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age 23
visits member for 3 years, 6 months
seen Jun 30 '13 at 13:58

"An expert is a person who has made all the mistakes that can be made in a very narrow field." Niels Bohr


Feb
17
comment Injective and Surjective Functions
I edited my proof for a. Please confirm that this is valid.
Feb
17
revised Injective and Surjective Functions
added 409 characters in body
Feb
17
asked Injective and Surjective Functions
Feb
17
accepted Evaluate $\int \cos^3 x\;\sin^2 xdx$
Feb
14
comment Evaluate $\int \cos^3 x\;\sin^2 xdx$
good to know. thanks
Feb
11
awarded  Editor
Feb
11
revised Evaluate $\int \cos^3 x\;\sin^2 xdx$
added 4 characters in body
Feb
11
comment Evaluate $\int \cos^3 x\;\sin^2 xdx$
integral cos^3(x) sin^2(x) dx = 1/30 sin^3(x) (3 cos(2 x)+7)+constant
Feb
11
asked Evaluate $\int \cos^3 x\;\sin^2 xdx$
Feb
10
accepted Equivalence Relations
Feb
9
comment Equivalence Relations
How would I refer to the ordered pairs then?
Feb
9
asked Equivalence Relations
Jan
29
comment Equivalence Classes of this Relation on the integers : $a + b^2 \equiv 0\pmod{2}$.
I proved it by showing R to be reflexive, symmetric, and transitive. I just don't have much experience with equivalence classes.
Jan
28
comment Equivalence Classes of this Relation on the integers : $a + b^2 \equiv 0\pmod{2}$.
I concur. I probably should have been more specific and noteed that I had already proven that R was an equivalence relation.
Jan
27
awarded  Scholar
Jan
27
accepted Equivalence Classes of this Relation on the integers : $a + b^2 \equiv 0\pmod{2}$.
Jan
27
comment Equivalence Classes of this Relation on the integers : $a + b^2 \equiv 0\pmod{2}$.
But that's not what I asked for
Jan
27
comment Equivalence Classes of this Relation on the integers : $a + b^2 \equiv 0\pmod{2}$.
thanks. so the equivalence class would be the set of all odd integers.
Jan
27
awarded  Student
Jan
27
asked Equivalence Classes of this Relation on the integers : $a + b^2 \equiv 0\pmod{2}$.