I just did this question on my Chemistry 11 exam. I'm on my phone so sorry I can't help at the moment.
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I just did this question on my Chemistry 11 exam. I'm on my phone so sorry I can't help at the moment.
1 mole Na2CO3 is about 106g so (14.35g/106g) = 0.135 mol
molarity=moles of solute/liters of solution
M = 0.135 mol/.75L = 0.18 M
using same molarity equation solve for moles with new 150ml volume
.18M = moles Na2CO3/.15L = .027 mol
use the same equation again to find molarity with new volume.
M = 0.027 mol/0.6L = 0.045 M
There's probably some sig fig stuff I messed up somewhere that accounts for the next decimal place but I think it's more or less correct.
youngwerther
Banned
One last chemistry one -
My school uses this online, third-party thing called Sapling Learning for our quizzes, and sometimes we get quizzed on stuff we were not taught how to solve. This is one such case.
"Determine the concentration of each of the individual ions in a 0.550M Ba(OH)2."
My first though was to simply write the chemical equation, then balance it so I can get the molar ratio, but I don't know how that would help me get concentration if I wasn't given volume. But I'm guessing I don't need volume...
Am I to just take the ratio of the molarity I was given? I don't really get it. I feel like I'm really over-thinking it.
Immediate help would be appreciated if possible, I have to work at 6. I have to finish this tonight though because it will be due before I get off. This is my last problem.
My school uses this online, third-party thing called Sapling Learning for our quizzes, and sometimes we get quizzed on stuff we were not taught how to solve. This is one such case.
"Determine the concentration of each of the individual ions in a 0.550M Ba(OH)2."
My first though was to simply write the chemical equation, then balance it so I can get the molar ratio, but I don't know how that would help me get concentration if I wasn't given volume. But I'm guessing I don't need volume...
Am I to just take the ratio of the molarity I was given? I don't really get it. I feel like I'm really over-thinking it.
Immediate help would be appreciated if possible, I have to work at 6. I have to finish this tonight though because it will be due before I get off. This is my last problem.
youngwerther
Banned
Ok cool, that would have been my guess but I wanted to make sure. Thanks.
First year of university calculus help,
1. How do I solve the inequality lx^2+x-8l less than/equal to l2-2xl algebraically, I've graphed both functions and gotten to the point where I have four different cases to work out but the answers I'm getting seem to be wrong.
2. How do I find the inverse of a rational function with a degree of 2, such as f(x)= (x^2+3)/(x-1). I'm trying to find the range of this function btw.
Thanks guys.
1. How do I solve the inequality lx^2+x-8l less than/equal to l2-2xl algebraically, I've graphed both functions and gotten to the point where I have four different cases to work out but the answers I'm getting seem to be wrong.
2. How do I find the inverse of a rational function with a degree of 2, such as f(x)= (x^2+3)/(x-1). I'm trying to find the range of this function btw.
Thanks guys.
Are you trying to find the range of the inverse or of the function? You could simplify the problem first by doing:
(x^2+3)/(x-1) = (x^2-x+x+3)/(x-1) = x + (x+3)/(x-1)= x + (x -1+4)/(x-1) = x + 1 +(4/(x-1)).
I don't think you need to find an inverse to find the range of the function though. Seeing as the function is continuous except at x= 1, you should be able to find the range quite easily.
TimesEunuch
Member
Probably too late, but here goes.
They will lie in a plane if and only if alpha(t1)-alpha(t0), alpha(t2)-alpha(t0), alpha(t3)-alpha(t0) are linearly dependent vectors. So calculate the determinant
|t1-t0 t1^2-t0^2 t1^3-t0^3|
|t2-t0 t2^2-t1^2 t2^3-t0^3|
|t3-t0 t3^2-t0^2 t3^3-t0^3|
You can add a multiple of any row to any other row without changing the determinant. Use this (and the fact that t1-t0<>0 etc) to write the determinant in echelon form. You should be able to show that it equals Product_{0<=i<j<=4} (tj-ti). Since all the ti's are distinct, this determinant is non-zero. Hence the vectors above can't be linearly dependent, and so the four points can't lie in a plane.
