Here it's another way. The triangles AhC and Bhf are similar. If you put $Bh=x$ and $Bf=y$ you get the relations $$x+\frac{4}{3}y=\frac{4} {3} x$$ $$9=x^2+y^2$$ From which you can obtain the length ...

Generally speaking no, in fact $\mathbb{Q} [X, Y]/(Y) \simeq \mathbb{Q} [X]$ is a PID. In this case the quotient isn't a PID.

$F$ is the union of adjoining of finitely many quadratic extension which have degree over $\mathbb{Q}$ a power of 2. Then your element must be in one of those finite adjoining, but it has degree 3 ...

If you chose $y_i=-x_i$ the claim is false.

I assume you want all the pairs $(y,x)$ such that $y$ is 5 digits, $x$ is 4 digits, $y-x=11111$ and every digit from $1$ to $9$ appears either in $y$ nor $x$. This implies that there is no digit $0$ ...

In both cases $x^2-1$ factorizes as $(x-1)(x+1)$, so the ideal it generates can't be maximal both if you are on $\mathbb{Q}$ or $\mathbb{R}$. Notice that in both cases the class of $x-1$ is a zero-...

I'll suppose that you want a surjective homomorphism or the answer is trivial. For the first question you just can choose $\mathbb{Z}_2 \times \mathbb{Z}_{10}$. For the second answer you can't have ...

Your first claim is wrong. Take the free abelian group $\mathbb{Z}$, which has basis $\left\lbrace 1\right\rbrace$. Then a subgroup is of the form $n\mathbb{Z}$ which is free, generated by $\left\... View answer 2 votes Given that$yx=y^2x$every element can be written$x^ny^m$with$n\leq 3$and$m\leq 4$. That's exactly$20$elements so it's enough to show that different elements have different writings. View answer Accepted answer 2 votes You can consider$\frac{a_n} {n^2}$which is absolutely convergent (consider$S''(0)$) instead of$a_n$and$S''(x) $instead of$S$which is the corrispondent to$S$. If you prove the statement for ... View answer 2 votes To show that$R[[X]]$is a UFD you can use Theorem 20.3 in Commutative Ring Theory by Matsumura which states that if$A$is a regular UFD so is$A[[X]]$and show that every PID is regular which is ... View answer Accepted answer 2 votes I seem to understand that the problem is the equality$\left( \bigcap A_n\right)^C=\bigcup A_n^C$. So let's take an element$x\in \left( \bigcap A\right)^C$then there is an$A_n$such that$x\notin ...

The words with three different letters are the rearrangement of "AEP" so are $3!=6$. The words with two different letters are rearrangement of AAP,AAE,EEA,EEP so 12. So i get a total of 18.

Let's call good a string that starts with $1$ and contains two consecutive ones. To obtain a good string of length $n$ we can certainly take a good string of length $n-1$ and add a $1$ on the left. We ...

The intuition of subtracting from the total number is correct. Now we need to compute how many arrangements there are that don't fullfill the condition. So, let's count how many arrangements where ...

Recall that $$P(B|A)=\frac{P(A\cap B)}{P(A)}$$ Now you want to compute the probability of not be selected for the first two questions (for simplicity). If $A$= "you have not been selected for the ...

Fix a root $\alpha$ of $P(X)$. Then every polynomial of that form with root $\alpha$ must be of the form $(x-\alpha)(x-\beta)$ and $b=-\alpha-\beta, c=\alpha\beta$. From this formula we get that the ...

Not 100% sure if that's really what you are asking, but you should pay attention to the fact that you may have different bases in the domain and codomain. If you only apply the change of basis matrix ...

If $j\in R$ and $\frac{1}{256}\in R$ (which is true if and only if $\text{char} k\neq 2)$ you have that $\lambda$ solves a polynomial with coefficient in $R$ so is in $R$ because $R$ is normal as you ...

Every $\alpha$ has a finite orbit under H $\{\sigma_1(\alpha),\ldots,\sigma_r(\alpha)\}$. Then the polynomial $$\prod (x-\sigma_i(\alpha))$$ Is in $L^H$. This shows that the extension is separable, ...

Your statement is wrong, that's why you can't prove it. In fact you are not proving the uniqueness (which is true). To prove the uniqueness suppose that we have $b^2=c^2=a$. As you wrote we have $b^{... View answer 1 votes If it wasn't for the down payment the division would be$\frac{5}{9}(T-E)$for$s$and$\frac{4}{9}(T-E)$for$m$. So just add B to the amount due to$s$(let's say that he take his down payment and ... View answer 1 votes There is an error in the second row of the second column. The inequality should be $$\left(\frac{3}{4}\right)^n <0.25$$ Edit: to be fair there are some other errors after that point, but that's ... View answer 1 votes It's not true. In fact if you set$n$the function $$x\rightarrow x^n$$ Tends to 1 if$x$tends to 1, so there are no$a<1$that satisfies your hypotesis View answer Accepted answer 1 votes **HINT:**Think about how many ways you have to choose the first letter, then the second, then the third and so on... View answer 1 votes In your solution you fix the position of the vowels as if, for example, they're always the first and second letters of the word. So you have to multiply 720 for 6 and then subtract two times the words ... View answer 1 votes Connected means that there are no non-trivial subsets which are both open and closed View answer 1 votes You can use known limits such has$\lim _{n\to \infty} \frac {k^n}{n^h}=\infty \quad \forall k>1$and$h\in \mathbb{R}$or$\lim _{n\to \infty} \frac {n^h}{\log^k{n}}=\infty \quad \forall k> O,h\...

The differential equalities gives $f(x)=e^{\lambda \frac{x^2}{2}+c}$ which is never equal to 0 or f constant which can't satisfy your requests.
You can define the Cantor set recursively. What I mean is that you take the interval $[0,1]$, divide it in 3 equal parts $[0,1/3]$, $[1/3,2/3]$ and $[2/3,1]$ and consider the rightmost an the leftmost....