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  1. castor28

    MHB How many miles did she drive

    You can see immediately that something is missing. Imagine you can find a solution $D$ for the distance. How would you known if this represents miles, kilometers, or some other unit ?
  2. castor28

    MHB Definition of onto function

    Point 1 was correct. $f(A)$ is the range, $B$ is the co-domain. $f$ is onto if the range equals the co-domain, i.e., if $f(A) = B$.
  3. castor28

    MHB Definition of onto function

    Regarding point 4, $f^{-1}(B) = \{x \in A \mid f(x)\in B\}$. Since B is the co-domain, this is true for any function $f:A\to B$.
  4. castor28

    MHB Group homomorphism

    Hi lemonthree, That is true, but you should also check that $\phi(x^{-1}) = \phi(x)^{-1}$ $\phi(1) = 1$
  5. castor28

    MHB Guide to Dealing with Double Subscripts

    Hi Peter, You should use braces, like $E_{k_1}$.
  6. castor28

    MHB Passengers on the bus

    Hi anemone, Is there information available somewhere about the Singapore method ?
  7. castor28

    MHB The Union of Two Open Sets is Open

    The point is that the argument is valid for every $x\in A_1\cup A_2$. If $C = A_1\cup A_2$, we have proved that, for every $x\in C$, there is an open ball $B(x,r)\subset C$ (where $r>0$ depends on $x$). That is precisely the definition of an open set.
  8. castor28

    MHB absolute value domain

    The distance between $y$ and $-3$ is at most $4$. That means $y$ is between $-3-4=-7$ and $-3+4=1$.
  9. castor28

    MHB Find the diameter of one circle

    I would say that the sums of the gaps are 24 (above) and 60 (below). As there is one more semicircle above, its diameter is equal to the difference 36.
  10. castor28

    MHB Proving Z[x] and Q[x] is not isomorphic

    Hi again, In fact, it is even simpler. If $\theta:\mathbb{Z}[x]\to\mathbb{Q}[x]$ is an isomorphism, then $\theta(1) = 1$, because any ring homomorphism must map $1$ to $1$. Now, in $\mathbb{Q}[x]$, we have $1 = \dfrac12+\dfrac12$. If $f(x)=\theta^{-1}(\dfrac12)$, we must have $f(x)+f(x) = 1$...
  11. castor28

    MHB Proving Z[x] and Q[x] is not isomorphic

    Hi Cbarker1, I don't see what is the point of your function $\phi$: it is not even defined on the whole of $\mathbb{Z}$. In reference to the title of you post, to prove that $\mathbb{Z}[x]$ and $\mathbb{Q}[x]$ are not isomorphic, you could use the fact that $\mathbb{Q}[x]$ is a Euclidean...
  12. castor28

    MHB Find the total number of red and blue beads

    Hi anemone, Are you sure the problem is correctly stated ? As I read it, you should still end up with $\dfrac25$ blue beads in container A. As the proportion does not change, you must move $3$ red beads for every $2$ blue. However, this can only decrease the proportion of blue beads in...
  13. castor28

    MHB Find the last digit of a series

    We have: \begin{align*} S_1 &= 1 + \cdots + n = \dfrac{n(n+1)}{2}\\ S_3 &= 1^3 + \cdots + n^3 = \dfrac{n^2(n+1)^2}{4} \end{align*} This shows that $S_3 = S_1^2$. Therefore, if $S_3\equiv1\pmod{10}$, then $S_1\equiv\pm1\pmod{10}$. It is rather obvious that $S_1\equiv S_3\pmod2$. We may write...
  14. castor28

    MHB Real Roots of Polynomial Minimization Problem

    Since $x=\dfrac{n+1}{2}$ is an axis of symmetry, the point $x=\dfrac{n+1}{2}$ is either a minimum of a maximum, depending on the shape of the quartic. However, the derivative $f'(x) = 4\left((x-1)^3+\cdots+(x-n)^3\right)$ is an increasing function (since it is a sum of increasing functions)...
  15. castor28

    MHB Numbers with a quadratic property

    As a matter of fact, I was interested in that very question. The question is about finding points with integer coordinates on a hyperbola, and this is a classical problem on representation by quadratic forms. I wrote something about it here. Sorry, it's in French, but ‶the equations speak for...
  16. castor28

