January « 2008 « jkwiens.com

Archive for January, 2008

Question: Voltage and Current of a Circuit

January 26th, 2008

Question: Calculate the current and voltage difference between points D and B in the following circuit.

Comments(3) - Electricity and Magnetism, Questions

Answer: Prove function has even values

January 25th, 2008

Question: Prove that the function n^2 -n has even values for all natural numbers ( $\mathbb{N} = \{0, 1, 2, 3 \ldots\}$ ).

* Updated question for clarity.

Answer: Solving this problem is really trivial. The easiest way (by far) to solve this problem is to use Mathematical induction.

Wikipedia describes Mathematical Induction as:

The simplest and most common form of mathematical induction proves that a statement holds for all natural numbers n and consists of two steps:

The basis: showing that the statement holds when n = 0.

The inductive step: showing that if the statement holds for n = m, then the same statement also holds for n = m + 1.

This is exactly what we need to do to solve the problem. First, we need to prove that f(0) is even. This can easily be done:

Therefore, since 0 is an even number, we have satisfied the first step.

The second step is to show that if f(n) is even than f(n + 1) is also even.

If we assume f(n) is even, we know that f(n) + 2n has to be even because even + even = even .

Therefore, the function f(n) = n^2 -n is even for all $n \ge 0$ because:

is even.
If is even, it implies that is even.

Comments(8) - Algebra, Answers

Latex Text Editor

January 23rd, 2008

I found a cool widget that writes Latex at sitmo. I have gotten pretty good at writing latex since I started this blog, but I still seem to look a lot of things up. This works a lot better than a cheat sheet.

Comments(4) - Webapp

Question: Prove function has even values

January 19th, 2008

Question: Prove that the function n^2 -n has even values for all natural numbers ( $\mathbb{N} = \{0, 1, 2, 3 \ldots\}$ ).

* Updated question for clarity.

Comments(2) - Algebra, Questions

Answer: Binary operation Problem

January 19th, 2008

Question: Assume that a binary operation $\Box$ on a set has a left unit and satisfies the identity $x \Box (y \Box z) = (x \Box z) \Box y$ $\forall x,y,z \in X$ . Prove that $\Box$ is associative and commutative.

Answer – Proving that $\Box$ is commutative:
First, let x be equal to the left unit, x = u . Since the identity $x \Box (y \Box z) = (x \Box z) \Box y$ is true $\forall x \in X$ and since $u \in X$ because of the definition of a left unit, therefore $u \Box (y \Box z) = (u \Box z) \Box y$ .

Since $z \in X$ , which will mean $z = u \Box z$ and therefore

$u \Box (y \Box z) = (u \Box z) \Box y \Longrightarrow u \Box (y \Box z) = z \Box y$

Because $\Box$ is a binary operation, therefore $\Box : X \times X \rightarrow X$ . This means that $y \Box z \in X$ . Therefore, we know from the definition of a left unit that $u \Box (y \Box z) = y \Box z$ .

This means:

$u \Box (y \Box z) = z \Box y \Longrightarrow y \Box z = z \Box y$

$\therefore \Box$ is commutative.

Answer – Proving that $\Box$ is associative:
Since the identity $x \Box (y \Box z) = (x \Box z) \Box y$ is true $\forall x,y,z \in X$ , we can rename the variables while maintaining the validity of the identity.