This is a simple application of Green's Theorem. The area is \int\int_R du dv. You will need to pick different functions in Green's Theorem for each of the three cases.
It looks like the twisted cubic is the curve alpha(t) = (t, t^2, t^3). If that's the case, suppose you have four distinct points alpha(t0), alpha(t1), alpha(t2), alpha(t3) in the curve.
They will lie in a plane if and only if alpha(t1)-alpha(t0), alpha(t2)-alpha(t0), alpha(t3)-alpha(t0) are linearly dependent vectors. So calculate the determinant
|t1-t0 t1^2-t0^2 t1^3-t0^3|
|t2-t0 t2^2-t1^2 t2^3-t0^3|
|t3-t0 t3^2-t0^2 t3^3-t0^3|
You can add a multiple of any row to any other row without changing the determinant. Use this (and the fact that t1-t0<>0 etc) to write the determinant in echelon form. You should be able to show that it equals Product_{0<=i<j<=4} (tj-ti). Since all the ti's are distinct, this determinant is non-zero. Hence the vectors above can't be linearly dependent, and so the four points can't lie in a plane.
I don't really understand the logic of this question. The length is \int |alpha'| dt = \int(r'^2+r^2)^.5 dt. Part (d) is probably that the improper integral on the rhs exists. Since \int r dt <= \int (r'^2+r^2)^.5 dt, \int r dt must also exist, and this requires r(t)->0 as t>infinity (part (e)). I don't understand what is being looked for in part (a). I could just be missing something really obvious on this one. Does this class have a TA? Probably best to consult with them.
I could really use some help with some Graph Theory:
Prove the following: G= (V,E) has a Hamiltonian Cycle => ∀ V' ⊆ V the number of components of G' = (V - V', E') is at most |V'| + 1. (E' is the set E with all edges incident on vertices of V' removed.)
I'm just not sure if I'm fully understanding the question, so if someone could help me with it I could understand better if I knew the solution
edit: typo
Prove the following: G= (V,E) has a Hamiltonian Cycle => ∀ V' ⊆ V the number of components of G' = (V - V', E') is at most |V'| + 1. (E' is the set E with all edges incident on vertices of V' removed.)
I'm just not sure if I'm fully understanding the question, so if someone could help me with it I could understand better if I knew the solution
edit: typo
I guess you could do it this way then:
y =x + 1 +(4/(x-1)). set x-1 = u.
y = u + 2 +(4/(u)).
0 = u^2 +(2-y)u + 4
And just solve using the quadratic formula.
oxrock
Gravity is a myth, the Earth SUCKS!
Yet again I'm in need of someone to check my work to make sure I'm not making an ass out of myself. Currently working on inverse functions
3 part question
a) find an equation for f-1 (x)
b) graph f(x) and f-1(x) in the same coordinate system
c) Use interval notation to give the domains for both f(x) and f-1(x).
Equation: f(x) = (x-2)^3
What I have:
a) f-1(x) = cube root of x+2
b) f(x) coords: (0,-2) (1,-1) (8,0) / f-1(x) coords: (-2, 0) (-1,1) (0,8)
c) f(x) Domain: [0, infinity) Range: [-2, infinity) / f-1(x) Domain: [-2, infinity) Range: [0, infinity)
Comments/help appreciated. Nowhere near as sure of myself on these things as I wish I was.
3 part question
a) find an equation for f-1 (x)
b) graph f(x) and f-1(x) in the same coordinate system
c) Use interval notation to give the domains for both f(x) and f-1(x).
Equation: f(x) = (x-2)^3
What I have:
a) f-1(x) = cube root of x+2
b) f(x) coords: (0,-2) (1,-1) (8,0) / f-1(x) coords: (-2, 0) (-1,1) (0,8)
c) f(x) Domain: [0, infinity) Range: [-2, infinity) / f-1(x) Domain: [-2, infinity) Range: [0, infinity)
Comments/help appreciated. Nowhere near as sure of myself on these things as I wish I was.
oxrock
Gravity is a myth, the Earth SUCKS!