    MHB [ASK]Show that the sum of the fifth powers of these numbers is divisible by 5

    Hi Monoxdifly, The distraction is the fact that there are ten integers. The fact that the sum is $0$ is essential. More precisely, if the integers are $a, b, c, \ldots$, Fermat's theorem tells us that $a^5\equiv a, b^5\equiv b\ldots\pmod5$. We have therefore $a^5 + b^5 + \cdots\equiv a + b +...
  17. castor28

    MHB [ASK]Show that the sum of the fifth powers of these numbers is divisible by 5

    Hi Monoxdifly, The number of integers does not matter. This would be a great time to use Fermat's Little Theorem : for any integer $a$, $a^5\equiv a\pmod5$, since $5$ is prime.
  18. castor28

    [ASK] Mathematical Induction: Prove 7^n-2^n is divisible by 5.

    Hi Monoxdifly, $12\cdot2^k$ is certainly not divisible by $5$ (look at the prime factors). You made a mistake in your calculation: I get: $$ 7^k\cdot7 - 2^k\cdot2 = 7(7^k-2^k) + 5\cdot2^k $$ and this should clear things up.
  19. castor28

    MHB [ASK} Prove (cos2x+cos2y)/(sin2x−sin2y)=1/tan(x−y)

    Hi Monoxdifly, You could start with the LHS and the identities: \begin{align*} \cos a + \cos b &= 2\cos\frac{a+b}{2}\cos\frac{a-b}{2}\\ \sin a - \sin b &= 2\sin\frac{a-b}{2}\cos\frac{a+b}{2} \end{align*}
  20. castor28

    MHB Sort elements of matrix

    Hi evinda, I see a few errors in what you did. I understand that 'matrix' is the given matrix; its size should be $n$, not $100$. I would suggest to make the count array 101 elements long, and leave count[0] unused. In this way, the index of the count array corresponds directly to the value...
  21. castor28

    MHB Complex Derivative .... Remark in Apostol, Section 16.1 .... ....

    Hi Peter, With $z=x+iy$, you have $f(z)=u + iv$, with $u=x^2+y^2$ and $v=0$. What do the Cauchy-Riemann equations tell you ?
  22. castor28

    MHB Proof of the Equality of Supremums (Or Something Like That Anyway :) )

    Hi AutGuy98, Here is a simple example: take $A=\{1\}$ and $B=\{0,1\}$. You have $\sup A = \sup B = 1$.
  23. castor28

    MHB Multiple consists of only 1's

    Let $N$ be the number. As $\gcd(N,10)=1$, $10$ has finite order, say $k$, modulo $N$. This means that $10^k-1$ is a multiple of $N$; that number consists of $k$ digits $9$. Let $M = \dfrac{10^k-1}{9}$ (a number consisting of $k$ digits $1$). We have $10^k-1 = 9M = aN$ for some integer $a$. If...
  24. castor28

    MHB Compact Metric Spaces and Inverse Functions .... Apostol, Example After Theorem 4.29 ....

    Hi Peter, Your idea is correct. Now, the sequence $\{x_n\}$ converges to $1$ in $\mathbb{R}$, but, as $S$ does not contain $1$, that sequence does not converge in $S$. Intuitively, what happens is that, when a point $P$ travels counterclockwise in circles in $f(S)$, the image $f^{-1}(P)$ jumps...
  25. castor28

    MHB Compact Topological Spaces .... Stromberg, Theorem 3.36 .... ....

    Hi Peter, You need that to ensure that $U$ is a neighborhood of $x$. In a topological space, any union or any finite intersection of open sets is open, but that does not hold for infinite intersections. For example, in $\mathbb{R}$, the intersection of the open intervals $\{(-x,x)\mid x\ne0\}$...
  26. castor28

    MHB Compact Topological Spaces .... Stromberg, Theorem 3.36 .... ....

    Hi Peter, If $U\not\subset S'$, there is an element $u\in U\cap S$. Since the $V_y$ cover $S$, $u\in V_{y_n}$ for some $n$. On the other hand, as $u\in U$, $u\in U_{y_k}$ for all $k$. In particular, $u\in U_{y_n}$ and this contradicts the fact that $U_{y_n}$ and $V_{y_n}$ are disjoint. Note...
  27. castor28

    MHB Closed and Bounded Intervals are Compact .... Sohrab, Propostion 4.1.9 .... ....

