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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 ?

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 $f(A) = B$.

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 $f:A\to B$.

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

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

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

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.

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

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.

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 $f(x)+f(x) = 1$...

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...

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...

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...

Hi evinda, That looks correct

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)...

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...

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 +...

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.

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.

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*}

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...

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 ?

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

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...

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...

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 $\{(-x,x)\mid x\ne0\}$...

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...

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 $x$...

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...

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.

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) <...

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...

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.

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.

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...

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...

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 $F$.

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...

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...

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...

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 $y<x$.

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

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

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.

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...

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...

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...

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.

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...

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...