^{-1} cdot b^{-1}$.
So as a basic algebra lemma for all of this: if $(G, ast,^{-1}, e)$ is a group, inverses are unique: if $a ast b=e$ then $b=a^{-1}$, where $a^{-1}$ is promised by the group axioms. In a field $F$, $(F, + ,-,0)$ forms a group and so does $(Fsetminus {0},cdot,^{-1}, 1)$, so we can apply this.
answered Jan 26 at 22:49
Henno BrandsmaHenno Brandsma
113k348123
$endgroup$
$begingroup$
"Why not: (−a)b+ab=(−a+a)b=0⋅b=0" Well..... if one is REALLY picky one can point out that we know $-(ab)$ exists but not that it is unique. So $-(ab) + ab = 0$ and $(-a)b + ab = 0$ doesn't actually mean that $-(ab) = (-a)b$ until you con prove it. But that's pretty picky.....
$endgroup$
– fleablood
Jan 26 at 23:01
$begingroup$
@fleablood as remarked, in a group all inverses are unique (pretty easy lemma). And the OP also needs to show $0x=0$ which follows from unicity of the unit element in a group etc.
$endgroup$
– Henno Brandsma
Jan 26 at 23:04
add a comment |
Your Answer
StackExchange.ifUsing("editor", function () {
return StackExchange.using("mathjaxEditing", function () {
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
});
});
}, "mathjax-editing");
StackExchange.ready(function() {
var channelOptions = {
tags: "".split(" "),
id: "69"
};
initTagRenderer("".split(" "), "".split(" "), channelOptions);
StackExchange.using("externalEditor", function() {
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled) {
StackExchange.using("snippets", function() {
createEditor();
});
}
else {
createEditor();
}
});
function createEditor() {
StackExchange.prepareEditor({
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
imageUploader: {
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
},
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
});
}
});
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3088858%2fprove-that-a-b-a-b-a-1-a-1-and-a-b-1-a-1-b-1%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Your proofs are mostly good.
But before you can prove 1) you must prove:
$0cdot b = 0$ for all $b$.
Pf: $0cdot b = (0 + 0)cdot b$ (as $0 = 0 + 0$)$
$= 0cdot b + 0cdot b$ (via distribution)
So $0 = 0cdot b + (-0cdot b) $ (as every number has an additive inverse)
$= (0cdot b + 0cdot b) + (-0cdot b) $ (by substitution)
$= 0cdot b + (0cdot b + (-0cdot b))$ (associativity)
$= 0cdot b + 0$ (definition of inverse)
$= 0cdot b$ (definition of 0)$
In your prove of 2) you substituted $1$ with $(-a)(-a^{-1})$ which I don't think is justified without proof.
But you have proven in 1) that $(-a)(-a^{-1}) = -(a(-a^{-1}))$ and by commutativity $-(a(-a^{-1})) = -((-a^{-1})a)$ and by 1 again $= -(-(a^{-1}a)) = -(-1)$.
But we still need to prove that $-(-a) = a$ which can be done as
$-(-a) + (-a) =0$
$-(-a) + (-a) + a = 0 + a$
$-(-a) + 0 = a$
$-(-a) = a$.
Your 3) is just fine
answered Jan 26 at 23:10
fleabloodfleablood
73k22789
$endgroup$
add a comment |
$begingroup$
Your proofs are mostly good.
But before you can prove 1) you must prove:
$0cdot b = 0$ for all $b$.
Pf: $0cdot b = (0 + 0)cdot b$ (as $0 = 0 + 0$)$
$= 0cdot b + 0cdot b$ (via distribution)
So $0 = 0cdot b + (-0cdot b) $ (as every number has an additive inverse)
$= (0cdot b + 0cdot b) + (-0cdot b) $ (by substitution)
$= 0cdot b + (0cdot b + (-0cdot b))$ (associativity)
$= 0cdot b + 0$ (definition of inverse)
$= 0cdot b$ (definition of 0)$
In your prove of 2) you substituted $1$ with $(-a)(-a^{-1})$ which I don't think is justified without proof.
But you have proven in 1) that $(-a)(-a^{-1}) = -(a(-a^{-1}))$ and by commutativity $-(a(-a^{-1})) = -((-a^{-1})a)$ and by 1 again $= -(-(a^{-1}a)) = -(-1)$.
But we still need to prove that $-(-a) = a$ which can be done as
$-(-a) + (-a) =0$
$-(-a) + (-a) + a = 0 + a$
$-(-a) + 0 = a$
$-(-a) = a$.
Your 3) is just fine
answered Jan 26 at 23:10
fleabloodfleablood
73k22789
$endgroup$
add a comment |
$begingroup$
Your proofs are mostly good.
But before you can prove 1) you must prove:
$0cdot b = 0$ for all $b$.