    Hi Peter, My answer of question 2 is not quite correct. In fact, we do not necessarily have a finite cover of the interval $[a,c)$ (which is not compact). We do not know if $c\in S$. What we do know is that, for any $x$ with $a<x<c$, we have a finite cover $U_x$ of $[a,x]$. We can take $x$...
  28. castor28

    MHB Closed and Bounded Intervals are Compact .... Sohrab, Propostion 4.1.9 .... ....

    Hi Peter, For question 1, since $c<b$ and $\epsilon>0$, you can take $d$ between $c$ and $\min(c+\epsilon,b)$. For question 2, since $c<d<c+\epsilon$, and $(c-\epsilon,c+\epsilon)\subset O_\lambda$, $d\in O_\lambda$, and you can add $O_\lambda$ to the finite cover of $[a,c)$ to get a finite...
  29. castor28

    MHB The Number e .... Sohrab Proposition 2.3.15 ....

    Hi Peter, $n!=1\cdot2\cdot3\cdots n$ is the product of $n$ factors. As each of the $n-2$ factors after $2$ is at least equal to $3$, we have: $$ n!\ge1\cdot2\cdot3^{n-2}=2\cdot3^{n-2} $$ and the result follows by taking the inverse.
  30. castor28

    MHB Evaluate ⌊ 1/a_1+1/a_2+....+1/a_{2008} ⌋

    Note: I assume that the recurrence holds for $k\ge\bf1$. Let us write $ b_n = \dfrac{1}{a_n}$, $S(n) = \sum_{k=1}^nb_k$, and $T(n) = \sum_{k=n}^{2008}b_k$. As $a_{n+1} > a_n^2$, we have $b_{n+1} < b_n^2$. If $b_n < 1$, comparison with the geometric progression of ratio $b_n$ gives: $$T(n) <...
  31. castor28

    MHB Elementary Number Theory proof

    Hi Cbarker1, $s$ and $t$ are arbitrary integers in the question, you should not use them as you do in the proof, where they are fixed integers that depend on $a$, $b$, and $k$. You can say that there are integers $m$ and $n$ such that $a=km$ and $b=kn$. Now, you have: $$ as+bt = (km)s + (kn)t...
  32. castor28

    MHB Linear Combination - missing data ?

    You must have $(k-3)b = b$; since $b\ne0$, this implies $k-3 = 1$ and $k=4$. This is the solution you mentioned as correct.
  33. castor28

    MHB Linear Combination - missing data ?

    Hi Yankel, You have: $$ A(ku-3v) = kAu - 3Av = (k-3)b $$ On the other hand, as $ku-3v$ is also a solution of the system, you have: $$ A(ku-3v) = b $$ Since $b\ne0$ by hypothesis, the conclusion follows.
  34. castor28

    MHB Set vs Class: Exploring the Differences

    Hi highmath, Each set can be viewed as a class, but the converse is not true. You can have sets of sets, but not necessarily sets of classes. A class is basically any collection you can think of; this was the approach used by Cantor in his initial theory (the naive set theory). However, it...
  35. castor28

    MHB Converse, Contrapositive and Negation for multiple Quantifiers

    Hi Sandra, The first two propositions are correct. For the third one, the natural language is correct too. For the symbolic form, note that you already have (from the first two parts) the expression of the two parts of the statement: Every printer is busy: $\forall p B(p)$ There is no job in...
  36. castor28

    MHB Intermediate field extensions

    Hi mathmari All that is correct. For the second part, we could be a little more precise and modify the second paragraph as: Since the basis consists of some elements of the basis of $L_2$ over $F$, we get that $[L_1L_2:L_1]\leq$ dimension of $L_2$ over $F$.
  37. castor28

    MHB Intermediate field extensions

    Hi mathmari, It is true that $L_1L_2=L_1(b_1,\dots,b_m)$, but the $b_i$ are not necessarily independent over $L1$; we only know that they are independent over $F$. In a vector space, any spanning set contains a basis. By removing redundant elements, you can extract form the $b_i$ a basis of...
  38. castor28

    MHB Trigonometric inequality: sin (1/(n+1934))<1/1994

    For $0<x<1$, we have $\sin(x)<x$. This shows that the inequality is satisfied for $n=60$. Taking $n=59$, we find: $$ \sin\left(\frac{1}{1993}\right)\approx 0.00050176> \frac{1}{1994}\approx 0.00050150 $$ As $\sin\left(\dfrac{1}{n+1934}\right)$ is a decreasing function of $n$ for $n>0$, this...
  39. castor28

    MHB Well Orders and Total Orders .... Searcoid Definition 1.3.10 ....

