TranceAddict Forums (www.tranceaddict.com/forums)
- Chill Out Room
-- mcdonalds: now for black people!
Pages (5): « 1 2 [3] 4 5 »
| quote: |
| Originally posted by Domesticated What, so because some people were enslaved three generations ago they deserve a condescending, trite website offering them 'opportunities' and only reminding them once again that the white population sees them as 'different' from themselves? Wake the fuck up. The only way to end discrimination and oppression is to do away with bullshit like this that convinces Caucasians that the Negroes living in their midst are any different from them and deserve special treatment and/or recognition. |
mcdonalds is fully aware many black people frequent their "restaurants" so they are doing more to speak to their consumer. you people are ridiculous riling yourselves up for nothing. i really don't see this as any different from mlb or mls teams having hispanic nights or really any other company choosing to reach out to a particular group they want to target; it could very well be low income mothers.
people are laughing at the site and getting riled up about the related issue. for the record.
McDonalds' decision to butter up a certain group of their customers isn't anything to get annoyed about, but the way they've done it certainly is.
| quote: |
| Originally posted by SYSTEM-J McDonalds' decision to butter up a certain group of their customers isn't anything to get annoyed about, but the way they've done it certainly is. |
| quote: |
| Originally posted by inconspicuous it usually irritates me when people on this forum from other countries try to comment on what another one should do (yes, usually the us, but it goes both ways), but you're actually right on this one. bullshit like this website and affirmative action do everybody a disservice--black, white, aquamarine, w/e. and before someone bitches about it, my issue with affirmative action is that it does two things: a) gives spots of more capable candidates to those who bring 'diversity' instead of 'merit.' b) puts under-qualified people in a position in which they're doomed to fail when they realize they're in over their heads. /.02 |
| quote: |
| Originally posted by chimera66 mcdonalds is fully aware many black people frequent their "restaurants" so they are doing more to speak to their consumer. you people are ridiculous riling yourselves up for nothing. i really don't see this as any different from mlb or mls teams having hispanic nights or really any other company choosing to reach out to a particular group they want to target; it could very well be low income mothers. |
| quote: |
| Originally posted by chimera66 do you honestly believe they hire under-qualified people just because of affirmative action? |
| quote: |
| i can think if plenty of people who have jobs/opportunties they shouldn't because of an assortment of things. |
Affirmative action != quotas.
well, yeah... if you have to have a certain percentage of minorities occupying your employment force, is that not having to fill a quota?
| quote: |
| Originally posted by The17sss well, yeah... if you have to have a certain percentage of minorities occupying your employment force, is that not having to fill a quota? |
| quote: |
| Originally posted by MrJiveBoJingles Not all affirmative action policies mandate percentages. Affirmative action may simply be allowing employers to consider minority origin in hiring. |
| quote: |
| Originally posted by chimera66 that i can agree with do you honestly believe they hire under-qualified people just because of affirmative action? |
| quote: |
| Originally posted by chimera66 it is meant to get minorities, which does not translate just to black people opportunities but they have to be qualifed for the opportunity. how many under-qualified people with connections get jobs they shouldn't? i can think if plenty of people who have jobs/opportunties they shouldn't because of an assortment of things. |
| quote: |
| Originally posted by chimera66 anyhow, what pisses me off about people talking about affirmative action is they instantly think it is all about helping black people when honestly it's to help a wide group of people including women who probably benefit more so than any other group. the only group who doesn't benefit from it are white males. is that fair, no but don't point the finger at black people specifically as if we are the only ones to benefit. |
| quote: |
| Originally posted by The17sss Yes... for sure. When you have to fill a certain quota that doesn't rely on merit/qualification first, by definition you're trending the wrong way... in terms of job performance. |
| quote: |
| Originally posted by inconspicuous I'm against anyone getting opportunities they shouldn't. Affirmative action just compounds the number of people who do, though, exponentially. |
The common theme is that a homomorphism is a function between two algebraic objects that respects the algebraic structure.
For example, a group is an algebraic object consisting of a set together with a single binary operation, satisfying certain axioms. If G and H are groups, a homomorphism from G to H is a function �: G → H such that f(g_1 * g_2) = f(g_1) * f(g_2)\,\! for any elements g1, g2 ∈ G, where ∗ denotes the respective binary operations (the first ∗ denoting the operation in G, and the second ∗ denoting the operation in H).
