NBHM 2020 PART A Question 10 Solution: Possible rank of a matrix obtained by changing the entries $(1, 3)$, $(1, 4)$, $(3, 3)$, $(3, 4)$, and $(5, 4)$

Let $M$ be a $7×6$ real matrix. The entries of $M$ in the positions $(1, 3)$, $(1, 4)$, $(3, 3)$, $(3, 4)$, and $(5, 4)$ are changed to obtain another $7×6$ real matrix $\widetilde{M}$. Suppose that the rank of $\widetilde{M}$ is 4. What could be the rank of $M$? List all possibilities.
Solution:
First, we will derive a connection between the rank of $M$ and the rank of $\widetilde{M}$. We have,
$$rank(M) = rank(M-\widetilde{M}+\widetilde{M}) \le rank(M-\widetilde{M})+rank(M)$$ Similarly
$$rank(\widetilde M) = rank(\widetilde{M}-M+M) \le rank(\widetilde{M}-M)+rank(M)$$
This shows that $rank(\widetilde{M}) - rank(\widetilde{M}-M) \le rank(M) \le rank(M-\widetilde{M})+rank(\widetilde{M}) $
The matrices $M - \widetilde{M}$ and $\widetilde{M}-M$ can have non-zero entries only in the columns $3$ and $4$ and hence they are of rank atmost $2$. Also, it is given that the rank of $\widetilde{M}$ is equal to $4$. Hence we have $$2 \le rank(M) \le 6.$$
Using $0,1$ matrics one can check that all these possibilities are occurring indeed.
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NBHM 2020 PART A Question 12 Solution

Let $k$ be the field obtained by adjoining to the field $\Bbb Q$ of rational numbers the roots of the polynomial $x^4−2$. Let $k^{'}$ be the field obtained by adjoining to $k$ the roots of the polynomial $x^4 + 2$. What is the degree of $k^{'}$ over $k$?
Solution:
The roots of the polynomial $x^4 - 2 = (x^2-\sqrt 2)(x^2+\sqrt 2)$ are $\pm \sqrt[4]2$ and $\pm \sqrt[4]2\,i$. Similarly the roots of the polynomial $x^4 + 2 = (x^2-\sqrt 2\,i)(x^2+\sqrt 2\,i)$ are $\pm \sqrt[4]2 \sqrt i$ and $\pm \sqrt[4]2\,i\,\sqrt i (= \pm \sqrt[4]2\,i^{\frac{3}{2}})$.

We claim that $[k^{'}:k] = 1$. Equivalently, we will prove that $$k = k^{'}.$$ 
To prove this, we will prove that all the roots of $x^4 + 2$ are in $k$. To prove this, from the above-given list of roots of $x^4 + 2$, it is enough to prove that $\sqrt i \in k$. [Since we already have $i, \sqrt[4]2 \in k$.]
Now, one can check that $\sqrt i = \frac{1}{\sqrt 2} + i \frac{1}{\sqrt 2}$ which is a primitive 8th root of unity. Since $\sqrt[4] 2 \in k$, its square $\sqrt 2 \in k$ and the reciprocal $\frac{1}{\sqrt 2} \in k$. Since, we already have $i \in k$, this shows that $\sqrt i = \frac{1}{\sqrt 2} + i \frac{1}{\sqrt 2} \in k$  and the proof is done. 
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NBHM 2020 PART A Question 11 Solution

What are the maximum and minimum values of $x + y$ in the region $S = \{(x, y) : x^2 +4y^2 ≤ 1\}$? 
SolutionA convex function on a compact convex set attains its maximum and minimum at an extreme point of the set. So the required maximum and minimum should attain on the boundary $T = \{(x,y): x^2 + 4y^2 = 1\}$. We now use the Lagrange's multiplier to find the maximum and minimum of the given function in the boundary $T$. 

Let $f(x,y) = x + y$ and $g(x,y) = x^2 + 4y^2 -1$. Lagrange's multiplier $\lambda$ is given by $\mathcal L(x,y,\lambda) = f(x,y) + \lambda g(x,y) = (x+y) + \lambda (x^2+4y^2-1)$. To find the required maximum and minimum, we need to find the gradient of $\mathcal L$ and equate it to zero. The gradient of $\mathcal L$ is equal to
$$\nabla(\mathcal{L}) =(\frac{\partial \mathcal {L}}{\partial x}, \frac{\partial \mathcal{L}}{\partial y}, \frac{\mathcal{L}}{\partial \lambda}) = (1+2\lambda x, 1+8\lambda y, x^2 + 4y^2-1).$$
$\nabla(\mathcal{L}) = 0$ implies that $$1+2\lambda x = 0,$$ $$1+8\lambda y = 0,$$ $$x^2 + 4y^2 -1 = 0.$$ From first two equations we get, $$x = \frac{-1}{2\lambda},$$ $$y = \frac{-1}{8\lambda}$$ Substitute this value in the third equation we get $$\lambda = \pm \frac{\sqrt 5}{4}.$$ Now substitue the values of $\lambda$ in the above two equations we get $$x = \frac{-2}{\sqrt 5},$$ $$y = \frac{-1}{2\sqrt 5}$$
Hence the stationary points are $$(\frac{-2}{\sqrt 5}, \frac{-1}{2\sqrt{5}},\frac{\sqrt{5}}{4})$$ and $$(\frac{2}{\sqrt 5}, \frac{1}{2\sqrt{5}}, \frac{-\sqrt 5}{4}).$$

Substitute the $x$ and $y$ coordinates of stationary points in $f(x,y)$  we get $x+y = \pm \frac{\sqrt 5}{2}$ which is the required maximum and minimum.