    Hi Joppy, A singleton set is totally ordered, because the definition is vacuously true. The definition of a totally ordered set $S$ is equivalent to: if $\{x,y\}\subset S$ and $x\ne y$, then $x<y$ or $y<x$. If $S$ is a singleton, the condition $x\ne y$ is always false. As the antecedent of...
  40. castor28

    MHB Well Orders and Total Orders .... Searcoid Definition 1.3.10 ....

    Hi Peter, You just have to apply the definition to subsets of two elements. More explicitly, if $S$ is well ordered and $\{x,y\}\subset S$ with $x\ne y$, then $\{x,y\}$ has a smallest element. If that element is $x$, then $x<y$; if that element is $y$, then $y<x$.
  41. castor28

    MHB Finding the Greatest Common Divisor of Two Integers

    Hi, They use Euclid's algorithm, described here. (Look in particular at section 2, Description).
  42. castor28

    MHB Prime and Irreducible Elements in Principal Ideal Domains .... Bland - AA - Theorem 7.2.14 .... ....

    Hi Peter, That comes from the assumption that $p\mid ab$ (on line 3 of the proof).
  43. castor28

    MHB Solves Theorem 3.2.19 in Bland's Abstract Algebra

    Hi Peter, We know that $x\ne0$ because the proof says that $xR$ is a nonzero prime ideal. To be completely correct, you should say that $x=xyb$ implies $x(1-yb)=0$, and this implies $yb=1$ because $x\ne0$ and $R$ is an integral domain.
  44. castor28

    MHB Maximal Ideal .... Bland - AA - Example 2, Section 3.2.12 .... ....

    Hi Peter, Actually, Bland claims that $5\mathbb{Z}$ (not $I$ as you wrote) is a maximal ideal of $\mathbb{Z}$. The two statements are equivalent to ‘‘if $5\mathbb{Z}\subseteq I \subseteq\mathbb{Z}$, then $I=5\mathbb{Z}$ or $I=\mathbb{Z}$’’ (this is the definition in the book). To see this...
  45. castor28

    MHB Principal Ideal Domains .... ....

    Hi Peter, If $(a)=(b)$ then $a\in(b)$ and $a=rb$ from some $r\in D$; a similar argument shows that $b=sa$ for some $s\in D$. We have therefore $a = rb = (rs)a$ and $a(rs-1)=0$. As we are in an integral domain, this implies $rs=1$, which means that $r$ and $s$ are units and $a$ and $b$ are...
  46. castor28

    MHB Challenge problem #5 [Olinguito]

    As $\zeta_n^{[1]}$ is closed under multiplication, $\zeta_n^{[r]}\subset\zeta_n^{[1]}$. On the other hand, as $1\in\zeta_n^{[1]}$, $\zeta_n^{[1]}\subset\zeta_n^{[r]}$. The conclusion is that $\zeta_n^{[r]}=\zeta_n^{[1]}$ for all $r$. The sum of the elements of that set is the sum of the roots of...
  47. castor28

    MHB Calculate Area of Tree Structure (Wondering)

    Hi mathmari, Yes, that is correct, and it also gives the answer $3S_1$ for the total area. This is even simpler than what I had in mind: after all, the fact that the shapes are similar does not matter, the only thing that matters is the ratio of the areas.
  48. castor28

    MHB Calculate Area of Tree Structure (Wondering)

    Hi mathmari, That is correct so far. Now that I think a bit more about it, I realize that what I wrote is not correct (although the final result is correct by chance;)); sorry for that :o. The horizontal rectangles are not similar to the vertical rectangles, contrary to what I first thought...
  49. castor28

    MHB Calculate Area of Tree Structure (Wondering)

    Hi mathmari, Note that all those trees are similar. Specifically, each of the two subtrees on top of the initial rectangle has an area equal to $\dfrac13$ of the whole tree. If $T$ is the area of the whole tree and $S$ the area of the initial rectangle, we have: $$T = S + \frac23 T$$ which...