Hey everyone,

I posted a new page called Projects which describes the subjects that I'm learning during my spare time. You can locate the page at the top of this site (it's the third tab),

Enjoy!

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Hey everyone,

I posted a new page called Projects which describes the subjects that I'm learning during my spare time. You can locate the page at the top of this site (it's the third tab),

Enjoy!

Posted in: Logs

Hey everyone,

Today's problem is a proof derived from mathematical fallacy. See if you can spot the error! (Hint: you're not Chuck Norris).

Find the error in the following "proof." Let x = y. Then:

$latex displaystyle begin{aligned} x^2&=xy,\ x^2-y^2&= xy-y^2,\ (x+y)(x-y)&= y(x-y),\ x+y&=y,\ 2y&=y,\ 2&=1. end{aligned}$

We know that something is obviously wrong with this proof since we know that 2 ≠ 1. It took me a while to solve this, but after a few minutes of staring at the problem I think I've found the solution. Consider the following lines:

$latex displaystyle (x+y)(x-y) = y(x-y),$

$latex displaystyle x+y=y$

Here, the proof writer *attempts to divide by zero.* In other words:

$latex displaystyle (x+y)(x-y)(x-y)^{-1} = y(x-y)(x-y)^{-1}$

However, x = y, x - y = 0, so:

$latex displaystyle (x-y)^{-1}=text{undefined}.$

I don't have a solutions manual for these problems, so feel free to disagree if you're not convinced. I am, however, quite certain that this is a correct solution.

Note: I used a new command today, the begin{aligned} environment to align the equal signs in my first box of $latex LaTeX$ equations, and the target="blank" attribute to open up links in a new window.

Posted in: Logs

Hey everyone,

I didn't write a problem yesterday, so today I'm going to write about two famous unsolved problems in mathematics: the Twin Prime Conjecture and Goldbach's Conjecture. These two problems are so famous because they're simply stated, yet so seemingly difficult to solve.

A twin prime is a prime number that differs from another prime number by two. A twin prime can refer to a single prime number with this property, or it can refer to a pair of prime numbers that differ by two. Are there an infinite number of twin primes? The **Twin Prime Conjecture** states that there are infinitely many primes p such that p+2 is also prime. Is the conjecture true?

As of today, there are no proofs of the Twin Prime Conjecture accepted by the mathematical society. This is one of the oldest unsolved problems in mathematics and no one (that we know of) has been able to solve it for thousands of years! It's my wild guess that there are an infinite number of twin primes. Mathematicians have made some progress, and have shown that there are infinitely many n such that at least two of n, n+2, n+4, and n+6 are prime.

**Goldbach's Conjecture** states that *every even number greater than two can be written as the sum of two primes. *Is Goldbach's Conjecture true?

Again, no one knows the answer to this conundrum. There's a large prize of $1,000,000 (I fathom much more has been spent trying to solve the problem) offered to the solver! The term "solver," however, is erroneous, because I suspect the final proof will have been the cumulative result of several mathematicians' works over several centuries.

I don't want to close this post without solving a problem, so today I'll write a bonus problem proving the fact that the product of two negative numbers is a positive number.

Prove that the product of two negative numbers is a positive number.

If *a* and *b* are any numbers, then:

$latex displaystyle (-a) cdot (-b) +[-(a cdot b)] = (-a) cdot (-b) + (-a)cdot b$

(Associative Property)

$latex displaystyle (-a) cdot (-b) +[-(a cdot b)] = (-a) cdot [(-b)+(b)]$

(Distributive Property)

$latex displaystyle (-a) cdot (-b) +[-(a cdot b)] = (-a) cdot 0$

(Since -*b* + *b* = 0)

$latex displaystyle (-a) cdot (-b) +[-(a cdot b)] = 0$

(Since -*a* x 0 = 0)

$latex displaystyle (-a) cdot (-b) = (a cdot b)$

(Add *a* x *b* to both sides)

That's why the product of two negative numbers is a positive number! Ever since kindergarten I've taken this fact for granted, but I'm grateful for coming across this proof now, for the sake of better understanding.

Note: I used a new command today, cdot, which specifies the dot commonly used to denote multiplication.

Posted in: Logs

On January 1, 2010, Sergei signed a one-year Cable TV contract with Comcast, which specifies that he provide payments of $45 at the end of each month for 12 months, with the first payment beginning at the end of the month. If he had instead deposited these 12 monthly payments into a savings account at an annual effective interest rate of 4%, with interest paid at the end of every month, how much money would he have in the account by the end of the year?

The problem gives us an annual effective interest rate *i*=4%, but we need to know the monthly effective interest rate *i*^{(12)}/12 because the payments occur every month. To find the monthly effective interest rate, we need to use the relation:

$latex (1+i)=(1+frac{i^{(m)}}{m})^{(m)}$

The left side of the equation tells us how much an account governed by an annual effective rate of interest *i* grows in one year. For instance, an account governed by an annual effective interest rate i will grow by (1+*i*)x(100%) each year. The right side of the equation tells us how much an account governed by an interest rate of *i*^{(m)}/*m* per *m*-th of a year grows in one year. Using algebra to rearrange the equation, we have:

$latex displaystyle (1+i)^{frac{1}{m}}-1=frac{i^{(m)}}{m}$

By setting *m*=12, we can use the relationship to find the annual effective interest rate equivalent to the monthly effective rate:

$latex displaystyle (1+i)^{frac{1}{12}}-1=frac{i^{(12)}}{12}$

Setting the annual effective rate *i*=4%=.04 yields us:

$latex displaystyle (1+.04)^{frac{1}{12}}-1=frac{i^{(12)}}{12}approx 0.00327374$

Now that we have the monthly effective rate of interest, we can find the accumulated value of the twelve payments. Each of these payments is governed by the general accumulation formula, which yields us the total accumulated value of:

$latex displaystyle 45(1+frac{i^{(12)}}{12})^{11}+45(1+frac{i^{(12)}}{12})^{10}+dotsm+45(1+frac{i^{(12)}}{12})^1+45(1+frac{i^{(12)}}{12})^0$

Factoring out the 45 gives us:

$latex displaystyle 45[(1+frac{i^{(12)}}{12})^{11}+(1+frac{i^{(12)}}{12})^{10}+dotsm+(1+frac{i^{(12)}}{12})^1+(1+frac{i^{(12)}}{12})^0]$

We can solve this sum right now if we wanted, but the calculation is tedious if we do it now. However, there is a way to simplify the terms inside the square brackets. Notice that the sum inside the square brackets:

$latex displaystyle (1+frac{i^{(12)}}{12})^{11}+(1+frac{i^{(12)}}{12})^{10}+dotsm+(1+frac{i^{(12)}}{12})^1+(1+frac{i^{(12)}}{12})^0$

Is equal to a geometric sum beginning with 1, with common ratio (1+*i*^{(12)}/12):

$latex displaystyle sum_{k=0}^{11} (1+frac{i^{(12)}}{12})^{k}=frac{(1+frac{i^{(12)}}{12})^{12}-1}{frac{i^{(12)}}{12}}$

Multiplying by 45, and substituting the value we found for *i*^{(12)}/12 gives us the final result:

$latex displaystyle 45sum_{k=0}^{11} (1+frac{i^{(12)}}{12})^{k}=45left(frac{(1+frac{i^{(12)}}{12})^{12}-1}{frac{i^{(12)}}{12}}right)approx 549.8298948$

Note: I used a few new commands here, such as dotsm for center ellipsis, sum for summation, and p style="text-align:center" since the new version of html will get rid of the old align="center" tag.

Posted in: Logs

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