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NBHM 2020 PART A Question 8 Solution

You are given $20$ identical balls and $5$ bins that are colored differently (so that any two of the bins can be distinguished from each other). In how many ways can the balls be distributed into the bins in such a way that each bin has at least two balls?
Solution:
First, take $10$ balls in the hand. Consider the $5$ bins and put $2$ balls in each of these bins so that each bin will have at least $2$ balls. Since the balls are identical this can be done in only one way. Now, we are left with $10$ identical balls which have to be distributed in $5$ bins without any constraints. We know that the number of ways to put $n$ identical objects into $k$ labeled bins is equal to the binomial coefficient $$\binom{n+k-1}{k-1}$$ by stars and bars (CHECK THIS LINK). In our case, this number is equal to $$\binom{10+5-1}{5-1} = 1001$$.
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NBHM 2020 PART A Question 7 Solution

From the collection of all permutation matrices of size $10×10$, one such matrix is randomly picked. What is the expected value of its trace?  (A permutation matrix is one that has precisely one non-zero entry in each column and in each row, that non-zero entry being 1).
Solution:
Direct method: Expectation is nothing but the average. We will calculate the avarage. Let $P_n$ be the set of all $n \times n$ permutation matrices, then there are $n!$ many permuation matrices one for each permuation in $S_n$ (see below for the precise definition). We get the average (expected) trace as, 
$$\begin{align*}\frac{1}{10!}\sum_{A\in P_{10}}\text{tr}(A) &= \frac{1}{10!}\sum_{A\in P_{10}}\sum_{i=1}^{10}A_{ii} \\&= \frac{1}{10!}\sum_{i=1}^{10}\sum_{A\in P_{10}}A_{ii} \\&= \frac{1}{10!}\sum_{i=1}^{10}9! \\&= \frac{10\cdot9!}{10!} \\&= 1\end{align*}$$
Probabilistic method
Let $\sigma$ be a permutation in $S_n$ and $A_{\sigma}$ be the associated permutation matrix. For example, consider the permutation $\sigma = (1)(234)(5)$ in $S_5$, then the associated permutation matrix is obtained by permuting the rows (or colums (fix a convention)) of the $5 \times 5$ identity matrix. Hence $$A_{\sigma} = \begin{bmatrix} 1&0&0&0&0\\0&0&0&1&0\\ 0&1&0&0&0\\0&0&1&0&0\\0&0&0&0&1\\\end{bmatrix}.$$ Note that the second row of the identity matrix moved to third row, third row moved to fourth row and fourth row moved to second row as in the permutation $\sigma$. Other rows are not disturbed bacause they are not permuted by the permutation $\sigma$.
Observation: $\text{Trace } A_{\sigma} = \text{the number of fixed points of } \sigma$.
Proof:  $i$ th row contribute to trace iff $i$ th row is not disturbed iff $\sigma(i)=i$.
So we can rephrase the given question as, from the set of all $10 \times 10$ permutations $S_{10}$ one permutation is randomly picked. What is the expected value of its number of fixed points?  
Consider $1 \le i \le 10$, then the set $\{\sigma \in S_{10} : \sigma(i) = i\}$ has $9!$ number of elements. We get that the probability $p_i$ of $i$ being fixed is $\frac{9!}{10!} = \frac{1}{10}$. This probability is same for any $1 \le i \le 10$. Now, the required expected number of fixed points is equal to $\sum_{i=1}^{10} p_i = \sum_{i=1}^{10} \frac{1}{10} = 1$ (By the linearity of the expectation).
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NBHM 2020 PART A Question 5 Solution

Let $p(x)$ be the minimal polynomial of $\sqrt{2}+\sqrt{-2}$ over the field $\Bbb Q$ of rational numbers. Evaluate $p(\sqrt 2)$.
Solution:
We shall find the minimal polynomial of $\sqrt{2}+i\sqrt{2}$ explicitly.
Let $x = \sqrt{2}+i\sqrt{2}$ then $x-\sqrt 2= i\sqrt 2$  and squaring both sides we get $x^2  + 4 = -2 \sqrt 2 x$. This not a polynomial over $\Bbb Q$, so again we square both sides we get $$p(x) = x^4 + 16 = 0.$$ This is a monic polynomial of degree 4 over $\Bbb Q$ satisfied by $\sqrt{2}+i\sqrt{2}$ and hence it has to be the minimal polynomial of $\sqrt{2}+i\sqrt{2}$ . Now substituting $x = \sqrt 2$ in the above polynomial we get $p(\sqrt 2) = 20$.

Why the degree of the minimal polynomial has to be $4$?
We have $\sqrt{2}+\sqrt{-2} = \sqrt{2}+i\sqrt{2}$ and this element belong to $\Bbb{Q}(\sqrt 2,i)$ which is a degree 4 extension. Because $$[\Bbb{Q}(\sqrt 2,i): \Bbb Q] = [\Bbb{Q}(\sqrt 2,i): \Bbb Q(\sqrt 2)] \cdot [\Bbb{Q}(\sqrt 2): \Bbb Q] = 2 \times 2 = 4.$$
The proper subfields of $\Bbb{Q}(\sqrt 2,i)$ are $\Bbb{Q}(\sqrt 2)$, $\Bbb{Q}(i)$, $\Bbb{Q}(\sqrt 2i)$ and $\Bbb{Q}$. So $\Bbb{Q}(\sqrt 2,i)$ is the smallest field containing $\sqrt{2}+i\sqrt{2}$ and hence $$\Bbb{Q}(\sqrt 2,i) = \Bbb{Q}(\sqrt{2}+i\sqrt{2})$$ and the degree of minimal polynomial of $\sqrt{2}+i\sqrt{2}$ is equal to $4$.

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NBHM 2020 PART A Question 4 Solution $$\int_{-\infty}^{\infty}(1+2x^4)e^{-x^2} dx$$

Evaluate: $$\int_{-\infty}^{\infty}(1+2x^4)e^{-x^2} dx$$
Solution: $$\int_{-\infty}^{\infty}(1+2x^4)e^{-x^2} dx = \int_{-\infty}^{\infty}e^{-x^2} dx +2 \int_{0}^{\infty}(2x^4)e^{-x^2} dx = \int_{-\infty}^{\infty}e^{-x^2} dx +4\int_{0}^{\infty}x^4e^{-x^2} dx .$$ It is well-known that $$\int_{-\infty}^{\infty}e^{-x^2} dx = \sqrt{\pi}.$$
We will calculate the second integral $$4\int_{0}^{\infty}x^4e^{-x^2}dx.$$
Substitute $u = x^2$ then the integral transforms to $2 \int_{0}^{\infty}u^{\frac{3}{2}}e^{-u}du = 2 \Gamma(\frac{3}{2}) = 2 \cdot \frac{3}{4}\cdot \sqrt{\pi} = \frac{3}{2} \sqrt{\pi}$
Hence, $$\int_{-\infty}^{\infty}(1+2x^4)e^{-x^2} dx = \sqrt{\pi} + \frac{3}{2} \sqrt{\pi} = \frac{5}{2} \sqrt{\pi}.$$
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NBHM 2020 PART A Question 3 Solution

A point is moving along the curve $y = x^2$ with unit speed. What is the magnitude of its acceleration at the point $(1/2, 1/4)$?
Solution:
Fact: If a particle is in motion along a curved path at a constant speed then its acceleration at any point of its path is inversely proportional to the radius of curvature of the path at that point.  First, we find the curvature $K$ of $y=x^2$ at the point $(1/2, 1/4)$. Now, $$K = \frac{y{''}(x)}{((1+y^{'}(x))^2)^{\frac{3}{2}}} = \frac{2}{(1+4x+4x^2)^{\frac{3}{2}}}.$$Substituting the given point $x=\frac{1}{2}$ gives $K = \frac{1}{\sqrt{2}}$. Now, the radius of curvature = $\frac{1}{\text{curvature}} = \sqrt{2}$.
The required acceleration = radius of curvature = $\frac{1}{\text{curvature}} = \frac{1}{\sqrt{2}}$.