When an algebraic structure includes more than one operation, homomorphisms are required to preserve each operation. For example, a ring possesses both addition and multiplication, and a homomorphism between two rings is a function such that
f(r+s) = f(r) + f(s)\qquad\text{and}\qquad f(rs) = f(r)\,f(s)
for any elements r and s of the domain ring. In most contexts, a homomorphism will map identity elements to identity elements, inverse elements to inverse elements, and so forth.
The notion of a homomorphism can be given a formal definition in the context of universal algebra, a field which studies ideas common to all algebraic structures. In this setting, a homomorphism �: A → B is a function between two algebraic structures of the same type such that
f(\mu_A(a_1, \ldots, a_n)) = \mu_B(f(a_1), \ldots, f(a_n))\,
for each n-ary operation μ and for all elements a1,...,an ∈ A.
The real numbers are a ring, having both addition and multiplication. The set of all 2 � 2 matrices is also a ring, using matrix addition and matrix multiplication. Define a function between these rings by
f(r) = \begin{pmatrix} r & 0 \\ 0 & r \end{pmatrix}
where r is a real number. Then � is a homomorphism of rings, since � preserves both addition:
f(r+s) = \begin{pmatrix} r+s & 0 \\ 0 & r+s \end{pmatrix} = \begin{pmatrix} r & 0 \\ 0 & r \end{pmatrix} + \begin{pmatrix} s & 0 \\ 0 & s \end{pmatrix} = f(r) + f(s)
and multiplication:
f(rs) = \begin{pmatrix} rs & 0 \\ 0 & rs \end{pmatrix} = \begin{pmatrix} r & 0 \\ 0 & r \end{pmatrix} \begin{pmatrix} s & 0 \\ 0 & s \end{pmatrix} = f(r)\,f(s).
For another example, the nonzero complex numbers form a group under the operation of multiplication, as do the nonzero real numbers. (Zero must be excluded from both groups since it does not have a multiplicative inverse, which is required for elements of a group.) Define a function � from the nonzero complex numbers to the nonzero real numbers by
f(z) = |z|.\,\!
That is, �(z) is the absolute value (or modulus) of the complex number z. Then � is a homomorphism of groups, since it preserves multiplication:
f(z_1 z_2) = |z_1 z_2| = |z_1|\,|z_2| = f(z_1)\,f(z_2).
Note that � cannot be extended to a homomorphism of rings (from the complex numbers to the real numbers), since it does not preserve addition:
|z_1 + z_2| \ne |z_1| + |z_2|.
| quote: |
| Originally posted by winston The common theme is that a homomorphism is a function between two algebraic objects that respects the algebraic structure. For example, a group is an algebraic object consisting of a set together with a single binary operation, satisfying certain axioms. If G and H are groups, a homomorphism from G to H is a function �: G → H such that f(g_1 * g_2) = f(g_1) * f(g_2)\,\! for any elements g1, g2 ∈ G, where ∗ denotes the respective binary operations (the first ∗ denoting the operation in G, and the second ∗ denoting the operation in H). When an algebraic structure includes more than one operation, homomorphisms are required to preserve each operation. For example, a ring possesses both addition and multiplication, and a homomorphism between two rings is a function such that f(r+s) = f(r) + f(s)\qquad\text{and}\qquad f(rs) = f(r)\,f(s) for any elements r and s of the domain ring. In most contexts, a homomorphism will map identity elements to identity elements, inverse elements to inverse elements, and so forth. The notion of a homomorphism can be given a formal definition in the context of universal algebra, a field which studies ideas common to all algebraic structures. In this setting, a homomorphism �: A → B is a function between two algebraic structures of the same type such that f(\mu_A(a_1, \ldots, a_n)) = \mu_B(f(a_1), \ldots, f(a_n))\, for each n-ary operation μ and for all elements a1,...,an ∈ A. The real numbers are a ring, having both addition and multiplication. The set of all 2 � 2 matrices is also a ring, using matrix addition and matrix multiplication. Define a function between these rings by f(r) = \begin{pmatrix} r & 0 \\ 0 & r \end{pmatrix} where r is a real number. Then � is a homomorphism