In part (c) in line 3 the inequality should be (1-frac{1}{p}le 1-frac{1}{sqrt{n}}) not  (1-frac{1}{p}le 1-sqrt{n})…..

(a) Let (n=2^{k_0}p_1^{k_1}p_2^{k_2}cdots p_r^{k_r}) . So, (phi(n)=2^{k_0-1}p_1^{k_1-1}p_2^{k_2-1}cdots p_r^{k_r-1}(2-1)(p_1-1)(p_2-1)cdots (p_r-1)) . Now, we use the inequalities (k-frac{1}{2}ge frac{k}{2})  and (p-1>sqrt{p}) for (p>2). So, we have (phi(n)le 2^{k_0-1}p_1^{k_1/2}p_2^{k_2/2}cdots p_r^{k_r/2}ge frac{1}{2}sqrt{n})  .

Now, also (…Read More

(b) Let (n=prod_{k=1}^r p_k^{a_k}=p_1^{a_1}p_2^{a_2}p_3^{a_3}cdots p_r^{a_r}) . So, (phi(n) =nleft(1-frac{1}{p_1}right) left(1-frac{1}{p_2}right)cdots left(1-frac{1}{p_r}right)) .

For any prime (p)  the inequality $$2p-2geq p$$ or, $$frac{p-1}{p}geq frac{1}{2}$$ So, we have (r) prime number for which the inequality is true. Multiplying them gi…Read More

Notice that (|Atimes B|=mn). Also, (|C_i|=n) for all (i=1,2,3,cdots ,m). hence (sum_{i=1}^m |C_i|=sum_{i=1}^m n=mn). Similarly, we can prove that (sum_{j=1}^n |D_j|=mn).

And hence, we get our desired result.

This is just proving that:

Theorem: 

Given any three non-collinear points (A, B, C)  there exists a unique circle passing through (A, B, C).

Proof: 

Let us consider three vertices of (Delta ABC) i.e. (A, B, C). Suppose the perpendicular bisectors of (BC) and (CA) meet at (S). Then (S) lies on the perpendicular bisector of (BC) implies that (S…Read More