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NBHM 2020 PART A Question 2 Solution

Let $u$ and $v$ be the real and imaginary parts respectively of the function $f(z) = \frac{1}{z^2 −6z +8}$ of a complex variable $z = x + i y$. Let $C$ be the simple closed curve $|z| = 3$ oriented in the counter clockwise direction. Evaluate the following integral: $$\int_C (udy + vdx).$$
Solution: Let $f(z) = u + iv$ and $dz = dx+idy$, then $$f(z)dz = (u + iv)(dx+idy) = (udx-vdy)+i(udy+vdx).$$ Also, $$\int_C f(z)dz = \int_C ((udx-vdy)+i(udy+vdx))$$ which is equal to $$\big(\int_C udx-vdy\big) + i \big(\int_C udy+vdx\big).$$ So the required integral is the imaginary part of the value of the integral $\int_C \frac{1}{z^2 −6z +8}dz$ which can be calculated using the residue theorem.
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NBHM 2020 PART A Question 1 Solution

Find rational numbers a,b,c such that $(1+\sqrt[3]2)^{-1} = a + b\sqrt[3]2 + c \sqrt[3]2^2$.
Solution: Consider the field extension $\Bbb F = \Bbb{Q}(\sqrt[3]2))$, then $\Bbb F$ is a degree 3 extension of $Q$ with the associated irreducible polynomial $x^3-2$. The set $\{1, \sqrt[3]2), \sqrt[3]2)^2\}$ form a $\Bbb Q$-basis of $\Bbb F$. Hence, it is possible to write $(1+\sqrt[3]2)^{-1} = a + b\sqrt[3]2 + c \sqrt[3]2^2$. Now we have $(1+\sqrt[3]2)(1+\sqrt[3]2)^{-1} = 1$ in $\Bbb F$ and hence $$(1+\sqrt[3]2)(a + b\sqrt[3]2 + c \sqrt[3]2^2) = 1.$$
Expand the left-hand side, we get $$a + b\sqrt[3]2 + c \sqrt[3]2^2 + a\sqrt[3]2 + b\sqrt[3]2^2 + 2c = 1.$$
Now, by equating the coefficients of left and right-hand side we get,
$$a+2c=1, a+b=0, b+c=0$$
In the last step, we have used the fact that the set $\{1, \sqrt[3]2), \sqrt[3]2)^2\}$ form a $\Bbb Q$-basis of $\Bbb F$.
Solving this equations, we get $$a = \frac{1}{3}, b=-\frac{1}{3}, c=\frac{1}{3}.$$
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CSIR JUNE 2011 PART B QUESTION 25 SOLUTION (SUBSCRIBE MY BLOG TO GET REGULAR UPDATES))

Let $A = \{x^2 : 0<x<1\}$ and $B = \{x^3 : 1<x<2\}$. Which of the following statements is true?
1. There is a one to one, onto function from $A$ to $B$.
2.  There is no one to one, onto function from $A$ to $B$ taking rationals to rationals.
3. There is no one to one function from $A$ to $B$ which is onto.
4. There is no onto function from $A$ to $B$ which is one-one.

Solution: We observe that $A = [0,1]$ and $B = [1,8]$. This is because of  $f(x) = x^2$ and $g(x) = x^3$ are continuous strictly increasing functions on $[0,1]$ and $[1,2]$ and we have $A = f([0,1])$ and $B = g([1,2])$.

 1. (true) Both $A$ and $B$ are uncountable sets and the function defined by $h(x) = 8x$ defines a bijection between $A$ and $B$.

2. (false) If $x$ is a rational number then $8x$ is also a rational number. Hence the bijection $h(x)$ from $A$ to $B$ (given above) takes rationals to rationals.

3. (false) Let $f$ be a one to one function which is onto. This means that it is a bijection. So option 3 says that there is no bijection between $A$ and $B$. But we have constructed a bijection between $A$ and $B$ in option (1).

4. (false) Similar to (3).

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CSIR JUNE 2011 PART B QUESTION 22 SOLUTION (SUBSCRIBE MY BLOG TO GET REGULAR UPDATES)

The number of 4 digit numbers with no two-digit common is
1. 4536,
2. 3024,
3.5040,
4. 4823.
Solution: We will explicitly calculate the number of four-digit numbers with no two-digits common (repeated).

The first digit can be any number between $0$ to $9$ except it cannot be $0$. Because if the first digit is zero then it is no more a four-digit number. Hence the first digit has $9$ possibilities $1$ to $9$.

The second digit can be any number between $0$ to $9$  except the one used in the first digit. But this digit can be 0. Hence the possibilities are $\{0,1,2,\dots,9\} \backslash \{first digit\}$ hence there are $9$ possibilities.

The third digit can be any number between $0$ to $9$  except the numbers used in the first and the second digit. Hence the possibilities are $\{0,1,2,\dots,9\} \backslash \{first digit, second digit\}$ hence there are $8$ possibilities. Note that, by our choice first and the second digit was different.

The fourth digit can be any number between $0$ to $9$  except the numbers used in the first, second and third digit. Hence the possibilities are $\{0,1,2,\dots,9\} \backslash \{first digit, second digit, third digit\}$ hence there are $7$ possibilities.

Hence the total number of possibilities is equal to 9 x 9 x 8 x 7 = 4536!!!.