of rings, since � preserves both addition: f(r+s) = \begin{pmatrix} r+s & 0 \\ 0 & r+s \end{pmatrix} = \begin{pmatrix} r & 0 \\ 0 & r \end{pmatrix} + \begin{pmatrix} s & 0 \\ 0 & s \end{pmatrix} = f(r) + f(s) and multiplication: f(rs) = \begin{pmatrix} rs & 0 \\ 0 & rs \end{pmatrix} = \begin{pmatrix} r & 0 \\ 0 & r \end{pmatrix} \begin{pmatrix} s & 0 \\ 0 & s \end{pmatrix} = f(r)\,f(s). For another example, the nonzero complex numbers form a group under the operation of multiplication, as do the nonzero real numbers. (Zero must be excluded from both groups since it does not have a multiplicative inverse, which is required for elements of a group.) Define a function � from the nonzero complex numbers to the nonzero real numbers by f(z) = |z|.\,\! That is, �(z) is the absolute value (or modulus) of the complex number z. Then � is a homomorphism of groups, since it preserves multiplication: f(z_1 z_2) = |z_1 z_2| = |z_1|\,|z_2| = f(z_1)\,f(z_2). Note that � cannot be extended to a homomorphism of rings (from the complex numbers to the real numbers), since it does not preserve addition: |z_1 + z_2| \ne |z_1| + |z_2|. |
| quote: |
| Originally posted by winston The common theme is that a homomorphism is a function between two algebraic objects that respects the algebraic structure. For example, a group is an algebraic object consisting of a set together with a single binary operation, satisfying certain axioms. If G and H are groups, a homomorphism from G to H is a function �: G → H such that f(g_1 * g_2) = f(g_1) * f(g_2)\,\! for any elements g1, g2 ∈ G, where ∗ denotes the respective binary operations (the first ∗ denoting the operation in G, and the second ∗ denoting the operation in H). When an algebraic structure includes more than one operation, homomorphisms are required to preserve each operation. For example, a ring possesses both addition and multiplication, and a homomorphism between two rings is a function such that f(r+s) = f(r) + f(s)\qquad\text{and}\qquad f(rs) = f(r)\,f(s) for any elements r and s of the domain ring. In most contexts, a homomorphism will map identity elements to identity elements, inverse elements to inverse elements, and so forth. The notion of a homomorphism can be given a formal definition in the context of universal algebra, a field which studies ideas common to all algebraic structures. In this setting, a homomorphism �: A → B is a function between two algebraic structures of the same type such that f(\mu_A(a_1, \ldots, a_n)) = \mu_B(f(a_1), \ldots, f(a_n))\, for each n-ary operation μ and for all elements a1,...,an ∈ A. The real numbers are a ring, having both addition and multiplication. The set of all 2 � 2 matrices is also a ring, using matrix addition and matrix multiplication. Define a function between these rings by f(r) = \begin{pmatrix} r & 0 \\ 0 & r \end{pmatrix} where r is a real number. Then � is a homomorphism of rings, since � preserves both addition: f(r+s) = \begin{pmatrix} r+s & 0 \\ 0 & r+s \end{pmatrix} = \begin{pmatrix} r & 0 \\ 0 & r \end{pmatrix} + \begin{pmatrix} s & 0 \\ 0 & s \end{pmatrix} = f(r) + f(s) and multiplication: f(rs) = \begin{pmatrix} rs & 0 \\ 0 & rs \end{pmatrix} = \begin{pmatrix} r & 0 \\ 0 & r \end{pmatrix} \begin{pmatrix} s & 0 \\ 0 & s \end{pmatrix} = f(r)\,f(s). For another example, the nonzero complex numbers form a group under the operation of multiplication, as do the nonzero real numbers. (Zero must be excluded from both groups since it does not have a multiplicative inverse, which is required for elements of a group.) Define a function � from the nonzero complex numbers to the nonzero real numbers by f(z) = |z|.\,\! That is, �(z) is the absolute value (or modulus) of the complex number z. Then � is a homomorphism of groups, since it preserves multiplication: f(z_1 z_2) = |z_1 z_2| = |z_1|\,|z_2| = f(z_1)\,f(z_2). Note that � cannot be extended to a homomorphism of rings (from the complex numbers to the real numbers), since it does not preserve addition: |z_1 + z_2| \ne |z_1| + |z_2|. |
| quote: |
