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Qn 1. Say you choose a sequence that has as its limit.

Obviously none of the denominators will be even tending to zero.

Qn 2. See, wehn you take LCM, every numerator except the last will be multiplied by an even number. The last numerator is 1. So, the numerator is odd.

the denom will be in general 2^kq where k is large. So, the sequence of denominators does not tend to q.

$n>N^2 \Rightarrow \sqrt {n+1} > \sqrt n > N \\ \\<br/>\therefore \sqrt {n+1}+\sqrt n > 2N \\ \\<br/>\therefore \sqrt {n+1} - \sqrt n = \frac{1}{\sqrt {n+1}+\sqrt n } < \frac{1}{2N}$

PS: Your question is tough to read. Make use of the editor please

This is a shot in the dark, but consider that a rational number p/q is the limit of the sequence

$\frac{p}{2q} \left( 1 + \frac{1}{2} + \frac{1}{2^2} + ...\infty} \right)$

The numerator of the bracketed expression is an odd number and the denominator is 2ⁿ.

So, again, we see that the denominator does not tend to q for the members of the sequence.

So, for rational points also it is discontinuous.

The basic concept is that Dedekind (if I remember right) defined irrational numbers as being the limit of a series of rational numbers.

Next, there is a definition of Heine Continuity of a function, that a function f is continuous if for every sequence $x_1, x_2, x_3,..., x_n \rightarrow x$ , then $f(x_1), f(x_2), f(x_3),..., f(x_n) \rightarrow f(x)$

So, from this, if you consider a sequence of rational numbers that in the limit is an irrational number, the limit of the sequence is definitely non-zero. However we are given that f(x) = 0 if x is an irrational number.

From the above definition, the function becomes discontinuous at irrational points.

The rational points one, I am not sure, gotta think

Correction: the first ordered pair is (1,1) and not (1,11)

This is a nice problem. Only this is not the typical JEE kind. Not, that I discourage you guys from trying out such problems, but you are much better off doing Calculus and Coord Geometry than these.

I am solving for x and y both positive integers here.

Ok, we are given that both $x^2 + 3y \ \text{and} \ y^2 + 3x$ are perfect squares.

Now, with x² of course being a square, adding 3y makes it a perfect square. That means 3y must be of the form 2mx + m^2 where m is a positive integer as x^2 + 2mx + m^2 = (x+m)^2 is a perfect square. Similarly we have the relation 3x = 2ny + n^2

Thus, we have the simulataneous equations

3x = 2ny + n^2

3y = 2mx + m^2

Eliminating y we get $x = \frac{m(3m+2n^2)}{9-4mn}$

m and n are positive. Since x>0, we must have 9>mn

Hence, the possibilities are

1. m = n =1

2. m=2, n=1

3.m=1, n=2

For these we get the ordered pairs (1,11), (16,11) and (11,16) as the only solutions among positive integers.

take that back: i am not sure anymore that x and y have to be relatively prime

I think he means that all numbers involved are integers. If x= y, you can prove that the only solutions is x = y =1.

You can even prove that x and y have got to be relatively prime.

First you have to establish the inequality

$\sin A+ \sin B + \sin C \le \frac{3 \sqrt 3}{2}$

This easily done as sine is a concave function in the interval $[0, \pi]$ as f"(x) <0 in this interval

Hence, by Jensen's Inequality which states that for a concave function,

$\omega_1f(x_1)+\omega_2 f(x_2)+...+ \omega_n f(x_n) \le f(\omega_1 x_1 + \omega_2 x_2+...+\omega_n x_n) \\ \\<br/>\text{where} \ \sum_{i=1}^n \omega_i = 1$

So, $\frac{\sin A + \sin B + \sin C}{3} \le \sin {\left (\frac{A+B+C}{3} \right )}$

or $\sin A+ \sin B + \sin C \le \frac{3 \sqrt 3}{2}$ as A+B+C = $\pi$

Now, from AM-GM Inequality, $\left (\sin A + \sin B + \sin C \right ) \left ( \frac{1}{\sin A} +\frac{1}{\sin B}+\frac{1}{\sin C} \right) \ge 9$

$\Rightarrow \left ( \frac{1}{\sin A} +\frac{1}{\sin B}+\frac{1}{\sin C} \right) \ge \frac{9}{\sin A + \sin B + \sin C} \ge \frac{9} { \frac{3 \sqrt 3}{2}} = 2 \sqrt 3$

That's a valid question which I have failed to address. Thank you for pointing it out.

Again go back to the equation

9a^2+14b^2 = 9abn

Suppose b = 9. Then you can divide by 9 throughout and you would get

a²+14x9 = 9an

a is not a multiple of 9, but the other terms are. So you obtain a contradiction.

Sorry, my post is long overdue.

Ok, you must first rewrite the equation as

9a^2+14b^2 = 9abn

Now, 14b² and 9abn are both divisible by b. Hence 9a² must also be divisible by b

Now since gcd(a,b) = 1, b divides 9a² means, b divides 9. Also, 9 must divide b². The only possibility is b = 9.

Similarly you conclude that a divides 14.

So a = 1,2, 7 or 14 and b = 3.

Plugging in these you see that all the pairs (1,3), (2,3), (7,3) and (14,3) are solutions and no other solutions exist.

A small clarification: I said: We can eliminate the second quadratic by taking b = c = 0.

By this I meant considering the given quadratic to be n²

Then n²(n+1)² = [n(n+1)]²which does not hold true if you choose the second quadratic.

While of course, the brute force method is available to us to find out the answer, there is an elegant way of solving:

(n^2+bn+c)((n+1)^2+b(n+1)+c) = m^2+bm+c where m is a in integer.

This has to be true for any n.

That means we must have m = Q(n). So, we rewrite the equation as

(n^2+bn+c)((n+1)^2+b(n+1)+c) = Q^2(n)+bQ(n)+c

By comparing the polynomials on both sides, we can see that Q(n) must be a monic quadratic polynomial (i.e. a quadratic with coefficient 1).

Since, the expression is true for infinitely many integers, it is true for any real number. So, we can replace n by x where x is a real number. So, we rewrite as:

(x^2+bx+c)((x+1)^2+b(x+1)+c) = Q^2(x)+bQ(x)+c

Suppose, $\alpha \ \text{and} \ \beta$ are the roots of the given quadratic x²+bx+c = 0. Then, the RHS must disappear for precisely those numbers.

i.e. $Q(\alpha) \ \text{and} \ Q(\beta)$ are the roots of the equation x²+bx+c = 0. But $\alpha \ \text{and} \ \beta$ are the only roots of the quadratic