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CSIR JUNE 2011 PART B QUESTION 21 SOLUTION (SUBSCRIBE MY BLOG TO GET REGULAR UPDATES)

Let $S = \{A \in M_5(\Bbb{R}): a_{ij} = 0 \text{ or } 1 \text{ for all } i,j \text{ and } \\ \text{ ith row sum of $A$ = ith column sum of $A$ =  $1$} \forall i \}$.
Then the number of elements in $S$ is
1. $5^2$,
2. $5^5$,
3. $5!$,
4. 55.
Solution:  We will explicitly calculate the cardinality of $S$. We note that every matrix in $S$ is 0,1 matrix and has the row sum and column sum 1. Hence each row and each column has exactly one 1.  Now, observe that any such matrix can be obtained from the $5 \times 5$ identity matrix by permuting the rows. This defines a bijection between the sets $S$ and $S_5$ of permutations on $\{1,2,3,4,5\}$. Such matrices are called permutation matrices and have cardinality $5!$. 
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CSIR JUNE 2011 PART B QUESTION 23 SOLUTION (SUBSCRIBE MY BLOG TO GET REGULAR UPDATES)

Let $D$ be a non-zero $n \times n$ real matrix with $n \ge 2$, Which of the following implications are valid?
1. $det(D) = 0 \implies rank(D)=0$.
2. $det(D) = 1 \implies rank(D)\ne1$.
3. $rank(D)=1 \implies det(D) \ne 0$ and
4. $rank(D) = n \implies det(D) \ne 1$.

Solution: TO SAY AN OPTION IS WRONG IT IS ENOUGH TO GIVE AN EXAMPLE

1.  (false) Consider the matrix $D = \begin{bmatrix} 1&0 \\ 2&0 \end{bmatrix}$ then $det(D)= 0$ but $rank(D) = 1 \ne 0$.

Result : $D$ is invertible iff $det(D) \ne 0$ iff $rank(D) = n$. Equivalently $det(D) = 0$ iff $rank(D) < n$.

2. (True) Given that $det(D) = 1$ and hence by the above result $rank(D) = n$ and it is given that $n \ge 2$. Therefore $rank(D) \ne 1$.

3. (false) Given $rank(D) = 1$ which is strictly less than $n$. From the above-given result we have $det(D) = 0$.

4. (false) Consider the matrix $D = \begin{bmatrix} 1&0 \\ 0&1 \end{bmatrix}$ then $rank(D)= 2 = n$ but $det(D) = 1$.

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A sequence of uniformly continuous functions that converge to a non-uniformly continuous function (CSIR) (INTERVIEW)

Give an example of a sequence of uniformly continuous functions that converges to a non-uniformly continuous function,

Solution:

Case 1: The limit function is not even continuous

Consider $f_n(x) = x^n$ on $[0,1]$ then $f_n$ converge pointwise to $f$ where $f(x) = \begin{cases} 0 \text{ if } x \in [0,1) \\ 1 \text{ if } x = 1\end{cases}$. Now, each $f_n$ is a continuous function on complact set and hence uniformly continuous but its pointwise limit is not even continuous.

Case 2: The limit function is continuous but not uniformly continuous

Consider the function $f(x) = x^2$ on $[0,\infty)$. We know that $x^2$ is uniformly continuous only on finite intervals and hence $f$ is not uniformly continuous on $[0,\infty)$. Define $f_n(x) = \begin{cases} f(x) \text{ if } 0 \le x \le n \\ f(n) \text{ if } x > n \end{cases}$, then clearly each $f_n$ is uniformly continuous and converges pointwise to the function $f$. This complete the proof.

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Sequence whose limit points are all the points of $[0,1]$ (CSIR)

Prove that there exists a sequence of real numbers whose limits points are points of $[0,1]$.
Proof :
Let $\{x_n\}$ be a sequence of real numbers and a real number $\alpha$ is said to be a limit point of $\{x_n\}$ if there exists a subsequence $\{y_n\}$ of $x_n$ such that $\{y_n\}$ converges to $\alpha$.
To answer the given problem, we need to find a sequence $\{x_n\}$ satisfying "given a point $\alpha$ in $[0,1]$ the sequence $\{x_n\}$ should have a subsequence $\{y_n\}$ that converges to $\alpha$". We shall construct one such sequence.

Fact : Rationals in $[0,1]$ form a countable dense subset of $[0,1]$.

Since $[0,1] \cap \Bbb Q$ is countable, we can enumerate them and write it as a sequence say $\{x_n\}$. Let $\alpha \in [0,1]$, since $[0,1] \cap \Bbb Q$ is dense in $[0,1]$ every point of $[0,1]$ is a limit point of $[0,1] \cap \Bbb Q$. Hence there is a sequnce of points of $[0,1] \cap \Bbb Q$ which converges to $\alpha$. This is clearly a subsequence of $\{x_n\}$ and the proof is complete.
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$I = \{f \in C[0,1] : f(\alpha) = 0\}$ is a maximal ideal in $C[0,1]$ (NBHM) (CSIR)

Let $C[0,1]$ be the set of all continuous functions from $[0,1]$ to $\Bbb{R}$ which is a commutative ring with 1 with respect to the pointwise addition and multiplication: $$(f+g)(x) = f(x)+g(x)$$ and $$(fg)(x) = f(x)g(x)..$$ Fix $\alpha \in [0,1]$ and define $I = \{f \in C[0,1] : f(\alpha) = 0 \}$.
Prove that $I$ is a maximal ideal in $C[0,1]$.
Solution:
Clam: $I$ is an ideal in $C[0,1]$
Let $f,g \in I$, then $f(\alpha) = 0 = g(\alpha)$. Now, $$(f+g)(\alpha) = f(\alpha) + g(\alpha) = 0 + 0 = 0$$
Therefore $I$ is closed under addition.
Let $f \in C[0,1]$ and $g \in I$. Now, $g \in I$ implies that $g(\alpha) = 0$.  We claim that $fg \in I$.
$$(fg)(\alpha) = f(\alpha) \cdot g(\alpha) = f(\alpha) \cdot 0 = 0$$
Therefor $I$ is an ideal in $C[0,1]$.

Clam: $I$ is a maximal ideal.
An ideal $I$ is said to be a maximal if whenever there exists an ideal $J$ such that $I \subseteq J \subseteq C[0,1]$ we have either $J = I$ or $J = C[0,1]$.
There is a characterization for the maximal ideals which we will use here.
Result: Let $R$ be a commutative ring with 1 and $I$ be an ideal in $R$ then $I$ is maximal in $R$ if and only if the quotient $\frac{R}{I}$ is a field.
We prove that $\frac{C[0,1]}{I} \cong \Bbb{R}$ which is a field and proof follows.

