layout(matrix(c(1,2), 1,2, byrow=TRUE))

# http://www.stat.umn.edu/geyer/5102/examp/reg.html
qftn <- function(x) x^2 * 3
curve(qftn, 1, 3, main = "Function: y = x^2 * 3") 

x <- c(1:3)
y <- c(3, 12, 27)
plot(x, y, main="Fitting model: y = x^2 * 3")
model <- lm(y ~ I(x^2 * 3))
curve(predict(model, newdata = data.frame(x = x)), add = TRUE)
# May also draw the coordinate pairs, like so:
text(1,4, "(1,3)")
text(2,13, "(2,12)")
text(2.9,27, "(3,27)")

# posted by rot13(Unafba Pune) @ 6:48 PM 0 comments

Prove that if every even natural number greater than 2 is a sum of two primes (the Goldbach Conjecture), then every odd natural number greater than 5 is a sum of three primes.

This is equivalent to asking if every even natural number greater than 2 is a sum of two primes,

$$\begin{aligned} \forall{(n \in \mathbb{N} \land n > 1)} \exists{(p,q \in \mathbb{P})}[p + q & = 2n] \end{aligned}$$

is it true that:

$$\begin{aligned} \forall{(m \in \mathbb{N} \land m > 2)} \exists{(r,s,t \in \mathbb{P})}[r + s + t & = 2m + 1] \end{aligned}$$

Exploiting m > 2 and n > 1:

$$\begin{aligned} \forall{(m \in \mathbb{N} \land m > 2)} \exists{(x \in \mathbb{N})}[x & = m - 1 > 1] \\ m & = x + 1 > 2 \end{aligned}$$

Now,

$$\begin{aligned} 2m + 1 & = 2(x + 1) + 1 \\ & = 2x + 3 \\ \end{aligned}$$

But x is greater than 1 which means 2x is an even number greater than 2, and therefore is a sum of two primes. Since 3 is prime, (2m + 1), an arbitrary odd natural number greater than 5, is therefore a sum of three primes.

# posted by rot13(Unafba Pune) @ 8:30 AM 0 comments

Prove that there is a quadratic $f(n) = n^2 + bn + c$, with positive integers coefficients b, c, such that $f(n)$ is composite (i.e, not prime) for all positive integers n, or else prove that the statement is false.

# posted by rot13(Unafba Pune) @ 11:26 PM 0 comments

Prove or disprove the statement “An integer n is divisible by 12 if and only if $n^3$ is divisible by 12.”

If integer n is divisible by 12, for some integer m:

$$\begin{aligned} n & = 12m \\ n^3 & = 12^3m^3 \end{aligned}$$

Clearly, $n^3$ is divisible by 12. How about the converse ? Is n divisible by 12 if $n^3$ is divisible by 12 ? Note $12 = 2^2 \times 3$, so one counter example would be:

$$\begin{aligned} n^3 & = 2^3 \cdot 3^3 = 12 \cdot 18 \\ n & = 2 \cdot 3 = 6 \end{aligned}$$

In other words, $n^3$ is divisible by 12, but not n. Hence the converse is false.

# posted by rot13(Unafba Pune) @ 9:16 PM 0 comments

Prove that $\sqrt{3}$ is irrational.

First, let's prove the lemma that if the square of an integer n is divisible by 3, n must be divisible by 3. Why ? A integer can be written in only one of 3 forms:

$$\begin{aligned} Case \space 1: n & = 3x \\ Case \space 2: n & = 3x + 1 \\ Case \space 3: n & = 3x + 2 \\ \end{aligned}$$

where x is an integer. If we take the square of n in each case:

$$\begin{aligned} Case \space 1: n^2 & = (3x)^2 = 9x^2 \\ Case \space 2: n^2 & = (3x + 1)^2 = 9x^2 + 6x + 1 = 3(3x^2 + 2x) + 1 \\ Case \space 3: n^2 & = (3x + 2)^2 = 9x^2 + 12x + 4 = 3(3x^2 + 4x + 1) + 1 \end{aligned}$$

Note that only in case 1 when n is divisible by 3 would $n^2$ be divisible by 3. This completes the proof of the lemma.

Now suppose $\sqrt{3}$ is rational,

$$\exists{p,q \in \mathbb{N}}[\sqrt{3} = \frac{p}{q}]$$

such that p and q has no common factors. Squaring both sides,

$$\begin{aligned} 3 & = \frac{p^2}{q^2} \\ 3q^2 & = p^2 \end{aligned}$$

This means $p^2$ is divisible by 3. By the above lemma, this means p must be divisible by 3. Or p = 3r for some integer r.

$$\begin{aligned} 3q^2 & = p^2 = (3r)^2 = 9r^2 \\ q^2 & = 3r^2 \end{aligned}$$

This means $q^2$ is divisible by 3, and by the lemma above, q must be divisible by 3. To summarize, both p and q are divisible by 3, but this contradicts the initial assumption that p and q has no common factor!

Hence $\sqrt{3}$ is irrational.

# posted by rot13(Unafba Pune) @ 10:49 PM 0 comments

Prove or disprove the claim that for any positive integer m there is a positive integer n such that mn + 1 is a perfect square.

First, the claim can be expressed as:

$$\forall{(m,n \in \mathbb{N})} \exists {(r \in \mathbb{Z})} [m \cdot n + 1 = r^2]$$

Observe that:

$$\begin{aligned} m \cdot n + 1 & = r^2 \\ m \cdot n & = r^2 - 1 \\ m \cdot n & = (r + 1)(r - 1) \\ n & = \frac{(r + 1)(r - 1)}{m} \\ \end{aligned}$$

This means if the claim is true, m must divide either (r + 1) or (r - 1):

$$\begin{aligned} \exists{(x \in \mathbb{N})} [(m \cdot x = r + 1) \lor (m \cdot x = r - 1)] \end{aligned}$$

Or,

$$\begin{aligned} \exists{(x \in \mathbb{N})} [(r = m \cdot x - 1) \lor (r = m \cdot x + 1)] \end{aligned}$$

For example, when m = 1, the smallest value of r such that n > 0 would be when x = 3. So:

$$\begin{aligned} r & = 1 \cdot 3 - 1 & = 2 \\ n & = \frac{(2 + 1)(2 - 1)}{1} & = 3 \\ m \cdot n + 1 & = 1 \cdot 3 + 1 & = 2^2 \end{aligned}$$

Given m >= 1, we can always pick a value x such that:

$$\begin{aligned} r = m \cdot x - 1 >= 2 \end{aligned}$$

This is when:

$$\begin{aligned} x >= \frac{3}{m} \end{aligned}$$

which means we can always find a value r (from m and x) and n (from r and m) that satisfies the claim. This completes the proof.

# posted by rot13(Unafba Pune) @ 4:22 PM 2 comments

Ask the above question ! Or, in short, recursion :)

# posted by rot13(Unafba Pune) @ 9:22 AM 0 comments