Recent content by IKonquer
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What is the identity element in abstract algebra groups?
A + A' is the symmetric difference, and not by means of normal addition.- IKonquer
- Post #3
- Forum: Calculus and Beyond Homework Help
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Lets say you have a set S = {1, 2, 3}
Lets say you have a set S = {1, 2, 3} The subsets would be the following: {} (nullset) {1}, {2,}, {3} {1, 2}, {2, 3}, {1, 3} {1, 2, 3} Would {1, 2} be considered the same subset as {2, 1}?- IKonquer
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- Set
- Replies: 2
- Forum: Calculus and Beyond Homework Help
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What is the identity element in abstract algebra groups?
The .pdf can be ignored. Let A + B = (A - B) U (B - A) also known as the symmetric difference. 1. Look for the identity and let e be the identity element A + e = A (A - e) U (e - A) = A Now there are two cases: 1. (A - e) = A This equation can be interpreted as removing from A all elements...- IKonquer
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- Abstract Abstract algebra Algebra Groups
- Replies: 4
- Forum: Calculus and Beyond Homework Help
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Solving for Identity of x*y = x + 2y - xy
I see, |x+0| = |x| = +x or -x. As a result there should be no identity. So is it usually the case that most operations don't have an identity?- IKonquer
- Post #5
- Forum: Calculus and Beyond Homework Help
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Solving for Identity of x*y = x + 2y - xy
Wow. Thanks for the quick response micromass! :smile: What about this one? x*y = |x+y| To find the identity, x*e = x x*e = |x + e| = x The absolute value can be broken up into two cases: 1) x + e = x e = 0 In order to be an identity x*e =x and e*x = x x*0 = |x+0| = x 0*x...- IKonquer
- Post #3
- Forum: Calculus and Beyond Homework Help
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Solving for Identity of x*y = x + 2y - xy
I'm having trouble finding the identity of an operation. Could someone check my work? I'm trying to find the identity of x*y = x + 2y - xy In order to find the identity, I need to solve x*e = x for e \begin{align*} x*e &= x\\ x + 2e - xe &= x\\ 2e - xe &= 0\\ e(2-x) &= 0\\...- IKonquer
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- Identity
- Replies: 5
- Forum: Calculus and Beyond Homework Help
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Becoming Fluent in Math Proofs - Tips & Advice
What a coincidence! I actually have both Velleman's book and Abstract Algebra by Pinter. I actually started reading Velleman's book from the beginning and stopped. I didn't really understand the importance of unions, intersections, truth tables, and the whole operation on sets. Could you explain...- IKonquer
- Post #5
- Forum: STEM Academic Advising
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Becoming Fluent in Math Proofs - Tips & Advice
Hi micromass, thanks for the advice. I'm going to be taking abstract algebra next semester and I have had some exposure to the ideas of proof by induction, contradiction, truth tables, and the like in high school. But I've never taken an introduction to proofs class in college. Do you think it...- IKonquer
- Post #3
- Forum: STEM Academic Advising
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Becoming Fluent in Math Proofs - Tips & Advice
I'm starting to learn how to write proofs, and I am wondering how to become fluent in proofs. Is it necessary to do problems that are IMO/Putnam? Can anyone give me some advice? Thanks in advance.- IKonquer
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- Fluent Proofs
- Replies: 6
- Forum: STEM Academic Advising
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Understand Momentum & Energy Conservation in Ballistic Pendulum Problems
In solving a ballistic pendulum problem, you can break it up into two parts: (1) A bullet is fired and it lodges into a block. Momentum is conserved because there is no net external force. Energy is not conserved because there is friction between the block and bullet. (2) Once the bullet is...- IKonquer
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- Ballistic Ballistic pendulum Pendulum
- Replies: 1
- Forum: Introductory Physics Homework Help
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Checking operations over a set
"Yes. First use x * e = x to solve for the identity, then you check whether this identity is valid in e * x = x. If it is not, then you have shoved that the operation does not form a group structure on the set, and you're done." Right this is what my book tells me to do. But my question is why...- IKonquer
- Post #6
- Forum: Precalculus Mathematics Homework Help
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Checking operations over a set
Wingeer I am trying to show that the operation forms a group structure over the set. When finding the identity, why should it be x * e = x and not e * x = x? I know I have to check both, but when finding the identity, my book tells me to use x * e = x to solve for the identity. x * y = x +...- IKonquer
- Post #4
- Forum: Precalculus Mathematics Homework Help
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Checking operations over a set
I am trying to the following operation * is closed under R (the real numbers). x * y = x + 2y + 4 Check commutability - NOT commutative x * y = x + 2y + 4 y * x = y + 2x + 4 Check associativity - NOT associative x * (y * z) = x * (y + 2z + 4) = x + 2(y + 2z + 4) + 4 = x + 2y + 4z +...- IKonquer
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- Operations Set
- Replies: 5
- Forum: Precalculus Mathematics Homework Help
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How Effective Are Proof Books for Students New to Abstract Algebra?
Has anyone read How to Prove It: A Structured Approach by Daniel Velleman or Introduction to Mathematical Thinking: Algebra and Number Systems by Gilbert and Vanstone? If so, how well do they prepare a student who has had no exposure to proofs to take classes such as abstract algebra? Thanks...- IKonquer
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- Books Introduction Proof
- Replies: 2
- Forum: Science and Math Textbooks
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Atoms decay via alpha emission have a half-life of 150 min
Nice catch! Does it look right now? Also is there a way of using integral calculus to solve this? \begin{flalign*} 150 &= \frac{\ln 2}{\lambda}\\ \lambda &= 0.0046 \\ \\ K &= K_{0}e^{(-\lambda)(t)}\\ &= (2.3 \cdot 10^{10})e^{(-0.0046)(30)}\\...- IKonquer
- Post #3
- Forum: Introductory Physics Homework Help