We will construct an onto ring homomorphism from $C[0,1]$ to $\Bbb{R}$ whose kernel $I$, then by the fundamental theorem of a ring homomorphism, we have $\frac{C[0,1]}{I} \cong \Bbb R$

Define a ring homomorphism $\phi: C[0,1] \to \Bbb{R}$ by $\phi(f) = f(\alpha)$ (the evaluation map). Now, $$\phi(f+g) = (f+g)(\alpha) = f(\alpha) + g(\alpha) = \phi(f) + \phi(g)$$ Similarly
$$\phi(fg) = (fg)(\alpha) = f(\alpha) \cdot g(\alpha) = \phi(f) \cdot \phi(g)$$
Therefore $\phi$ is a ring homomorphism.

We will calculate the Kernal of $\phi$:
Ker $\phi = \{f \in C[0,1]: \phi(f) = 0 \} = \{f \in C[0,1]: f(\alpha) = 0 \} = I$.

We left with verifying onto for $\phi$. Let $a$ be a real number. We have to find $f \in C[0,1]$ such that $\phi(f) = a$. But $\phi(f) = f(\alpha)$. So we can take $f$ to be the linear polynomial $f = x + (a - \alpha)$ (which has constant term $a - \alpha$). Now, for this $f$, $\phi(f) = f(\alpha) = \alpha + (a - \alpha) = a$ and the proof is complete.

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$\{f \in C[0,1] : f(\frac{1}{3}) = f(\frac{1}{2}) = 0 \}$ is not a prime ideal (NBHM) (CSIR)

Let $C[0,1]$ be the set of all continuous functions from $[0,1]$ to $\Bbb{R}$ which is a commutative ring with 1 with respect to the pointwise addition and multiplication: $$(f+g)(x) = f(x)+g(x)$$ and $$(fg)(x) = f(x)g(x).$$ Define $I = \{f \in C[0,1] : f(\frac{1}{3}) = f(\frac{1}{2}) = 0 \}$.  (Note that $\frac{1}{3}$ and $\frac{1}{2}$ can be replaced with any two distinct numbers in $[0,1]$ and the result still holds true)
Prove that $I$ is an ideal which is neither prime nor maximal.
Solution:
Clam: $I$ is an ideal in $C[0,1]$
Let $f,g \in I$, then $f(\frac{1}{3}) = f(\frac{1}{2}) = 0 = g(\frac{1}{3}) = g(\frac{1}{2}) $. Now, $$(f+g)(\frac{1}{3}) = f(\frac{1}{3}) + g(\frac{1}{3}) = 0 + 0 = 0$$ Similarly
$$(f+g)(\frac{1}{2}) = f(\frac{1}{2}) + g(\frac{1}{2}) = 0 + 0 = 0$$
Therefore $I$ is closed under addition.
Let $f \in C[0,1]$ and $g \in I$. $g \in I$ implies that $g(\frac{1}{3}) = g(\frac{1}{2}) = 0$.  We claim that $fg \in I$.
$$(fg)(\frac{1}{3}) = f(\frac{1}{3}) \cdot g(\frac{1}{3}) = f(\frac{1}{3}) \cdot 0 = 0$$ Similarly
$$(fg)(\frac{1}{2}) = f(\frac{1}{2}) \cdot g(\frac{1}{2}) = f(\frac{1}{2}) \cdot 0 = 0.$$
Therefor $I$ is an ideal in $C[0,1]$.
We will show that this is not a prime ideal and hence it cannot be maximal also.
Clam: $I$ is not a prime ideal.
An ideal $I$ is said to be a prime ideal if $fg \in I$ implies either $f \in I$ or $g \in I$.
So to prove an ideal is not prime, we need to find $fg \in I$ such that $f \notin I$ and $g \notin I$.
Consider the polynomials $f = x - \frac{1}{3}$, $g = x - \frac{1}{2}$ and $fg = (x - \frac{1}{3})(x - \frac{1}{2})$. Polynomials are continuous and hence $f,g,fg \in C[0,1]$. Note that
$$(fg)(\frac{1}{3}) = 0 \text{ and } (fg)(\frac{1}{2})= 0.$$
Therefore $fg \in I$. But $f(\frac{1}{2}) \ne 0$ and $g(\frac{1}{3}) \ne 0$. Hence neither $f$ nor $g$ in $I$. This proves the claim.

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$f(x) = sin(\frac{1}{x})$ is not uniformly continuous on $(0,1)$ (NBHM)

Prove that the function $f: (0,1) \to \Bbb{R}$ defined by $f(x) = sin(\frac{1}{x})$ is not uniformly continuous. 
Proof :  A function $f : (0,1) \to \Bbb{R}$ is said to be uniformly continuous if  given $\epsilon > 0$ there exists a $\delta>0$ such that $|x-y| < \delta$ implies that $|f(x) - f(y)| < \epsilon$ for all $x,y \in (0,1)$.  We will show that, for $\epsilon = \frac{1}{2}$ finding such a $\delta$ (which works for all $x,y \in (0,1)$) is not possible. 
On the contrary we assume there is a $\delta$ and we derive a contradiction. $a_n = \frac{1}{2n\pi}$ and $b_n = \frac{1}{(2n+\frac{1}{2})\pi}$. We observe that $|\frac{1}{2n\pi} - \frac{1}{(2n+\frac{1}{2})\pi}| = |\frac{1}{2n\pi} - \frac{2}{(4n+1)\pi}| = \frac{1}{2n\pi(4n+1)}$ can be made arbitrarily small as the sequence $\{\frac{1}{2n\pi(4n+1)}\}$ converges to 0. Hence $|\frac{1}{2n\pi} - \frac{1}{(2n+\frac{1}{2})\pi}|$ can be made less that $\delta$ for large $n$. Now, for such $n$, $|f(\frac{1}{2n\pi})-f(\frac{1}{2n+\frac{1}{2})\pi})|$ should be less than $\epsilon = \frac{1}{2}$. But $$|f(\frac{1}{2n\pi})-f(\frac{1}{2n+\frac{1}{2})\pi})| = |sin(2n\pi) - sin((2n+\frac{1}{2})\pi)| = |0-1|=1$$ contradiction!.

The general strategy is, to prove a function is not uniformly continuous, find two sequences $a_n$ and $b_n$ such that $|a_n - b_n|$ is arbitrarily small for large $n$ but $|f(a_n) - f(b_n)|$ is a fixed constant for all $n$. Why it is enough to prove such sequences is explained above using $\epsilon$ and $\delta$ argument.

