#*16.

b) We are given that $\forall i \in \mathbb{Z}^+, \forall x_i \in \mathbb{R}, f(x_1 + \dots + x_n) = f(x_1) + \dots + f(x_n).$ We claim that $\exists c \in \mathbb{R}, \forall x \in \mathbb{Q}, f(x) = cx.$

First, note that $f(1 + 0) = f(1) = f(1) + f(0) \Rightarrow f(0) = 0.$ Thus, certainly any $c$ will work for $f(0) = c0 = 0.$ Assume that $k \in \mathbb{Z} \setminus \{ 0 \}.$ Notice that $\sum_{1}^k \frac{1}{k} = 1.$ Then we have that $f \left( \sum_1^k \frac{1}{k} \right) = \sum_{1}^k f \left( \frac{1}{k} \right) = k f \left( \frac{1}{k} \right) = f(1).$ Dividing by the nonzero …

*17.
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Assume on the contrary that $f(x)^2 \ge r^2$ and $f(x)^2 \le p^2$. Then $f(x)^2 \ge r^2 \Rightarrow (f(x) - r)(f(x) + r) \ge 0.$ Since $f$ is not the zero function and $r$ is positive, we have that $f(x) \in (-\infty, -r]\cup [r,\infty).$ Similarly, we have that $f(x)^2 \le p^2 \Rightarrow (f(x) - p)(f(x) + p) \le 0.$ This implies that $f(x) \in [-p,p].$ At the beginning, we assume that $0 < p < x < r \Rightarrow 0 < p^2 < x^2 < r^2.$ This tells us that $-p > -r,$ and so $f(x) \in [-p,-r],$ which implies that for any real positive $x$, the output will be negative.

If $f(x)=0$ for all $x$, then $f$ satisfies $f(x+y)=f(x)+f(y)$ for all $x$ and $y$ and also $f(x \cdot y)=f(x) \cdot f(y)$ for all $x$ and $y$. Now suppose that $f$ satisfies these two properties, but that $f(x)$ is not always 0 . Prove that $f(x)=x$ for all $x,$ as follows:
(a) Prove that $f(1)=1$.
(b) Prove that $f(x)=x$ if $x$ is rational.
(c) Prove that $f(x)>0$ if $x>0$. (This part is tricky, but if you have been paying attention to the philosophical remarks accompanying the problems in the last two chapters, you will know what to do.)

hey I have what feels like a super basic homework problem that I am stuck on.
This is taken from MIT OCW 18.02 pset 1.
I have a vector A, which is expressed as A = a*i + b*j
and I am trying to write an expression for A' in terms of a and b, where A' is A rotated 90deg.

I feel like I am most of the way there, (for the purpose of working out this problem I'm letting A' = u*i + v*j)

I have a/b = - v/u and seperatly a^2 + b^2 = v^2 + u^2

intuitivly, I can see that a,b can equal (-v,u) or (u,-v) [because I can see that in the diagrams I have drawn]....But is there a way to get there from the …