For B we're simply asked to draw both functions on the same graph. I just put down the coords for each function as I have no idea how I'd convey it otherwise on this forum. The coordinates aren't given, I'm supposed to obtain them from f(x) and graph and then graph f-1(x). So if I fucked them up (entirely possible) I'd appreciate an explanation if at all possible.
nope.gif
Sorry I can't help. I have nightmares about this subject (not really but I could).
What the hell is with this Graph Theory and proofs? They're too complex to derive on your own without considerable time and effort (at least I felt so), too convoluted to memorise, and though understandable, almost impossible to recreate on a page with pure logic without making some sort of error. Maybe I was just bad at them?
It didn't help that some of them were inductive when we hadn't even learned about mathematical induction yet and some of them were like 2 pages long! I'm glad to be rid of this subject. I think it was introduced to us way too soon. Is it normally a first year course?
oxrock
Gravity is a myth, the Earth SUCKS!
Thanks for taking the time to show me that. I have no problem transforming funtions once I know what they look like but graphing them from equations messes with me sometimes.The easiest way to do it - also known as what every mathematician does because we're all lazy - is to plot the function (e.g. using MATLAB or Wolfram Alpha). If you have to do it by hand, just write down enough coordinates to have a feeling how the function looks:
f(x)=(x-2)^3
f(-2)=-64
f(-1)=-27
f(0)=-8
f(1)=-1
f(2)=0
f(3)=8
f(4)=64
It's just a shifted version of x^3, this is what x^3 looks like.
f^-1(x)=³sqrt(x)+2 is exactly the opposite, i.e.
f^(-1)(-64)=-2
f^(-1)(-27)=-1
f^(-1)(-8)=0
f^(-1)(-1)=1
f^(-1)(0)=2
f^(-1)(8)=3
f^(-1)(64)=4
This is what ³sqrt(x) looks like, i.e. you only need to add 2.
Edit:
Before I forget it f(x) has range (-infinity,infinity) and therefore f^(-1)(x) has domain (-infinity,infinity). I don't know how I could miss that, sorry. That stuff happens if you read articles all day.
f(x)=y <=> ³sqrt+2 = x (which is defined for every y and vice-versa)
SailorCosmos
Member
Anyone here do number theory?
I'm taking a cryptography class and the professor asked the class two write the following two formulas into simplest form. I've never taken any number theory so I'm lost on what to do. Any help would be appreciated.
Problems: Write formulas in simplest form.
A. (g^a)^b mod p, where g, a, b and p are integer numbers.
B. (g^a x g^b) mod p, where g, a, b and p are integer numbers.
If the letters were numbers I sort of understand it but as is it now I don't have any idea on what to do.
I'm taking a cryptography class and the professor asked the class two write the following two formulas into simplest form. I've never taken any number theory so I'm lost on what to do. Any help would be appreciated.
Problems: Write formulas in simplest form.
A. (g^a)^b mod p, where g, a, b and p are integer numbers.
B. (g^a x g^b) mod p, where g, a, b and p are integer numbers.
If the letters were numbers I sort of understand it but as is it now I don't have any idea on what to do.
a. (g^a)^b mod p becomes g^(ab) mod p, then p|g^(ab) becomes g^(ab)=px for some x in the integers. The last part is just the def. of mod p, but the first part is a simple proof for integers that you can usually find online.
b.(g^a x g^b) mod p becomes g^(a+b) mod p becomes p|g^(a+b) becomes g^(a+b)=py for some y in the integers.
I hope this helps.
SailorCosmos
Member
Thank you, Troll and TimesEunuch.
TimesEunuch
Member
You're not told that p | g^(ab), so I doubt that's what the question is looking for. I dunno either: my guess would be that the professor wants you to say
(g^a)^b mod p = ((g mod p)^a mod p)^b mod p
(g^a.g^b) = ((g mod p)^a mod p).((g mod p)^b mod p) mod p
or something like that. In other words, mod commutes with multiplication and the taking of powers. `simplest form' is kind of a vague term, though, so who knows.