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$f(x) = \frac{1}{x}$ is not uniformly continuous on $(0,1)$ (NBHM)

Prove that the function $f: (0,1) \to \Bbb{R}$ defined by $f(x) = \frac{1}{x}$ is not uniformly continuous. 
Proof :  A function $f : (0,1) \to \Bbb{R}$ is said to be uniformly continuous if  given $\epsilon > 0$ there exists a $\delta>0$ such that $|x-y| < \delta$ implies that $|f(x) - f(y)| < \epsilon$ for all $x,y \in (0,1)$.  We will show that, for $\epsilon = \frac{1}{2}$ finding such a $\delta$ (which works for all $x,y \in (0,1)$) is not possible. 
On the contrary, we assume there is a $\delta$ and we derive a contradiction.
Let $a_n = \frac{1}{n}$ and $b_n = \frac{1}{n+1}$. We observe that $|\frac{1}{n} - \frac{1}{n+1}|$ can be made arbitrarily small as the sequence $\{\frac{1}{n}\}$ converges to 0. Hence $|\frac{1}{n} - \frac{1}{n+1}|$ can be made less that $\delta$ for large $n$. Now, for such $n$, $|f(\frac{1}{n})-f(\frac{1}{n+1})|$ should be less than $\epsilon = \frac{1}{2}$. But $$|f(\frac{1}{n})-f(\frac{1}{n+1})| = |n -(n+1)| = 1$$ contradiction!.

The general strategy is, to prove a function is not uniformly continuous, find two sequences $a_n$ and $b_n$ such that $|a_n - b_n|$ is arbitrarily small for large $n$ but $|f(a_n) - f(b_n)|$ is a fixed constant for all $n$. Why it is enough to prove such sequences is explained above using $\epsilon$ and $\delta$ argument.

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Generalized Lipschitz functions are uniformly continuous (NBHM)

Consider the set $\mathcal{F}= \{f:\Bbb{R} \to \Bbb{R}: |f(x)-f(y)| ≤ K|(x-y)|^a \}$ for some $a>0$ and $K>0$, then which of the following statements are true?
1. Every differentiable frunction $f : \Bbb{R} \to \Bbb{R}$ is in $\mathcal{F}$.
2. Every function $f \in \mathcal{F}$ is differentiable. (This is converse of (1))
3. Every function $f \in \mathcal{F}$ is uniformly continuous.
Solution:
1. (false) Consider the function $f(x) = x^2$, then $f$ is a differentiable function from $\mathbb{R}$ to $\mathbb{R}$. We claim that $f \notin \mathcal{F}$. We have for any $x,y \in \mathbb{R}$, $$|x^2 - y^2| = |x-y| |x + y|$$
Since $x,y \in \mathbb{R}$ are arbitrary $|x+y|$ can also be arbirtarily large. So there cannot exist the required $K>0$ and $f \notin \mathcal{F}$.

Remark:
If $f(x) = x^2$ is consider as function on $(a,b)$ a bounded interval (open, closed, semi open any kind), then the term $|x+y|$ can be bounded by  $2b$ since $x,y \in (a,b)$ and we can take $k = 2b$ and $a =1$ which will make $f \in \mathcal F$.

2. (false) Consider the function $f(x) = |x|$ and this function satisfies $$||x| - |y|| \le |x - y|.$$ Hence for $k=1$ and $a=1$ this function is in $\mathcal F$. But $|x|$ is not differentiable at origin.

3. (True)  Let $f$ be a function from $\mathcal{F}$, then $f$ satisfies $$|f(x)-f(y)| ≤ K|(x-y)|^a$$ for some $K>0$ and $a>0$. We claim that $f$ is uniformly continuous. Let $\epsilon > 0$, we find a $\delta$ which depends only on $\epsilon$ and hence the function will be uniformly continuous.  We take $\delta = (\frac{\epsilon}{K})^{\frac{1}{a}}$ then $|x-y| < \delta$ implies that $|f(x) - f(y)| \leq K |x-y|^ a < K \delta^a =  K ((\frac{\epsilon}{K})^{\frac{1}{a}})^a = \epsilon$. This completes the proof.
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Discrete metric spaces are complete

Result: Discrete metric spaces are complete.
Proof:
          Let $\{x_n\}$ be a Cauchy sequence in a discrete metric space $(X,d)$, then given $\epsilon > 0$ there exists a positive integer $p \in \mathbb{N}$ such that $d(x_n , x_m) < \epsilon$ for all $m,n \ge p$. Let $\epsilon = 1/2$ then there exists a positive integer $p \in \mathbb{N}$ such that $d(x_n , x_m) < 1/2$ for all $m,n \ge p$. But $d(x_n , x_m) < 1/2 \to d(x_n , x_m) = 0$ for all $m,n \ge p$ in discrete metric. Hence $x_n = x_m = a (say)$ for all $m,n \ge P$ and hence this cauchy sequence $\{x_n\}$ is an eventually constant sequence $a$ and converge to $a$. This completes the proof.

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Every continuous function $f : [0,1] \to [0,1]$ has a fixed point.

Result: Let $f : [0,1] \to [0,1]$ be a continuous function then there exists $c \in [0,1]$ such that f(c) = c.

Proof: 
If $f(0) = 0$ or $f(1) = 1$ then nothing to prove. We assume $f(0) \ne 0$ and $f(1) \ne 1$.
Consider $g(x) = f(x) - x$, then $g$ is a continuous function from $[0,1]$ to $[0,1]$.
Since $f$ is a function from $[0,1]$ to $[0,1$ we have $g(0) = f(0) - 0 > 0$ and $g(1) = f(1) - 1 < 0$. Now by intermediate value theorem there exists $x$ in $[0,1]$ such that $g(x) = 0 = f(x) - x$ and the proof is done.

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Application of Intermediate Value theorem (NBHM)

Suppose that $f:[0,1]→[0,2]$ is continuous then prove that there exists $c \in [0,1]$
such that $f(c)=2c^2$.

Proof : 

Intermediate value theorem :

If a continuous function has values of an opposite sign inside an interval, then it has a root in that interval.


Let $f$ be the function satisfying the given hypothesis. Consider the function $g(x) = \sqrt{\frac{f(x)}{2}}$. Then $g$ is continuous function from $[0,1]$ to $[0,1]$ then by the following result, there exists $c \in [0,1]$ such that $g(c) = c$. i.e., $\sqrt{\frac{f(c)}{2}} = c$ and the proof is done.

Result: Let $f : [0,1] \to [0,1]$ be a continuous function then there exists $c \in [0,1]$ such that f(c) = c.

