Recent content by castor28

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)...