[swastik pramanik]

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

[swastik pramanik]

(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 \(p-1<p\)  . So, \(\phi(n)\le 2^{k_0}p_1^{k_1}p_2^{k_2}\cdots p_r^{k_r}=n\)  .

Combining them up we get $$\frac{1}{2}\sqrt{n}\le \phi(n)\le n$$ as required.

 

(c) Let \(p\) be the smallest prime divisor of \(n\) so that, \(p\le \sqrt{n}\).  Then \(\phi(n) \le n\left(1-\frac{1}{p}\right)\) . We know that, \(p\le \sqrt{n}\) which implies \(1-\frac{1}{p}\le 1-\sqrt{n}\). So, \(\phi(n) \le n\left(1-\frac{1}{p}\right)\le n\left(1-\frac{1}{\sqrt{n}}\right)=n-\sqrt{n}\).

Hence, we are done!  🙂

[swastik pramanik]

(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, \(\phi(n) =n\left(1-\frac{1}{p_1}\right) \left(1-\frac{1}{p_2}\right)\cdots \left(1-\frac{1}{p_r}\right)\) .

For any prime \(p\)  the inequality $$2p-2\geq p$$ or, $$\frac{p-1}{p}\geq \frac{1}{2}$$ So, we have \(r\) prime number for which the inequality is true. Multiplying them gives $$\left(1-\frac{1}{p_1}\right) \left(1-\frac{1}{p_2}\right)\cdots \left(1-\frac{1}{p_r}\right)\geq \frac{1}{2^r}$$ multiplying both sides with \(n\) gives So, we want to show that $$n\left(1-\frac{1}{p_1}\right) \left(1-\frac{1}{p_2}\right)\cdots \left(1-\frac{1}{p_r}\right)\geq \frac{n}{2^r}$$ or concisely $$\phi(n)\geq \frac{n}{2^r}$$ as required.

[swastik pramanik]

Notice that \(|A\times B|=mn\). 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.

[swastik pramanik]

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 \(SB=SC\). Again, \(S\) is also a perpendicular bisector of \(CA\)  which implies that \(SC=SA\). Hence, we have \(SA=SB=SC\). Further more, any point equidistant from \(A, B, C\)  should lie on the perpendicular bisectors of \(BC, CA\). Therefore, \(S\) is the only point equidistant from \(A, B, C\) and so the circle with centre \(S\) and radius \(SA\) is the unique circle passing through \(A, B, C\).

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