Pf: $0cdot b = (0 + 0)cdot b$ (as $0 = 0 + 0$)$
$= 0cdot b + 0cdot b$ (via distribution)
So $0 = 0cdot b + (-0cdot b) $ (as every number has an additive inverse)
$= (0cdot b + 0cdot b) + (-0cdot b) $ (by substitution)
$= 0cdot b + (0cdot b + (-0cdot b))$ (associativity)
$= 0cdot b + 0$ (definition of inverse)
$= 0cdot b$ (definition of 0)$
In your prove of 2) you substituted $1$ with $(-a)(-a^{-1})$ which I don't think is justified without proof.
But you have proven in 1) that $(-a)(-a^{-1}) = -(a(-a^{-1}))$ and by commutativity $-(a(-a^{-1})) = -((-a^{-1})a)$ and by 1 again $= -(-(a^{-1}a)) = -(-1)$.
But we still need to prove that $-(-a) = a$ which can be done as
$-(-a) + (-a) =0$
$-(-a) + (-a) + a = 0 + a$
$-(-a) + 0 = a$
$-(-a) = a$.
Your 3) is just fine
answered Jan 26 at 23:10
fleabloodfleablood
73k22789
$endgroup$
Your proofs are mostly good.
But before you can prove 1) you must prove:
$0cdot b = 0$ for all $b$.
Pf: $0cdot b = (0 + 0)cdot b$ (as $0 = 0 + 0$)$
$= 0cdot b + 0cdot b$ (via distribution)
So $0 = 0cdot b + (-0cdot b) $ (as every number has an additive inverse)
$= (0cdot b + 0cdot b) + (-0cdot b) $ (by substitution)
$= 0cdot b + (0cdot b + (-0cdot b))$ (associativity)
$= 0cdot b + 0$ (definition of inverse)
$= 0cdot b$ (definition of 0)$
In your prove of 2) you substituted $1$ with $(-a)(-a^{-1})$ which I don't think is justified without proof.
But you have proven in 1) that $(-a)(-a^{-1}) = -(a(-a^{-1}))$ and by commutativity $-(a(-a^{-1})) = -((-a^{-1})a)$ and by 1 again $= -(-(a^{-1}a)) = -(-1)$.
But we still need to prove that $-(-a) = a$ which can be done as
$-(-a) + (-a) =0$
$-(-a) + (-a) + a = 0 + a$
$-(-a) + 0 = a$
$-(-a) = a$.
Your 3) is just fine
answered Jan 26 at 23:10
fleabloodfleablood
73k22789
answered Jan 26 at 23:10
fleabloodfleablood
73k22789
answered Jan 26 at 23:10
fleabloodfleablood
73k22789
answered Jan 26 at 23:10
fleabloodfleablood
73k22789
73k22789
add a comment |
add a comment |
$begingroup$
Why not: $(-a)b + ab = (-a + a)b= 0cdot b = 0$ (assuming you know the latter, which si not an axiom but a consequence of them, hopefully shown before), so that $ab$ is the additive inverse of $(-a)b$ and as inverses are unique in any group we have $(-a)b = -(ab)$.
For the second $acdot a^{-1}=1$ by definition. Assuming you have shown $(-x)cdot(-y)=xy$ we write $(-a) cdot -(a^{-1})=1$ and so $-(a^{-1})$ is the multiplicative inverse of $-a$ so by unicity again $(-a)^{-1} = -(a^{-1})$.
For 3, $(ab)(a^{-1}b^{-1}) =1 $ by applying commutativity and the definition of inverse twice. So again by unicity of multiplicative inverses: $(ab)^{-1} = a^{-1} cdot b^{-1}$.
So as a basic algebra lemma for all of this: if $(G, ast,^{-1}, e)$ is a group, inverses are unique: if $a ast b=e$ then $b=a^{-1}$, where $a^{-1}$ is promised by the group axioms. In a field $F$, $(F, + ,-,0)$ forms a group and so does $(Fsetminus {0},cdot,^{-1}, 1)$, so we can apply this.
answered Jan 26 at 22:49
Henno BrandsmaHenno Brandsma
113k348123
$endgroup$
$begingroup$
"Why not: (−a)b+ab=(−a+a)b=0⋅b=0" Well..... if one is REALLY picky one can point out that we know $-(ab)$ exists but not that it is unique. So $-(ab) + ab = 0$ and $(-a)b + ab = 0$ doesn't actually mean that $-(ab) = (-a)b$ until you con prove it. But that's pretty picky.....
$endgroup$
– fleablood
Jan 26 at 23:01
$begingroup$
@fleablood as remarked, in a group all inverses are unique (pretty easy lemma). And the OP also needs to show $0x=0$ which follows from unicity of the unit element in a group etc.
