Homework Statement
Prove that set of all onto mappings of A->A is closed under composition of mappings:
Homework Equations
Definition of onto and closure on sets.
The Attempt at a Solution
Say, ##f## and ##g## are onto mappings from A to A.
Now, say I have a set S(A) = {all onto mappings of A...
The question says that A= R-{0} and B =R. Then, that f:A ->B and I need to show whether they 1-1 and whether they are onto. Prove.
Thanks for the hint.
Homework Statement
I need to show that $$\frac{x}{x^2+1}$$ is either onto or not.
My domain is $$R-{0}$$ and range is $$R$$
Homework Equations
I have learn to do this to show that a function is surjective
y = $$\frac{x}{x^2+1}$$ and solve for x, but I am not sure how to proceed here.
The...
Homework Statement
Imagine that force for is atom was ## F= - \frac{\beta}{r^4}##, rather than ##F=- \frac{ke^2}{r^2}##, and consider only circular orbits, it would remain true that ##L_n= n \hbar##
a.) From Netwon's law find the relationship between ##T ##(Kinetic Energy) and ##V##,
b.) Find...
Homework Statement
Title of pie chart: New Construction in Daisy Hill Subdivision
Given a pie chart with sections:
Currently Completed: 26%
Currently Under Construction: 42%
Approved, but Not Yet Started: 32%
Question:
When construction is completed in Daisy Hill Subdivision there'll...
Homework Statement
Let's assume that the classical ideas of space and time are correct, so that there could only be one frame, "ether", in which light traveled with same speed in all directions.
Assume that the earth's speed relative to the ether frame is our orbital speed around the sun...
Homework Statement
At time t=0, a block is released from point O on the slope shown in the figure.
The block accelerates down the slope, overcoming sliding friction.
a.) Choose axes 0xy as shown, and solve the equation ##\Sigma F = m a## into its x and y components.
Hence find the block's...
I agree micromass, I never chose mediocre classes before for a GPA boost and I have never dropped a class.
But, I thought that since the end of the undergraduate program is near and there are requirements for graduate school, I should make an exception.
Austrian, the classes are Theory of...
Hello all,
I have a bit of a dilemma.
I have two free electives to take and I am considering taking two "advanced" math classes and get a math minor or two semi-good classes from whatever.
If I take two math classes and I get A's, it would be lovely. But, If I get low grades, my GPA will...
Homework Statement
Prove that
## lim_{x\implies 1} \frac{2}{x-3} = -1 ##
Use delta-epsilon.
The Attempt at a Solution
Proof strategy:
## | { \frac{ 2}{x-3} +1 } | < \epsilon ##
## \frac{x-1}{x-3} < \epsilon ##
, since delta have to be a function of epsilon alone and not include x. I...
Wait, so my R is an equivalence relation then? Supposedly, it partitions the set into disjoint classes. I guess that my classes would be [1], [2], [3]?
HallsofIvy, thank you for the clarification. I should have stated that in this case, it means the same.
Nevermind. I just read somewhere that reflexive statements don't count towards symmetry. Apparently, it involves something like a diagonal class; I guess they pair this combinations in a matrix like form.
Anyway. Thanks.
Now, I am wondering... do they have to be equal? a =2 and b =3. If the only restriction is that a,b ## \in \mathbb{N}##
What would be an argument against what I just wrote?
Sorry, I got confused.
Let's see:
## a\not | c \wedge a\not | b \implies a \not |bc ##
## pr \not = pqx \wedge qr \not = pqy \implies pq(r^2) \not = pqz##
## T \wedge T \implies F \equiv F##
So, the statement is false.
To show that ## g## is injective:
## g(a) = g(b ) \\ (a,1 ) = (b ,1 )##
Then, ##a## must be equal to ## b## and ## g ## is injective.
Thank you for the unique factorization explanation.
Homework Statement
a.) Prove: If an integer ##a## does not divide ##bc##, then ##a## does not divide ##b## and ##a## does not divide ##c##.
b.) State and either prove or disprove the converse of the above statement.
The Attempt at a Solution
a.) Proof by contrapositive
## a|c \vee a|b...
So, d | n
d =3
does 3 |n ?
Well, the stipulations say that ##n## is not divisible by 3. ## n\not | 3 ##
Sorry, but again I am lost.
## (n = 3k )\not = (3 \not = np), k,p \in \mathbb{Z} ##
I see what you are doing.
About what I wrote prior to this...
I was talking about this supposedly property of absolute value
## |a-b| < c ##, then
## -c< a-b <c ##
uh, are you serious?
## (|a-b|)^2 = (a^2 - 2ab + b^2 ) \leq (\sqrt{a^2 +b^2})^2 = a^2 +b^2##
Then the LHS will always be lesser for any ## \mathbb{R^+}##