TimesEunuch
Member
"a mod p" means take the remainder of a after dividing by p. So if a=np+r, where n, r are integers and 0<=r<p, then a mod p = r. It doesn't mean a is divisible by p. I think the whole point is that mod p commutes with multiplication and powers. For example, if a1=n1.p + r1, a2=n2.p + r2, then
a1 mod p = r1
a2 mod p = r2
and
a1.a2 mod p
= (n1.p+r1)(n2.p+r2) mod p
= (n1.n2.p + n1.r2+n2.r1)p+r1.r2 mod p
= r1.r1 mod p
= (a1 mod p).(a2 mod p) mod p
But I could be misinterpreting the question.
youngwerther
Banned
Blanking out really hard right now, can anyone explain how
√3 - √3/2 = √3/2
Those square roots only go in the numerator
√3 - √3/2 = √3/2
Those square roots only go in the numerator
You are literally subtracting half of the square root of 3 from the square root of 3, leaving...half of the square root of 3.
x minus half of x equals half of x. Just an even split.
youngwerther
Banned
Derp!!
Thanks.
Thanks.
Need help with eigenvalues. Just learning it and the professor didn't go over this type of example.
x' = {{1-4},{1-3}}x + {{2t},{3}} .
All I've done are x' = Ax problems, so I'm not sure how to use the second half. I solved for lamba which is -1, but then I'm not sure how to get the vectors.
x' = {{1-4},{1-3}}x + {{2t},{3}} .
All I've done are x' = Ax problems, so I'm not sure how to use the second half. I solved for lamba which is -1, but then I'm not sure how to get the vectors.
Sorry, I need to prove that the above equation is equal to
x =e^-t{{2t-2},{t-1}} + {{6t+2},{2t-1}} with an initial condition of x(0) = {{1},{2}}
Every eigenvalue question I've done has been in the form of x' = Ax so I'm not sure how to apply it to an equation like x' = Ax+b.
All I've done so far is find Lamba which I believe is -1. After that I'm not sure how to find the vectors.
Ugh. My problem with analysis is definitely that I just end up thinking about a problem for too long and getting lost in my own head. I almost made it all the way through a homework assignment before that happened this time, but one problem escaped me. I'm going to take a break from working on it to do other work and hopefully come back to it suddenly able to ace it, but in case I can't then I'll put it here and see if anyone wants to push me in the right direction.
I don't know what to do for "etc" because there are a lot of proof methods, but for the other two...
Direct Proof:
This is a proof done simply by combining established facts to reach a conclusion of truth or falsehood. A common example would give you two facts or truths:
1) If P is true, then Q is true
2) P is true.
And ask you to prove that Q is true.
Here's an example with words.
Contrapositive:
For a contrapositive proof, you're largely reaching the same goal of a direct proof by traveling a different path. You'll be asked to prove some statement "If P is true, then Q is true." But you won't do this by combining the statement "P is true" with other facts to obtain "Q is true." Instead, you begin with the statement "Q is false" and use that to prove that "P is false."
We can do the proof this way because "If Q is false then P is false" and "If P is true then Q is true" are logically equivalent statements. To put it into a word problem: "If the pencil is mine, then it is purple" is equivalent to "If the pencil is not purple, then it is not mine."
Word problems help me less when it comes to contrapositive though, I've always found it easier to understand the math. So let's try a math example.
Hope that helps some. If you're looking for more clarification, wikipedia is often a surprisingly helpful resource for logic. You could start at the article on direct proofs and go from there to all the kinds of indirect proofs, like proof by contrapositive.
I don't know what to do for "etc" because there are a lot of proof methods, but for the other two...
Direct Proof:
This is a proof done simply by combining established facts to reach a conclusion of truth or falsehood. A common example would give you two facts or truths:
1) If P is true, then Q is true
2) P is true.
And ask you to prove that Q is true.
Here's an example with words.
(1) If we have chocolate, we can make cookies.
(2) We have chocolate.