$endgroup$
– Henno Brandsma
Jan 26 at 23:04
add a comment |
$begingroup$
Why not: $(-a)b + ab = (-a + a)b= 0cdot b = 0$ (assuming you know the latter, which si not an axiom but a consequence of them, hopefully shown before), so that $ab$ is the additive inverse of $(-a)b$ and as inverses are unique in any group we have $(-a)b = -(ab)$.
For the second $acdot a^{-1}=1$ by definition. Assuming you have shown $(-x)cdot(-y)=xy$ we write $(-a) cdot -(a^{-1})=1$ and so $-(a^{-1})$ is the multiplicative inverse of $-a$ so by unicity again $(-a)^{-1} = -(a^{-1})$.
For 3, $(ab)(a^{-1}b^{-1}) =1 $ by applying commutativity and the definition of inverse twice. So again by unicity of multiplicative inverses: $(ab)^{-1} = a^{-1} cdot b^{-1}$.
So as a basic algebra lemma for all of this: if $(G, ast,^{-1}, e)$ is a group, inverses are unique: if $a ast b=e$ then $b=a^{-1}$, where $a^{-1}$ is promised by the group axioms. In a field $F$, $(F, + ,-,0)$ forms a group and so does $(Fsetminus {0},cdot,^{-1}, 1)$, so we can apply this.
answered Jan 26 at 22:49
Henno BrandsmaHenno Brandsma
113k348123
$endgroup$
$begingroup$
"Why not: (−a)b+ab=(−a+a)b=0⋅b=0" Well..... if one is REALLY picky one can point out that we know $-(ab)$ exists but not that it is unique. So $-(ab) + ab = 0$ and $(-a)b + ab = 0$ doesn't actually mean that $-(ab) = (-a)b$ until you con prove it. But that's pretty picky.....
$endgroup$
– fleablood
Jan 26 at 23:01
$begingroup$
@fleablood as remarked, in a group all inverses are unique (pretty easy lemma). And the OP also needs to show $0x=0$ which follows from unicity of the unit element in a group etc.
$endgroup$
– Henno Brandsma
Jan 26 at 23:04
add a comment |
$begingroup$
Why not: $(-a)b + ab = (-a + a)b= 0cdot b = 0$ (assuming you know the latter, which si not an axiom but a consequence of them, hopefully shown before), so that $ab$ is the additive inverse of $(-a)b$ and as inverses are unique in any group we have $(-a)b = -(ab)$.
For the second $acdot a^{-1}=1$ by definition. Assuming you have shown $(-x)cdot(-y)=xy$ we write $(-a) cdot -(a^{-1})=1$ and so $-(a^{-1})$ is the multiplicative inverse of $-a$ so by unicity again $(-a)^{-1} = -(a^{-1})$.
For 3, $(ab)(a^{-1}b^{-1}) =1 $ by applying commutativity and the definition of inverse twice. So again by unicity of multiplicative inverses: $(ab)^{-1} = a^{-1} cdot b^{-1}$.
So as a basic algebra lemma for all of this: if $(G, ast,^{-1}, e)$ is a group, inverses are unique: if $a ast b=e$ then $b=a^{-1}$, where $a^{-1}$ is promised by the group axioms. In a field $F$, $(F, + ,-,0)$ forms a group and so does $(Fsetminus {0},cdot,^{-1}, 1)$, so we can apply this.
answered Jan 26 at 22:49
Henno BrandsmaHenno Brandsma
113k348123
$endgroup$
Why not: $(-a)b + ab = (-a + a)b= 0cdot b = 0$ (assuming you know the latter, which si not an axiom but a consequence of them, hopefully shown before), so that $ab$ is the additive inverse of $(-a)b$ and as inverses are unique in any group we have $(-a)b = -(ab)$.
For the second $acdot a^{-1}=1$ by definition. Assuming you have shown $(-x)cdot(-y)=xy$ we write $(-a) cdot -(a^{-1})=1$ and so $-(a^{-1})$ is the multiplicative inverse of $-a$ so by unicity again $(-a)^{-1} = -(a^{-1})$.
For 3, $(ab)(a^{-1}b^{-1}) =1 $ by applying commutativity and the definition of inverse twice. So again by unicity of multiplicative inverses: $(ab)^{-1} = a^{-1} cdot b^{-1}$.
