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Hint $:$ If $\frac {\sqrt 2} {4}$ is a rational number then so is $\sqrt 2 = \frac 1 2 \times \frac {4} {\sqrt 2}.$ Is $\sqrt 2$ a rational number?

I think that you mean $f$ is a differentiable function of $X$ and $Y$ and $X$ and $Y$ are themselves differentiable functions of $x$ and $y$ defined by $$X(x,y) = x, Y(x,y) = x+y,\ (x,y) \in D \... View answer Accepted answer 2 votes Observe that$$\sum\limits_{k=1}^{n} q^{2k} = \frac {q^2 (1 - q^{2n} )} {1 - q^2},\ \text {for any}\ n \geq 1.$$Also observe that q^{2n} \to 0 as n \to \infty for 0 < q < 1. View answer 1 votes First of all note that the graph of T is precisely$$\mathscr G(T) : = \{(f,f)\ :\ f \in S \} \subseteq \left (S,\|\cdot\|_2 \right) \times \left (S, \|\cdot\|_{\infty} \right ).To prove that \... View answer Accepted answer 1 votes By integration by parts we have \begin{align*} I & = \int_{0}^{1} (x^2 + 1)^7\ dx \\ & = x(x^2 + 1)^7 \ \bigg |_{0}^{1} - 14 \int_{0}^{1} \left [(x^2 + 1)^7 - (x^2 + 1)^6 \right ]\ dx \end{... View answer 1 votes Hint : Take \epsilon = 1. Suppose for that \epsilon (>0) you have \delta > 0 such that\left |\frac 1 x - c \right | < 1,\ \ \text {whenever}\ 0 < x < \delta.\ \ \ \ \ \ \ \...

Hint $:$ Break the entire problem in two cases. Where $m$ is odd and $m$ is even. If $m$ is odd. Then $m$ is of the form $2k+1$ for some $k \in \Bbb Z.$ Then observe that $\left \lfloor {\frac {m-1} ... View answer Accepted answer 1 votes I have started from the stage where I got stuck in proving the above lemma. It is easy to show what I just mentioned in the edit is that$\text {Ord}\ (ab)\ \big |\ \text {lcm}\ \left (\text {Ord}\ (a)...

Observe that $V \cong {\Bbb Z_3}^4$ and hence $|V| = 3^4 = 81.$

We need only to take any element $b \in G \setminus \left \langle a \right \rangle$ which has the minimal order amongst all the elements of $G \setminus \left \langle a \right \rangle.$ Then it will ...

If ${[p]_n}^m = [1]_n$ then $\text {Ord}_n\ p = \text {Ord}_n\ q = m$ since $\text {Ord}_n\ q = m\ \bigg |\ \text {Ord}_n\ p.$ Now since $q = p^e$ so $\langle q \rangle \subseteq \langle p \rangle.$ ...

Let $J : X \longrightarrow X^{**}$ be the natural embedding of $X$ into $X^{**}.$ Suppose that weak-topology and weak$^{*}$-topology on $X^*$ coincides i.e. $\sigma (X^*, J(X)) = \sigma (X^*, X^{**}).$...

This function has a removable discontinuity at $x=2.$ By defining $f(2) = \frac 1 5$ you can make $f$ continuous at $x=2.$ But the discontinuity at $x=-3$ is not removable and in fact $f$ has an ...

Observe that for any $n \in \Bbb N$ $$\begin{pmatrix} a_{2n} \\ a_{2n-1} \end{pmatrix} = {\begin{pmatrix} 7 & 6 \\ 2 & 3 \end{pmatrix}}^{n-1} \begin{pmatrix} a_2 \\ a_1 \end{pmatrix}$$ and ...

Hint $:$ The power of $2$ in the prime factorization of $n!$ is given by $$\sum\limits_{i = 1}^{k} \left \lfloor \frac {n} {2^i} \right \rfloor$$ where $k = \left \lfloor \frac {\ln n} {\ln 2} \right \... View answer 0 votes Bairnsdale is the King and Darlinghurst is the Treasurer. There are only twenty possibilities which are easy to figure out. All of them strike out except the one I have mentioned in my answer. Try ... View answer Accepted answer 0 votes With the help of Ragib Zaman I finally able to do it. Here it is$:$We first observe that $$\log (2) = - \log \left ( \frac 1 2 \right ) = \sum\limits_{n=1}^{\infty} \frac {1} {2^n \cdot n}.$$ So $$... View answer Accepted answer 0 votes Since L|K is a finite field extension then so is L|M where M is any intermediary field between K and L; so that any x \in L is algebraic over M. Let \mu_{x,M} be the minimal polynomial ... View answer 0 votes Observe that$$a_n - a_0 = \sum\limits_{k=1}^n (a_k - a _{k-1}) = 3 \sum\limits_{k=1}^{n} k - 5 n.$$View answer Accepted answer 0 votes We can view 4 drawings to be 4 different boxes and the number on the ticket drawn at i-th drawing is the number of balls in the i-th box after the balls are thrown to the boxes where 1 \leq i ... View answer Accepted answer 0 votes What I observed is that the actual problem lies in the definition of discriminant of a monic polynomial. Below is a way to prove the desired proposition by redefining the discriminant of a monic ... View answer 0 votes Observe that the open sets of [0,1] are precisely of the form U \cap [0,1] for some open set U of \Bbb R. Now take any a \in (-\infty,0) and consider the open interval I=\left ( a,\frac 1 2 ... View answer Accepted answer 0 votes I have found the answer. It's quite easy. Take for instance f=X_1^2X_2 \in K[X_1,X_2] and g=X_1 \in K[X_1,X_2]. Neither f nor g is symmetric. But \frac f g = X_1X_2 \in K [X_1,X_2] \subseteq ... View answer Accepted answer 0 votes Let \pi be any prime factor of \Phi_n in \Bbb F_q[X]. WLOG we may assume that \pi is monic. Let \zeta be a zero of \pi in \Bbb F_q^{(n)}. Then \pi is the minimal polynomial of \zeta ... View answer Accepted answer 0 votes I have an answer according to my idea. Let me give my argument regarding this. Let L' be the splitting field of F. I will prove that \chi_y\ \big |\ F in L'[X]. Let w \in Gy. Then \exists ... View answer Accepted answer 0 votes I got an example. \Bbb R^2 is not an integral domain but it is an algebra over \Bbb R. Take x=(1,0) \in \Bbb R^2. Then \mu_x = X(X-1) \in \Bbb R[X] which is not irreducible in \Bbb R[X]. View answer Accepted answer 0 votes There is a more elegant way to think that problem. First we note that S/J is integral over R/I. So the map \psi : \text {Spec} (S/J) \longrightarrow \text {Spec} (R/I) defined by$$\psi (\... View answer -1 votes If$f = g$almost everywhere on$[a,b]$then$|f - g| = 0$almost everywhere on$[a,b].$Let$\mu$denote the Lebesgue measure on$[a,b].$Then$$\int_{a}^{b} |f-g|\ dx = \int_{[a,b]} |f-g|\ d\mu = 0.$...
Hint $:$ Observe that $f(x) = \lim\limits_{n \rightarrow \infty} \left (1 + \frac x n \right )^n = e^x,$ for all $x \in \Bbb R.$