(3) Therefore, we can make cookies.
Lines (1) and (2) would be given to you. They would come from definitions or theorems. Here's a math example.(2) We have chocolate.
(3) Therefore, we can make cookies.
Prove that if two integers are odd, then their sum is even.
(1) x is an odd integer if and only if x=(2n+1) for some integer n. This is the definition of an odd integer.
(2) w is an even integer if and only if w=2j for some integer j. This is the definition of an even integer.
(3) The sum of integers is an integer. This is a property of the set of integers.
(4) Take any two odd integers y and z. Then we can say y=(2k+1) and z=(2m+1) for some integers m and k. Here we're simply using our given definition of an odd integer.
(5) Then y+z=(2k+1)+(2m+1)=2k+2m+2=2(k+m+1).
(6) Since k, m, and 1 are all integers, k+m+1 must be an integer. We know this because the sum of integers is also an integer.
(7) Since k+m+1 is an integer and y+z=2(k+m+1), y+z is an even integer.
Here, lines (1)-(3) were given to us. They are definitions and known properties of the sets we are dealing with. Lines (4)-(7) were the direct proof: combining those definitions to reach our desired conclusion.(1) x is an odd integer if and only if x=(2n+1) for some integer n. This is the definition of an odd integer.
(2) w is an even integer if and only if w=2j for some integer j. This is the definition of an even integer.
(3) The sum of integers is an integer. This is a property of the set of integers.
(4) Take any two odd integers y and z. Then we can say y=(2k+1) and z=(2m+1) for some integers m and k. Here we're simply using our given definition of an odd integer.
(5) Then y+z=(2k+1)+(2m+1)=2k+2m+2=2(k+m+1).
(6) Since k, m, and 1 are all integers, k+m+1 must be an integer. We know this because the sum of integers is also an integer.
(7) Since k+m+1 is an integer and y+z=2(k+m+1), y+z is an even integer.
Contrapositive:
For a contrapositive proof, you're largely reaching the same goal of a direct proof by traveling a different path. You'll be asked to prove some statement "If P is true, then Q is true." But you won't do this by combining the statement "P is true" with other facts to obtain "Q is true." Instead, you begin with the statement "Q is false" and use that to prove that "P is false."
We can do the proof this way because "If Q is false then P is false" and "If P is true then Q is true" are logically equivalent statements. To put it into a word problem: "If the pencil is mine, then it is purple" is equivalent to "If the pencil is not purple, then it is not mine."
Word problems help me less when it comes to contrapositive though, I've always found it easier to understand the math. So let's try a math example.
Prove that if the sum of two integers x and y is even, then they are either both even or both odd.
(1) The contrapositive of this is "If two integers x and y are not either both odd or both even (meaning one is odd and the other is even), then their sum is odd."
(2) Next we assume that of the two integers x and y, one is odd and the other is even. Let's say x is odd and y is even.
(3) If x is odd, then x=(2k+1) for some integer k.
(4) If y is even, then y=(2n) for some integer n.
(5) Then x+y=(2k+1)+(2n)=2k+2n+1=2(k+n)+1.
(6) Since k and n are both integers and x+y=2(k+n)+1, x+y is an odd integer..
(7) The contrapositive of our original statement is true. Therefore, our original statement is true.
(1) The contrapositive of this is "If two integers x and y are not either both odd or both even (meaning one is odd and the other is even), then their sum is odd."
(2) Next we assume that of the two integers x and y, one is odd and the other is even. Let's say x is odd and y is even.
(3) If x is odd, then x=(2k+1) for some integer k.
(4) If y is even, then y=(2n) for some integer n.
(5) Then x+y=(2k+1)+(2n)=2k+2n+1=2(k+n)+1.
(6) Since k and n are both integers and x+y=2(k+n)+1, x+y is an odd integer..
(7) The contrapositive of our original statement is true. Therefore, our original statement is true.
Hope that helps some. If you're looking for more clarification, wikipedia is often a surprisingly helpful resource for logic. You could start at the article on direct proofs and go from there to all the kinds of indirect proofs, like proof by contrapositive.
