Product property [closed]
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What is the name of the following property of two equal products?
If $ab = cd$, then $a(b-d)=(c-a)d$
universal-property
asked Dec 20 '18 at 11:56
Michael MuntaMichael Munta
97111
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closed as off-topic by Saad, Brahadeesh, Frpzzd, Widawensen, user593746 Dec 20 '18 at 15:03
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Saad, Brahadeesh, Frpzzd, Widawensen, Community
If this question can be reworded to fit the rules in the help center, please edit the question.
add a comment |
$begingroup$
What is the name of the following property of two equal products?
If $ab = cd$, then $a(b-d)=(c-a)d$
universal-property
asked Dec 20 '18 at 11:56
Michael MuntaMichael Munta
97111
$endgroup$
closed as off-topic by Saad, Brahadeesh, Frpzzd, Widawensen, user593746 Dec 20 '18 at 15:03
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Saad, Brahadeesh, Frpzzd, Widawensen, Community
If this question can be reworded to fit the rules in the help center, please edit the question.
2
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I don't suppose this has a name. There are many, many algebraic identities. Most of them lack names.
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– lulu
Dec 20 '18 at 11:58
add a comment |
$begingroup$
What is the name of the following property of two equal products?
If $ab = cd$, then $a(b-d)=(c-a)d$
universal-property
asked Dec 20 '18 at 11:56
Michael MuntaMichael Munta
97111
$endgroup$
What is the name of the following property of two equal products?
If $ab = cd$, then $a(b-d)=(c-a)d$
universal-property
universal-property
asked Dec 20 '18 at 11:56
Michael MuntaMichael Munta
97111
asked Dec 20 '18 at 11:56
Michael MuntaMichael Munta
97111
asked Dec 20 '18 at 11:56
Michael MuntaMichael Munta
97111
asked Dec 20 '18 at 11:56
Michael MuntaMichael Munta
97111
asked Dec 20 '18 at 11:56
Michael MuntaMichael Munta
97111
97111
closed as off-topic by Saad, Brahadeesh, Frpzzd, Widawensen, user593746 Dec 20 '18 at 15:03
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Saad, Brahadeesh, Frpzzd, Widawensen, Community
If this question can be reworded to fit the rules in the help center, please edit the question.
closed as off-topic by Saad, Brahadeesh, Frpzzd, Widawensen, user593746 Dec 20 '18 at 15:03
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Saad, Brahadeesh, Frpzzd, Widawensen, Community
If this question can be reworded to fit the rules in the help center, please edit the question.
2
$begingroup$
I don't suppose this has a name. There are many, many algebraic identities. Most of them lack names.
$endgroup$
– lulu
Dec 20 '18 at 11:58
add a comment |
2
$begingroup$
I don't suppose this has a name. There are many, many algebraic identities. Most of them lack names.
$endgroup$
– lulu
Dec 20 '18 at 11:58
2
2
$begingroup$
I don't suppose this has a name. There are many, many algebraic identities. Most of them lack names.
$endgroup$
– lulu
Dec 20 '18 at 11:58
$begingroup$
I don't suppose this has a name. There are many, many algebraic identities. Most of them lack names.
$endgroup$
– lulu
Dec 20 '18 at 11:58
add a comment |
2 Answers
2
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oldest
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$a(b-d)=(c-a)d iff ab-ad=cd-ad iff ab=cd$.
The reason ist the distributive property of addition and multiplication in $ mathbb R$.
answered Dec 20 '18 at 12:04
FredFred
48.7k11849
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add a comment |
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It is the multilinearity of the determinant (substracting the first column from the second one does not change the determinant), i.e.,
$$
ab-cd=det begin{pmatrix} a & c cr d & b end{pmatrix}=det
begin{pmatrix} a & c-a cr d & b-d end{pmatrix}=a(b-d)-(c-a)d.
$$
answered Dec 20 '18 at 12:08
Dietrich BurdeDietrich Burde
81.2k648106
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add a comment |
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
$a(b-d)=(c-a)d iff ab-ad=cd-ad iff ab=cd$.
The reason ist the distributive property of addition and multiplication in $ mathbb R$.
answered Dec 20 '18 at 12:04
FredFred
48.7k11849
$endgroup$
add a comment |
$begingroup$
$a(b-d)=(c-a)d iff ab-ad=cd-ad iff ab=cd$.
The reason ist the distributive property of addition and multiplication in $ mathbb R$.
answered Dec 20 '18 at 12:04
FredFred
48.7k11849
$endgroup$
add a comment |
$begingroup$
$a(b-d)=(c-a)d iff ab-ad=cd-ad iff ab=cd$.
The reason ist the distributive property of addition and multiplication in $ mathbb R$.
answered Dec 20 '18 at 12:04
FredFred
48.7k11849
$endgroup$
$a(b-d)=(c-a)d iff ab-ad=cd-ad iff ab=cd$.
The reason ist the distributive property of addition and multiplication in $ mathbb R$.
answered Dec 20 '18 at 12:04
FredFred
48.7k11849
answered Dec 20 '18 at 12:04
FredFred
48.7k11849
answered Dec 20 '18 at 12:04
FredFred
48.7k11849
answered Dec 20 '18 at 12:04
FredFred
48.7k11849
48.7k11849
add a comment |
add a comment |
$begingroup$
It is the multilinearity of the determinant (substracting the first column from the second one does not change the determinant), i.e.,
$$
ab-cd=det begin{pmatrix} a & c cr d & b end{pmatrix}=det
begin{pmatrix} a & c-a cr d & b-d end{pmatrix}=a(b-d)-(c-a)d.
$$
answered Dec 20 '18 at 12:08
Dietrich BurdeDietrich Burde
81.2k648106
$endgroup$
add a comment |
$begingroup$
It is the multilinearity of the determinant (substracting the first column from the second one does not change the determinant), i.e.,
$$
ab-cd=det begin{pmatrix} a & c cr d & b end{pmatrix}=det
begin{pmatrix} a & c-a cr d & b-d end{pmatrix}=a(b-d)-(c-a)d.
$$
answered Dec 20 '18 at 12:08
Dietrich BurdeDietrich Burde
81.2k648106
$endgroup$
add a comment |
$begingroup$
It is the multilinearity of the determinant (substracting the first column from the second one does not change the determinant), i.e.,
$$
ab-cd=det begin{pmatrix} a & c cr d & b end{pmatrix}=det
begin{pmatrix} a & c-a cr d & b-d end{pmatrix}=a(b-d)-(c-a)d.
$$
answered Dec 20 '18 at 12:08
Dietrich BurdeDietrich Burde
81.2k648106
$endgroup$
It is the multilinearity of the determinant (substracting the first column from the second one does not change the determinant), i.e.,
$$
ab-cd=det begin{pmatrix} a & c cr d & b end{pmatrix}=det
begin{pmatrix} a & c-a cr d & b-d end{pmatrix}=a(b-d)-(c-a)d.
$$
answered Dec 20 '18 at 12:08
Dietrich BurdeDietrich Burde
81.2k648106
answered Dec 20 '18 at 12:08
Dietrich BurdeDietrich Burde
81.2k648106
answered Dec 20 '18 at 12:08
Dietrich BurdeDietrich Burde
81.2k648106
answered Dec 20 '18 at 12:08
Dietrich BurdeDietrich Burde
81.2k648106
81.2k648106
add a comment |
add a comment |
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2
$begingroup$
I don't suppose this has a name. There are many, many algebraic identities. Most of them lack names.
$endgroup$
– lulu
Dec 20 '18 at 11:58