just for fun! assume there exist $a,b$ relative prime integers such that $\frac{a}{b}=\sqrt{2}$, we can assume $a$ odd (otherwise we can argue in a similar way with $b$) hence $a \;\text{mod} \;2 =1$....

You can use the $F$-test in order to calculate statistically significance, given you hypotheis you have that $$F_n= \frac{\rho^2}{1-\rho^2}*(n-2)$$ Hence you obtain the following $F_{100} = 0.0098$...

the last equation tell you that also $\mathbf{v_4}$ is not necessary since $$\mathbf{v_4} = -\frac{2 \mathbf{v_3} + 7 \mathbf{v_2} + 4 \mathbf{v_1}}{2}$$ hence it lies in $\mathcal{L} \{ \mathbf{... View answer 2 votes your hypothesis is that the equation$AB=B$holds for any matrix not only for the null matrix. Try to use different matrix such as for$B$take the identity matrix or any invertible one View answer Accepted answer 2 votes The expected sum after$k$rolls is not$k$but$6(k-1)+3$indeed if we stop at first roll it means that we get$1,2,3,4,5$and the average sum will the mean i.e.$3$, if we rolled twice means that ... View answer Accepted answer 2 votes In general it is false, just consider the cases in which$H$is abelian when two different elements cannot be conjugates. For a more concrete example let us consider$S_3$the group of permutation of ... View answer 2 votes Use double inclusion, clearly$[0,1]$is contained in the intersetion since each point$P\in [0,1]$lies in each open interval$(\frac{-1}n ; 1+\frac{1}n)$. For the other inclusion, pick a point$Q$... View answer Accepted answer 2 votes Without any other assumption the sentence seems to be false. Consider the topological space consisting of a line$l$and one point$P$external to the line. Then$P$is open in the total space, ... View answer 2 votes Have a look at https://en.wikipedia.org/wiki/Square_triangular_number So, the answere is yes! View answer 2 votes Written in a such way the problem is quite trivial, clearly $$ab = 2a\frac{b}{2},$$ hence if$ab=k$we get that$ 2a\frac{b}{2}=k$(unless you are in char =2, but I think it is not the case). So, ... View answer 1 votes In general the statement is not true. The problem is that$S_1T$is the identity on the image of$T$which could be smaller than$V. Think for example in the space of polynomial ring the ... View answer 1 votes For simple linear regression you the formula: \begin{align} \beta_1 &= \frac{\sum_{i=1}^N (x_i-\bar{x})(y_i-\bar{y})}{\sum_{i=1}^N (x_i-\bar{x})^2}\\ \beta_0 &= \bar{y}-\beta_1 \bar{x} \end{... View answer Accepted answer 1 votes let us consider any number in base 8,a_n 8^n +,\dots, + a_0$observe that if$n$is even then$ a^n \equiv 1 \;\text{mod} 9$and if$n$is odd then$ a^n \equiv -1\; \text{mod} 9$then write the ... View answer Accepted answer 1 votes If$Q= \emptyset$the$\tau$is the discrete topology i.e. each set is open hence$X$is regular. Otherwise each pair of open set$A$and$B$clearly have not trivial interception ($Q\subset A\cap B$) ... View answer 1 votes By definition eigenvector satisfies the equation $$A(v) = \lambda v$$ So you just have to find the kernel of the matricx$A-\lambda I$, where$\lambda$is one eigennvalue and$I$the identity ... View answer Accepted answer 1 votes The first element of the$n$th row will be$\frac{n(n-1)}{2}+1$, since it will be the sum of all diferences plus 1, then if you want the$k$the element of the$n$th row it will be$\frac{n(n-1)}{2}+...

You want to show that $|a_1+\dots+a_n+a_{n+1}|\leq |a_1|+\dots +|a_{n+1}|$. Using induction we have that $|a_1+\dots+a_n+|\leq |a_1|+\dots + |a_n|$, hence we get $$|a_1+\dots+a_n+a_{n+1}|\leq |a_1 +\... View answer 0 votes The t-score for a parameter \beta_i is defined as t_i = \frac{\hat{\beta_i}}{s_{\hat\beta_i}}, in your case you should find the t-score of the slpoe t=\frac{0.15}{0.0005} View answer 0 votes I G acts faithfully on a set X then we can define a new action on any set X\subset Y setting g(a)=a for any a\in Y\setminus X and Y can have any cardinality View answer Accepted answer 0 votes Since \dim\text{Ker}(N^4)= 13 and \dim\text{Ker}(N^5)= 15 the maximal Jordan block is of size 5, and you have exactly two Jordan blocks J_1 and J_2 of size 5, indeed, if there are no such ... View answer Accepted answer 0 votes If you want the 95% of the population in a gaussian you have two tails, so you need to consider the popultaion between 2.5% and 97.5% View answer Accepted answer 0 votes try to find what is missing from B to be a cube, i.e. p has exponent 1 so since it is a prime number you need to multply for p^2 so that the prodocut is p^3 that is a cube, in the same way you ... View answer 0 votes Observe that H is a cyclic gruop hence the order of x is n=|H| which implies that x^n=1_G hence x^{|G|}=1_G^{|G|/n}=1_G View answer 0 votes If m=1 then there is nothing to prove, otherwise let us assume they are linearly dependent, then we have:$$\alpha_0 v + \dots + \alpha_{m-1} T^{m−1}v =0$$Applying T^{m-1} on both sides we ... View answer 0 votes In order to find \beta_0 and \beta_1 you have to solve the linear system$$\begin{align} \frac{\partial L}{\partial \beta_0} &= -2\sum_{i=1}^N (y_i-(\beta_0 + \beta_1 x_i))=0,\\ \frac{\...

Start calculating the image of the basis $A$, for example for the first element of the basis we have $$T\left(\begin{bmatrix}1&0\\0&0\end{bmatrix}\right) = (1,0,0)$$ then write the element as ...

If it helps, you can start with a particular matrix, however in order to prove the statement you need to make it in general finding excplicty the vector $x$. For the first point if you pick $x=0$ ...
If you are looking for the OLS solution as best fit for your model, then the formula for finding the vector $\beta$ is given by \begin{equation} \beta =(XX^T)^{-1} X^T Y \end{equation} provided ...