Thank you for your help, I think I understand that better. The big problem I am having is finding stuff like, x is irrational and the proof to show it is x = a/b*c/d. That right there makes zero sense to me. My professor does not explain why an equation like that is needed for that proof other than some guy from eons ago says it works.Ugh. My problem with analysis is definitely that I just end up thinking about a problem for too long and getting lost in my own head. I almost made it all the way through a homework assignment before that happened this time, but one problem escaped me. I'm going to take a break from working on it to do other work and hopefully come back to it suddenly able to ace it, but in case I can't then I'll put it here and see if anyone wants to push me in the right direction.
I don't know what to do for "etc" because there are a lot of proof methods, but for the other two...
Direct Proof:
This is a proof done simply by combining established facts to reach a conclusion of truth or falsehood. A common example would give you two facts or truths:
1) If P is true, then Q is true
2) P is true.
And ask you to prove that Q is true.
Here's an example with words.
(1) If we have chocolate, we can make cookies.Lines (1) and (2) would be given to you. They would come from definitions or theorems. Here's a math example.
(2) We have chocolate.
(3) Therefore, we can make cookies.
Prove that if two integers are odd, then their sum is even.Here, lines (1)-(3) were given to us. They are definitions and known properties of the sets we are dealing with. Lines (4)-(7) were the direct proof: combining those definitions to reach our desired conclusion.
(1) x is an odd integer if and only if x=(2n+1) for some integer n. This is the definition of an odd integer.
(2) w is an even integer if and only if w=2j for some integer j. This is the definition of an even integer.
(3) The sum of integers is an integer. This is a property of the set of integers.
(4) Take any two odd integers y and z. Then we can say y=(2k+1) and z=(2m+1) for some integers m and k. Here we're simply using our given definition of an odd integer.
(5) Then y+z=(2k+1)+(2m+1)=2k+2m+2=2(k+m+1).
(6) Since k, m, and 1 are all integers, k+m+1 must be an integer. We know this because the sum of integers is also an integer.
(7) Since k+m+1 is an integer and y+z=2(k+m+1), y+z is an even integer.
Contrapositive:
For a contrapositive proof, you're largely reaching the same goal of a direct proof by traveling a different path. You'll be asked to prove some statement "If P is true, then Q is true." But you won't do this by combining the statement "P is true" with other facts to obtain "Q is true." Instead, you begin with the statement "Q is false" and use that to prove that "P is false."
We can do the proof this way because "If Q is false then P is false" and "If P is true then Q is true" are logically equivalent statements. To put it into a word problem: "If the pencil is mine, then it is purple" is equivalent to "If the pencil is not purple, then it is not mine."
Word problems help me less when it comes to contrapositive though, I've always found it easier to understand the math. So let's try a math example.
Prove that if the sum of two integers x and y is even, then they are either both even or both odd.
(1) The contrapositive of this is "If two integers x and y are not either both odd or both even (meaning one is odd and the other is even), then their sum is odd."
(2) Next we assume that of the two integers x and y, one is odd and the other is even. Let's say x is odd and y is even.
(3) If x is odd, then x=(2k+1) for some integer k.
(4) If y is even, then y=(2n) for some integer n.
(5) Then x+y=(2k+1)+(2n)=2k+2n+1=2(k+n)+1.
(6) Since k and n are both integers and x+y=2(k+n)+1, x+y is an odd integer..
(7) The contrapositive of our original statement is true. Therefore, our original statement is true.
Hope that helps some. If you're looking for more clarification, wikipedia is often a surprisingly helpful resource for logic. You could start at the article on direct proofs and go from there to all the kinds of indirect proofs, like proof by contrapositive.
A lot of stuff like that on math. Stroke of genius or just plain luck. Most of the time there's no way to explain it, and certainly there's no method other than hard work (Is proven theorem
). Sorry I can't help you. All we can do is explain you why different ideas work and give you the ones we know or find, so bring concrete questions.smokeandmirrors
Banned
In regards to Linear Algebra including solving linear systems of equations and linear dependence/independence, everything seems so goddamned straightforward that I'm worried I'll slip up somehow. It's extremely similar to other math I've done before (including solving integration by partial fractions where we did solve linear systems, or balancing chemical equations and the like).
