Tutor profile: Gerardo P.

Let $$V = \mathbb{R}^2$$ be the vector space of real tuples over the real field. Prove that any subspace of $$V$$ must be either the zero subspace $$\lbrace 0 \rbrace$$ , the entire space $$V$$, or a space of the form $$W_v = \lbrace cv : c \in \mathbb{R} \rbrace$$ for some $$v \in \mathbb{R}^2$$

We know $$V$$ has dimension two. If $$U$$ is a subspace of $$V$$ it must have dimension less or equal to the dimension of $$V$$. That is, $$\dim U \leq \dim V = 2$$ so $$\dim U = 0, 1 $$ or $$2$$. If $$\dim U = 0$$ then $$U$$ must be $$\lbrace 0 \rbrace$$ since it is the only vector space of dimension zero. On the other hand, if $$\dim U = 2$$ then $$U = V$$ since there are no subspaces of $$V$$ of dimension two and distinct to $$V$$. Finally, if $$\dim U = 1$$ then it must have a one-element basis, say, $$\lbrace v \rbrace$$, and all elements of $$U$$ must be linear combinations of this basis. That is, all elements of $$U$$ must be of the form $$cv$$ for real scalars $$c$$, so $$U = \lbrace cv : c \in \mathbb{R} \rbrace$$ for some $$v \in \mathbb{R}^2$$

Prove that for every $$n \geq 1$$ and every $$m \geq 1$$ the number of functions from $$\lbrace 1, 2, \dots, n \rbrace$$ to $$\lbrace 1, 2, \dots, m \rbrace$$ is $$m^n$$

Let A be the set $$\lbrace 1, 2, \dots, n \rbrace$$, and $$B = \lbrace 1, 2, \dots, m \rbrace$$. Any function from $$A$$ to $$B$$ is completely determined by where it maps the elements in the domain, so choosing $$m$$ elements from $$B$$ and assigning them to $$f(1), f(2), \dots, f(m)$$ describe any one of such functions. Note that first element of $$A$$ can mapped to any of the $$m$$ elements of $$B$$, and so can the second element of $$A$$, and the third, and so on. By the Multiplicative Principle, given we must choose between $$m$$ values $$n$$ times, the number of functions from $$A$$ to $$B$$ must be equal to $$m \cdot m \cdot \dots \cdot m = m^n$$

Two runners start a race at the same time and finish in a tie. Prove that at some time during the race they have the same speed.

Let $$g(t), h(t)$$ be the functions which indicate the position at time $$t$$ of runner one and runner two respectively. Consider a function $$F(t) = g(t) - h(t)$$. The runners start the race at time $$t = 0$$, and both finish at some time $$t = t_f$$ (since they finish in a tie). This means that $$g(0) = h(0)$$ and $$g(t_f) = h(t_f)$$, so $$F(0) = 0$$ and $$F(t_f) = 0$$ Now we can assume the position functions $$g$$ and $$h$$ are continuous in $$[0, t_f]$$ and differentiable in $$(0, t_f)$$. By the Mean Value Theorem, there must be a time $$c$$ between $$0$$ and $$t_f$$ where $(F'(c) = \frac{F(t_f) - F(0)}{t_f - 0} = 0$) So there's a time where $$F'(c) = g'(c) - h'(c) = 0$$, so $$g'(c) = h'(c)$$. Since the derivative of the position is the speed, there is a time $$c$$ where the speed of both runners is the same.