So as a basic algebra lemma for all of this: if $(G, ast,^{-1}, e)$ is a group, inverses are unique: if $a ast b=e$ then $b=a^{-1}$, where $a^{-1}$ is promised by the group axioms. In a field $F$, $(F, + ,-,0)$ forms a group and so does $(Fsetminus {0},cdot,^{-1}, 1)$, so we can apply this.
answered Jan 26 at 22:49
Henno BrandsmaHenno Brandsma
113k348123
edited Jan 26 at 22:54
answered Jan 26 at 22:49
Henno BrandsmaHenno Brandsma
113k348123
answered Jan 26 at 22:49
Henno BrandsmaHenno Brandsma
113k348123
answered Jan 26 at 22:49
Henno BrandsmaHenno Brandsma
113k348123
113k348123
$begingroup$
"Why not: (−a)b+ab=(−a+a)b=0⋅b=0" Well..... if one is REALLY picky one can point out that we know $-(ab)$ exists but not that it is unique. So $-(ab) + ab = 0$ and $(-a)b + ab = 0$ doesn't actually mean that $-(ab) = (-a)b$ until you con prove it. But that's pretty picky.....
$endgroup$
– fleablood
Jan 26 at 23:01
$begingroup$
@fleablood as remarked, in a group all inverses are unique (pretty easy lemma). And the OP also needs to show $0x=0$ which follows from unicity of the unit element in a group etc.
$endgroup$
– Henno Brandsma
Jan 26 at 23:04
add a comment |
$begingroup$
"Why not: (−a)b+ab=(−a+a)b=0⋅b=0" Well..... if one is REALLY picky one can point out that we know $-(ab)$ exists but not that it is unique. So $-(ab) + ab = 0$ and $(-a)b + ab = 0$ doesn't actually mean that $-(ab) = (-a)b$ until you con prove it. But that's pretty picky.....
$endgroup$
– fleablood
Jan 26 at 23:01
$begingroup$
@fleablood as remarked, in a group all inverses are unique (pretty easy lemma). And the OP also needs to show $0x=0$ which follows from unicity of the unit element in a group etc.
$endgroup$
– Henno Brandsma
Jan 26 at 23:04
$begingroup$
"Why not: (−a)b+ab=(−a+a)b=0⋅b=0" Well..... if one is REALLY picky one can point out that we know $-(ab)$ exists but not that it is unique. So $-(ab) + ab = 0$ and $(-a)b + ab = 0$ doesn't actually mean that $-(ab) = (-a)b$ until you con prove it. But that's pretty picky.....
$endgroup$
– fleablood
Jan 26 at 23:01
$begingroup$
"Why not: (−a)b+ab=(−a+a)b=0⋅b=0" Well..... if one is REALLY picky one can point out that we know $-(ab)$ exists but not that it is unique. So $-(ab) + ab = 0$ and $(-a)b + ab = 0$ doesn't actually mean that $-(ab) = (-a)b$ until you con prove it. But that's pretty picky.....
$endgroup$
– fleablood
Jan 26 at 23:01
$begingroup$
@fleablood as remarked, in a group all inverses are unique (pretty easy lemma). And the OP also needs to show $0x=0$ which follows from unicity of the unit element in a group etc.
$endgroup$
– Henno Brandsma
Jan 26 at 23:04
$begingroup$
@fleablood as remarked, in a group all inverses are unique (pretty easy lemma). And the OP also needs to show $0x=0$ which follows from unicity of the unit element in a group etc.
$endgroup$
– Henno Brandsma
Jan 26 at 23:04
add a comment |
Thanks for contributing an answer to Mathematics Stack Exchange!
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
Use MathJax to format equations. MathJax reference.
To learn more, see our tips on writing great answers.
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3088858%2fprove-that-a-b-a-b-a-1-a-1-and-a-b-1-a-1-b-1%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
$begingroup$
For 1) you first have to prove that $0.b = 0$ which usually is not a given.
$endgroup$
– fleablood
Jan 26 at 22:36
$begingroup$
In 2) you have no reason to assume $(-a)(-a^{-1}) = 1$. Unless you are assuming $(-a)(-b) = ab$ and so $(-a)(-a^{-1}) = a.a^{-1} = 1$. If so you should point that out, it wasn't clear. ANd to you do need to prove that $(-a)(-b) = ab$. It's not a given.
$endgroup$
– fleablood
Jan 26 at 22:39
$begingroup$
3) Is $.$ commutative? Was that an axiom? it isn't true if $.$ isn't commutative. So perhaps before either of had even gotten started I should have asked? What exactly are you doing? Is $.$ supposed to be multiplication on the reals? The rationals? An arbitrary field? What class is this for? What are your axioms? Are you just assuming the field axioms, etc? Or is this high school algebra? Or elementary school arithmetic? Context does matter.
$endgroup$
– fleablood
Jan 26 at 22:43
$begingroup$
Lest I seem too critical. Yes, your proofs and reasoning are perfectly correct and the way to go. But you must prove $0.b = 0$ for all $b$ and that $(-a)(-b) = ab$.
$endgroup$
– fleablood
Jan 26 at 22:46
$begingroup$
Thank you so much prof, I used the 9 axioms [in the question]. For $0.b=b$ my prof was prof if, I will prove that $(-a)(-b)=ab$,
$endgroup$
– Dima
Jan 26 at 22:50