Am I right in thinking this, or am I really missing something crucial? I'd rather not fuck up on a test I have this week, so I'm actually looking through everything again and it just seems... so straightforward..
Am I right in thinking this, or am I really missing something crucial? I'd rather not fuck up on a test I have this week, so I'm actually looking through everything again and it just seems... so straightforward..
TimesEunuch
Member
Ugh. My problem with analysis is definitely that I just end up thinking about a problem for too long and getting lost in my own head. I almost made it all the way through a homework assignment before that happened this time, but one problem escaped me. I'm going to take a break from working on it to do other work and hopefully come back to it suddenly able to ace it, but in case I can't then I'll put it here and see if anyone wants to push me in the right direction.
This is a pretty long question. The most efficient way through it is to prove that (a) => (b) => (c) => (d) => (e) => (a). I'll try to offer some hints on each step.
(a) => (b): f(A int B) is a subset of f(A) int f(B) for any function (you should show this first). Now assuming f is 1-1, you need to prove the reverse inclusion. Take t in f(A) int f(B), then t=f(s), s in A, and t=f(s'), s' in B. Now use the 1-1 property...
(b) => (c): A is a subset of f^{-1}(f(A)) for any f (show this). Now suppose A =/= f^{-1}(f(A)), so there exists s in f^{-1}(f(A)) - A. Take B={s}, and show assumption (b) is violated.
(c) => (d): suppose A int B = null, but f(A) int f(B) =/= null. So there exists t in f(A) int f(B), which implies that t=f(s) some s in A, and t=f(s'), some s' in B. s and s' must be distinct points, since A int B = null. Now take any one of these points, use it as the set `A' in assumption (c), and show that you get a contradiction.
(d) => (e): f(B)-f(A) subset f(B-A) always (show this). Now consider the sets (B -A) and A, and use assumption (d) to show that f(B-A) subset f(B)-f(A).
(e) => (a): suppose two points in S map to the same one in T. Take A={first point}, B={second point}, show that assumption (e) is violated.
Hope this helps (and that I've made no mistakes
). There might be slicker ways to move between the various parts, so look out for that too.This is just what I was looking for! See, in my tiredness I didn't even realize I could just prove a) => ... => e) => a). I was going to do a) implies each one, then b implies a, and so on. These hints should do it. Thanks a lot.This is a pretty long question. The most efficient way through it is to prove that (a) => (b) => (c) => (d) => (e) => (a). I'll try to offer some hints on each step.
(a) => (b): f(A int B) is a subset of f(A) int f(B) for any function (you should show this first). Now assuming f is 1-1, you need to prove the reverse inclusion. Take t in f(A) int f(B), then t=f(s), s in A, and t=f(s'), s' in B. Now use the 1-1 property...
(b) => (c): A is a subset of f^{-1}(f(A)) for any f (show this). Now suppose A =/= f^{-1}(f(A)), so there exists s in f^{-1}(f(A)) - A. Take B={s}, and show assumption (b) is violated.
(c) => (d): suppose A int B = null, but f(A) int f(B) =/= null. So there exists t in f(A) int f(B), which implies that t=f(s) some s in A, and t=f(s'), some s' in B. s and s' must be distinct points, since A int B = null. Now take any one of these points, use it as the set `A' in assumption (c), and show that you get a contradiction.
(d) => (e): f(B)-f(A) subset f(B-A) always (show this). Now consider the sets (B -A) and A, and use assumption (d) to show that f(B-A) subset f(B)-f(A).
(e) => (a): suppose two points in S map to the same one in T. Take A={first point}, B={second point}, show that assumption (e) is violated.