Proof: 
If $f(0) = 0$ or $f(1) = 1$ then nothing to prove. We assume $f(0) \ne 0$ and $f(1) \ne 1$.
Consider $g(x) = f(x) - x$, then $g$ is a continuous function from $[0,1]$ to $[0,1]$.
Since $f$ is a function from $[0,1]$ to $[0,1$ we have $g(0) = f(0) - 0 > 0$ and $g(1) = f(1) - 1 < 0$. Now by intermediate value theorem there exists $x$ in $[0,1]$ such that $g(x) = 0 = f(x) - x$ and the proof is done.

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Find all possible values of $i^{-2i}$ (NBHM 2013)

Find all possible values of $i^{-2i}$.

Solution:

Any complex number in the unit circle can be written as $e^{i\theta} = \cos(\theta) + i \sin(\theta)$ for some $\theta \in \Bbb{R}$. Since $i$ is in the imaginary axis we can take $i = e^{i \frac{\pi}{2}}$.

We have $e^{i\theta}$ is defined in terms of sin and cos and hence is $2\pi$ periodic. That is $e^{i(\theta+2n\pi)} = e^{i\theta}$ for $n \in \Bbb{Z}$. Therefore $i = e^{i \frac{\pi}{2}} = e^{(i(\frac{\pi}{2})+2n\pi))} = e^{i(\frac{(4n+1)\pi}{2})}$ and $i^{-2i} = e^{-2i \cdot i(\frac{(4n+1)\pi}{2})} = e^{2(\frac{(4n+1)\pi}{2})} = e^{(4n+1)\pi}$ where $n \in \Bbb{Z}$.

Hence, all possible values of $i^{-2i}$ are $\{\ldots, e^{-7\pi}, e^{-3\pi}, e^{\pi}, e^{5\pi}, \ldots \}$.

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Which of the following sets are countable (NBHM 2016)

Which of the following sets are countable (NBHM 2016)
  1. The set of all algebraic numbers
  2. the set of all strictly increasing infinite sequences of positive integers
  3. the set of all infinite sequences of integers that are in arithmetic progression.

Solution:

1. (Countable) Proof of the set of algebraic numbers is countable is standard. For example, a nice proof can be seen in the principles of mathematical analysis by Walter Rudin.

2. (Uncountable

Fact 1: The set of all subsets of $\mathbb{N}$ is uncountable.

Fact 2: The set of all finite subsets of $\mathbb{N}$ is countable. Therefore

Fact 3: The set of all infinite subsets of $\mathbb{N}$ is uncountable.

Now, the bijection between the set of all strictly increasing infinite sequences of positive integers and the set of all infinite subsets of $\mathbb{N}$ is immediate. For example, map the sequence$x_1,x_2,\dots$ to the infinite set $\{x_1,x_2,\dots\}$ This function is well defined because the sequences are strictly increasing and the bijection can be verified directly.

3. (Countable) Let $\{x_n\}$ be an infinite arithmetic progression of integers, then it can be uniquely represented as a pair $(x_0, d) \in \Bbb{Z} \times \Bbb{Z}$ where $x_0$ is the first term and $d$ is the common difference of the progression. Now mapping $\{x_n\}$ to the pair $(x_0, d) \in \Bbb{Z} \times \Bbb{Z}$ defines a well-defined map between the set of all infinite sequences of integers that are in arithmetic progression and $\Bbb{Z} \times \Bbb{Z}$. This map is injective because if we change the profession either its first term changes or its common difference changes.  Therefore the set of all infinite sequences of integers that are in the arithmetic progression is in bijection with a subset of $\Bbb{Z} \times \Bbb{Z}$ and hence countable.

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Find $\lim_{n\rightarrow \infty} \sin((2n\pi + \frac{1}{2n\pi}) \sin(2n\pi + \frac{1}{2n\pi}))$ (NBHM 2013)

Find the limit $$\lim_{n\rightarrow \infty} \sin((2n\pi + \frac{1}{2n\pi}) \sin(2n\pi + \frac{1}{2n\pi}))$$

Solution:

We have $\sin(2n\pi+\theta) = \sin(\theta)$ this implies $\sin(2n\pi + \frac{1}{2n\pi})) = \sin(\frac{1}{2n\pi}))$ and given limit becomes $$\lim_{n\rightarrow \infty} \sin((2n\pi + \frac{1}{2n\pi}) \sin(\frac{1}{2n\pi})).$$

Now, $$\sin((2n\pi + \frac{1}{2n\pi}) \sin(\frac{1}{2n\pi})) = \sin((2n\pi) \sin(\frac{1}{2n\pi})) + \frac{1}{2n\pi}\sin(\frac{1}{2n\pi})) $$

$|sin 2n\pi| \le 1$ and the sequence $\frac{1}{2n\pi}$ converges to zero and hence second term is converges to zero.

$\lim _{\theta \to 0}\frac{sin \theta}{\theta} = 1$ and so $(2n\pi) \sin(\frac{1}{2n\pi})$ converges to 1 as $n \to \infty$. Hence the required limit is $\sin 1$.

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Which of the following metric spaces are complete? (NBHM 2006)

  Which of the following metric spaces is complete?

1. $X_1=(0,1), d(x,y)=|\tan x-\tan y|$

2. $X_2=[0,1], d(x,y)=\frac{|x-y|}{1+|x-y|}$

3. $X_3=\mathbb{Q}, d(x,y)=1\forall x\neq y$

4. $X_4=\mathbb{R}, d(x,y)=|e^x-e^y|$

Solution :

1. (not complete) Consider the sequence $\{\frac{1}{2^n}\}$ in $X_1$. We claim that this is a Cauchy sequence which does not converge in $(X_1,d)$. We have $$d(\frac{1}{2^n},\frac{1}{2^m}) = |tan(\frac{1}{2^n})-tan(\frac{1}{2^m})| = |tan(\frac{1}{2^n}-\frac{1}{2^m})(1-tan(\frac{1}{2^n})tan( \frac{1}{2^m}))|.$$

The numbers $\frac{1}{2^n}-\frac{1}{2^m}$ and $\frac{1}{2^n}$ are close to zero for large $m,n$ in the usual metric and hence the numbers $tan(\frac{1}{2^n}-\frac{1}{2^m})$ and $tan(\frac{1}{2^n})$ are also close to zero in the usual metric because $\lim _{\theta \to 0} (tan \theta) = 0$. This shows that, from the above equation, $\{\frac{1}{2^n}\}$ is a cauchy sequence in $X_1$.

We have $\{\frac{1}{2^n}\}$ is a decresing sequence and hence if it converge in $(X_1,d)$ it should converge to 0.  Since $0$ is not an element of $X_1 = (0,1)$, the sequence $\{\frac{1}{2^n}\}$ is not converging in $X_1$. Therefore $X_1$ is not complete

2. (complete) because the given metric is equivalent to the usual metric on $[0,1]$.