| Originally posted by winston The common theme is that a homomorphism is a function between two algebraic objects that respects the algebraic structure. For example, a group is an algebraic object consisting of a set together with a single binary operation, satisfying certain axioms. If G and H are groups, a homomorphism from G to H is a function �: G → H such that f(g_1 * g_2) = f(g_1) * f(g_2)\,\! for any elements g1, g2 ∈ G, where ∗ denotes the respective binary operations (the first ∗ denoting the operation in G, and the second ∗ denoting the operation in H). When an algebraic structure includes more than one operation, homomorphisms are required to preserve each operation. For example, a ring possesses both addition and multiplication, and a homomorphism between two rings is a function such that f(r+s) = f(r) + f(s)\qquad\text{and}\qquad f(rs) = f(r)\,f(s) for any elements r and s of the domain ring. In most contexts, a homomorphism will map identity elements to identity elements, inverse elements to inverse elements, and so forth. The notion of a homomorphism can be given a formal definition in the context of universal algebra, a field which studies ideas common to all algebraic structures. In this setting, a homomorphism �: A → B is a function between two algebraic structures of the same type such that f(\mu_A(a_1, \ldots, a_n)) = \mu_B(f(a_1), \ldots, f(a_n))\, for each n-ary operation μ and for all elements a1,...,an ∈ A. The real numbers are a ring, having both addition and multiplication. The set of all 2 � 2 matrices is also a ring, using matrix addition and matrix multiplication. Define a function between these rings by f(r) = \begin{pmatrix} r & 0 \\ 0 & r \end{pmatrix} where r is a real number. Then � is a homomorphism of rings, since � preserves both addition: f(r+s) = \begin{pmatrix} r+s & 0 \\ 0 & r+s \end{pmatrix} = \begin{pmatrix} r & 0 \\ 0 & r \end{pmatrix} + \begin{pmatrix} s & 0 \\ 0 & s \end{pmatrix} = f(r) + f(s) and multiplication: f(rs) = \begin{pmatrix} rs & 0 \\ 0 & rs \end{pmatrix} = \begin{pmatrix} r & 0 \\ 0 & r \end{pmatrix} \begin{pmatrix} s & 0 \\ 0 & s \end{pmatrix} = f(r)\,f(s). For another example, the nonzero complex numbers form a group under the operation of multiplication, as do the nonzero real numbers. (Zero must be excluded from both groups since it does not have a multiplicative inverse, which is required for elements of a group.) Define a function � from the nonzero complex numbers to the nonzero real numbers by f(z) = |z|.\,\! That is, �(z) is the absolute value (or modulus) of the complex number z. Then � is a homomorphism of groups, since it preserves multiplication: f(z_1 z_2) = |z_1 z_2| = |z_1|\,|z_2| = f(z_1)\,f(z_2). Note that � cannot be extended to a homomorphism of rings (from the complex numbers to the real numbers), since it does not preserve addition: |z_1 + z_2| \ne |z_1| + |z_2|. |
| quote: |
| Originally posted by Spam Thank god someone's finally explained this to me in plain english. I don't know what I'd do without crackheads like winston explaining the world to me. |
lol.
| quote: |
| Originally posted by winston Du siehst den Wald vor lauter B�umen nicht. |
homme mort ne fait guerre
| quote: |
| Originally posted by MrJiveBoJingles Yes, because white people were enslaved in America for hundreds of years and legally oppressed for about a hundred years after being freed. The situation is exactly the same.  |
| quote: |
| Originally posted by Domesticated What, so because some people were enslaved three generations ago they deserve a condescending, trite website offering them 'opportunities' and only reminding them once again that the white population sees them as 'different' from themselves? Wake the fuck up. The only way to end discrimination and oppression is to do away with bullshit like this that convinces Caucasians that the Negroes living in their midst are any different from them and deserve special treatment and/or recognition. |
reffering to the Morgan Freeman interview, also:
It's a textbook 'us and them' paradigm IMO
Powered by: vBulletin
Copyright © 2000-2021, Jelsoft Enterprises Ltd.