Hope this helps (and that I've made no mistakes
TimesEunuch
Member
FWIW, I think proving (a) is equivalent to all the others (i.e. (a)<=>(b), (a)<=>(c) ... etc.) is conceptually easier, for me at least. I guess I think of 1-1ness (injectivity) as the most fundamental of the descriptions - I wouldn't be able to list off any of the other conditions without some thought - though maybe this is just years of conditioning. Doing things that way does almost double your work, however, so take your pick
Would data structures be considered part of a discrete mathematics curriculum?
The question uses the two as examples in a set problem. Given the sets of students studying discrete mathematics and students studying data structures, is either one a subset of the other? Obviously discrete is not a subset of structures, but I'm not sure what would be considered the components of a discrete math curriculum. Data structures is a separate course at my school, but I don't know if it's still considered part of discrete mathematics.
The question uses the two as examples in a set problem. Given the sets of students studying discrete mathematics and students studying data structures, is either one a subset of the other? Obviously discrete is not a subset of structures, but I'm not sure what would be considered the components of a discrete math curriculum. Data structures is a separate course at my school, but I don't know if it's still considered part of discrete mathematics.
Try rationalizing the numerator. or maybe denominator, play around with it.
Okay it's simple. Just find a common denominator of the numerator, which would be 9x , then you know make math. and you get (36-4x)/(9x)/(x-9) You can then factor the numerator and then you should know how to get the rest from here on out..
TimesEunuch
Member
Have you seen L'Hôpital's rule? If so, that's probably the intended way to solve the problem (and it takes one line).
krameriffic
Member
If I ever have to take a limit of an indeterminate expression, I simply fall back on l'Hospital's Rule:
http://en.wikipedia.org/wiki/L'Hôpital's_rule
hemtae
Member
I don't think he's got that far yet. That just looks like the definition of a derivative at a point. If he knew L'Hopital's rule then he could just take the derivative of f(x) and plug in 9.
TimesEunuch
Member
Good point <:-I
Partial Gamification
Banned
I need to prove this proposition using induction:
If n is greater than/equal to 2 and is an integer, then n can be written as a product of prime number.
So far I have P: n can be written as a product of prime numbers
But I'm struggling with the rest.
I am confused, "n can be written as a product of primes, or is itself prime[?]."
This is to say that prime numers are products of themselves and one (not a prime).
So n can be; prime, even, or odd-not-prime.
Looking at these integers,
2 , 3, 2*2, 5, 3*2, 7, 2*2*2, 3*3, 2*5, 11, 2*2*3, 13, ..., n, n+1
The base cases 2 and 3 are trivial; so, looking at n=4, it is known 2*2=4
The inductive case has n+1 being even or odd.
If you are still stuck or this doesn't point you in the right direction, the proof is available here. Do cite your work, even in mathematics.
So I've tried to get a hold of my math instructor a few times here in order to be prepped for my test Saturday. Basically I'm having issues knowing how to properly plug in numbers into a D=R*T formula.
Here's an example problem of which I'm having issues with:
Carol and Richard are travelling North in separate cars on the same highway. Carol is travelling at 65 miles per hour, Richard is travelling at 70 miles per hour. Carol passes Exit 102 at 1:30PM. Richard passes the same exit at 1:45PM. At what time will Richard catch up to Carol?
Another example would be:
In practicing maneuvers, two fighter jets fly toward each other. One flies east at 582 miles per hour and the other flies west at 625 miles per hour. If the two planes are 22 miles apart, how much time will it take for them to meet?
Once I can do these types of problems, then I'm prepared for my test. Thanks to anyone who can help out.
Here's an example problem of which I'm having issues with:
Carol and Richard are travelling North in separate cars on the same highway. Carol is travelling at 65 miles per hour, Richard is travelling at 70 miles per hour. Carol passes Exit 102 at 1:30PM. Richard passes the same exit at 1:45PM. At what time will Richard catch up to Carol?
Another example would be:
In practicing maneuvers, two fighter jets fly toward each other. One flies east at 582 miles per hour and the other flies west at 625 miles per hour. If the two planes are 22 miles apart, how much time will it take for them to meet?
Once I can do these types of problems, then I'm prepared for my test. Thanks to anyone who can help out.
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