3. (complete) because it is a discrete metric space.
Claim: Discrete metric spaces are complete.
Proof:
          Let $\{x_n\}$ be a Cauchy sequence in a discrete metric space $(X,d)$, then given $\epsilon > 0$ there exists a positive integer $p \in \Bbb{N}$ such that $d(x_n , x_m) < \epsilon$ for all $m,n \ge p$. Let $\epsilon = 1/2$ then there exists a positive integer $p \in Bbb{N}$ such that $d(x_n , x_m) < 1/2$ for all $m,n \ge p$. But $d(x_n, x_m) < 1/2 \to d(x_n , x_m) = 0$ $m,n \ge p$ in discrete metric. Hence $x_n = x_m = a (say)$ for all $m,n \ge P$ and hence this cauchy sequence $\{x_n\}$ is an eventually constant sequence $a$ and converge to $a$. This completes the proof.

4. (not complete) Consider the sequence $\{-n\}$ in $X_4$. We claim that this is a Cauchy sequence that does not converge in $(X_4,d)$.
We have for $k \le n$, $$d(-k,-n)=|e^{-k}-e^{-n}|=e^{-k}-e^{-n}<e^{-k}\le e^{-m_\epsilon}<\epsilon$$
Similarly for $n \le k$,  $$d(-k,-n)<\epsilon$$ by the symmetric property of the metric. This shows that the sequence $\{-n\}$ in $X_4$ is a cauchy sequence and it is clearly not convergent as $-\infty$ is not in $\Bbb{R}$.


Onto homomorphism from $\mathbb{R}^{*}$ to $\mathbb{Q}^{*}$ (CSIR)

Does there exist an onto group homomorphism from $(\mathbb R^*, .)$ (the multiplicative group of non-zero real numbers) onto $(\mathbb Q^*, .)$ (the multiplicative group of non-zero rational numbers)?

Solution :

We prove by the method of contradiction.

Let $f:\Bbb{R}^*\to\Bbb{Q}^*$ be one such homomorphism, then there exists $a \in \Bbb{R}^*$ such that $f(a)=2$.  Let $w = \sqrt[3]{a} \in \Bbb{R}^*$ be a cube root of $a$ and cube roots always exist in $\mathbb{R}^*$, then $(f(w))^3 = f(w^3) =  f(a) = 2$. This shows that $f(w)$ is a cube root of $2$ and also $f(w) \in \Bbb{Q}^*$. Which is a contradiction to the fact that $2$ has no cube root in $\mathbb{Q}^*$.

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Identify the quotient group $\frac{C^*}{R^+}$ (NBHM)

Let $C^*$ and $R^+$ denote the multiplicative group of non zero complex numbers and the subgroup of positive reals respectively. Identify the quotient group $\frac{C^*}{R^+}$.

Solution : 

The elements of the group $C^\ast/R^+$ are the cosets of $R^+$ in $C^\ast$ and $z_1, z_2 \in C^\ast$ are in the same coset if and only if $z_1 z_2^{-1} \in R^+$, i.e., if and only if $z_1 = rz_2$ for some $r \in R^+$. This means that $z_1$ and $z_2$ lie on the same ray through the origin, so we must have $\arg z_1 = \arg z_2$.  And so we guess that $C^\ast/R^+$ is the circle group $S^1$.


To prove it regorusly, write $z_j = r_j e^{i\theta_j}$, $j = 1, 2$, we see from $z_1 = rz_2$ or $r_1  e^{i\theta_1} = rr_2e^{i\theta_2}$ that $r_1 = r r_2$ and $e^{i\theta_1} = e^{i\theta_2}$; restricting $\theta_1, \theta_2 \in [0, 2\pi)$ we have $\theta_1 = \theta_2$; each coset is thus represented by a unique $\theta \in [0, 2\pi)$ or, equivalently, a unique unimodular complex number $e^{i\theta}$.

Furthermore, it is easy to see that since $z_1 z_2 = r_1 r_2 e^{i(\theta_1 + \theta_2)}$, the natural projection $C^\ast \to C^\ast/R^+ = S^1$ is a homomorphism taking $z_1z_2$ to $\theta_1 + \theta_2$, where we use the usual addition $\mod 2\pi$.  $C^\ast/R^+$ is the circle group indeed.

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Identify the quotient group (NBHM-2014)

Let $\mathbb{C}^{*}$ denote the multiplicative group of non-zero complex numbers and let $\mathbb{R}^*$ denote the subgroup of positive real numbers. Identify the quotient group.

Solution : 

Consider the function $f:\mathbb{C}^* \to \mathbb{C}^*$ defined by $f(r.e^{i\theta})= e^{i\theta}$  then $f$ is a homomorphism with kernel $\mathbb{R}^*$.


Therefor by the fundamental theorem of homomorphism, we can say that $\mathbb{C}^*/\mathbb{R}^*$ is isomorphic to the image $f(\mathbb{C}^*)=S^1$.

where $S^1=\{z\in\mathbb{C} | |z|=1\}$. Note that $f(z) = \frac{z}{|z|}$ is the above given map.




The fundamental theorem of homomorphism: Check my post regarding this for a wonderful article.


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CSIR DEC 2015 PART B SOLUTION


For a subset $A$ of the topological space $X$, let $A'$ denotes the union of set $A$ and all those connected components of $X-A $ which are relatively compact in $X$ (i.e. the closure is compact in $X$). Then for every subset $A$ of $X$, what can we say:

$1. A' \text{ is compact.}$

$2. A''=A'.$

$3. A' \text{ is connected.}$

$4.A'=X.$

Solution : 

1. (false) Let $X = (0,1)$ (any non-compact set will work) with usual topology and let $A = X$ then $X - A = \phi$ empty set. Therefore $A^{'} = X$ which is not compact.


3. (false) Let $X = (0,1) \cup \{2\}$ and $A = (0,1)$ then $\{2\}$ is both open and closed in $X$. Now $X - A = \{2\}$ and so $A^{'} = X$ but X is not connected.

4. (false) Let $X = (0,1) \cup (1,2)$ and $A = (1,2)$ then $X-A = (0,1)$ is not complact in X. Therefore $A^{'} = A \ne X$.

2. (True) Its a CSIR part B problem, so one option should be true. Try the proof by yourself. It is immediate from the definition. Any doubts, happy to help you in the comments section.

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