1 The language of mathematics and prompting

I am a student following the course based on the attached notes. Please guide me through Exercise 1.8. Emphasize my learning and do not give me the answers but only hints. You must only use material from the attached notes in the solution. Be sure to reference what you use. Please make a serious effort to render the mathematics in your output using KaTeX so that I can read it!

Please solve the equation $x^2 - x + 1 = 0$. Guide me through the steps. Make sure that the underlying logic in your arguments is correct.

Please act like a friendly tutor and teach me about the derivatives of simple functions. Test my understanding after each concept you explain.

I am doing weekly exercises in mathematics at the college level. Please suggest a very simple template in LaTeX for hand in of these exercises. Also, show me how to typeset an equation in LaTeX.

There seems to be a web interface to LaTeX called Overleaf. Please tell me how to access this so that I can enter a weekly exercise.

Please give feedback on the mathematics contained in the $\LaTeX$ below in triple quotes. Emphasize logic and precision. """
$$
x^2 = 1 \implies x = 1
$$
From this it follows that $1 + 1 = 3$. """

Please give me the complete solution to Exercise 3.44 in the attached pdf file. Be sure to use only mathematics from this attached file and referencing precisely the proper definitions, propositions, theorems, etc. Give your answer as the source code for perfectly formatted LaTeX.

What is
$$
-\sqrt{\frac{1}{2} (1 - \sqrt{2} + \sqrt{3 - 2\sqrt{2}})}
$$

I am taking a mathematics course, where I explore mathematics in the browser by running small snippets of python code using the libraries numpy, matplotlib and sympy. I have no/some/extensive prior programming experience. Give me a brief introduction to python and these three libraries. Finish your reply with a small exercise I can do. If no/some/extensive is present above in this prompt, remark this and only reply with Please select your programming experience.

Give me python code using sympy to compute pi with 100 decimals. I want a one line command.

The expression $(x_1 \implies ((\neg x_2) \lor x_3))$ is a proposition. Let us see how it is generated using the rules in Definition 1.14 .

1. First, $x_2$ is a proposition using (ⅰ) .
2. Then $(\neg x_2)$ is a proposition by using (ⅱ) with $P = x_2$, since we know by (1) that $P$ is a proposition.
3. Since $x_3$ is a proposition by (ⅰ) , it follows that $((\neg x_2) \lor x_3)$ is a proposition using (ⅲ) with $P = (\neg x_2)$ and $Q = x_3$, since we know by (2) that $P$ is a proposition.
4. Finally, since $x_1$ is a proposition by (ⅰ) it follows by (ⅲ) with $P = x_1$ and $Q = ((\neg x_2)\lor x_3)$ that
   $$(x_1 \implies ((\neg x_2) \lor x_3))$$ is a proposition, since we know by (3) that $Q$ is a proposition.

The truth tables corresponding to the propositions $(x_1 \land (x_2 \lor x_3))$ and $((x_1\land x_2) \lor (x_1\land x_3))$ are given below. $$\def\arraystretch{1.2} \begin{array}{c|c|c|c} x_1 & x_2 & x_3 & (x_1 \land (x_2 \lor x_3)) \\ \hline f & f & f & f \\ f & f & t & f \\ f & t & f & f \\ f & t & t & f \\ t & f & f & f \\ t & f & t & t \\ t & t & f & t \\ t & t & t & t \end{array}\qquad\qquad \begin{array}{c|c|c|c} x_1 & x_2 & x_3 & ((x_1\land x_2) \lor (x_1\land x_3)) \\ \hline f & f & f & f \\ f & f & t & f \\ f & t & f & f \\ f & t & t & f \\ t & f & f & f \\ t & f & t & t \\ t & t & f & t \\ t & t & t & t \end{array}$$ For example, if $x_1 = t, x_2 = f$ and $x_3 = t$, then $$(x_1 \land (x_2 \lor x_3)) = (t \land (f \lor t)) = (t \land t) = t$$ and $$((x_1\land x_2) \lor (x_1\land x_3)) = ((t \land f) \lor (t \land t)) = (f \lor t) = t.$$

If $p(n) = n\ \text{is a prime number}$, then $p(3)$ is true, whereas $p(6)$ is false. $$q(m, n) = p(m) \land (\neg p(n))$$ is a predicate in two variables $m$ and $n$. Here $q(3, 4)$ is true, whereas $q(5, 7)$ is false.

Here is a statement about real numbers $$x^2 = 1 \qquad \implies (x-1)(x+1)(x-2) = 0 \tag{1.4}$$ This statement reads: no matter which real number $x$ you pick, if $x^2 = 1$, then $(x-1)(x+1)(x-2) = 0$. We definitely want this to be true. Being true means that (1.4) must hold for all numbers $x$, also $x = 2$, which reads $$2^2 = 4 = 1 \qquad \implies \qquad (2-1)(2+1)(2-2) = 0 = 0$$ The above statement is an example of a false implies true statement, which we want to be true.In general terms, in proving the statement that $p(x) \implies q(x)$ holds for every $x$ in some set $S$, we are really only interested in $x\in S$ for which $p(x)$ is true, since $p(x)$ is our assumption. We still need $p(x)\implies q(x)$ to be true for $x\in S$ for which $p(x)$ is false. This is assured by the truth table for $\implies$, since $f \implies t$ and $f\implies f$ are both true.

Explain precisely in terms of propositions and logic why the empty set is a subset of any given set.

Suppose that $S = \{-2, -1, 2, 3\}$ and $$p(x) = x \text{\ is positive.}$$ Then $$\{x\in S\mid p(x)\} = \{2, 3\} \subseteq S.$$

If $A = \{1, 2\}$ and $B = \{1, 2, 3\}$, then $$A\times B = \{(1, 1), (1, 2), (1, 3), (2, 1), (2, 2), (2, 3)\}.$$

So there are many different ways of representing a rational number, such as $$\frac{3}{7} = \frac{6}{14} = \frac{9}{21}.$$ Here $$\frac{3}{7} = \frac{9}{21}\qquad\text{since}\qquad 3\cdot 21 = 7\cdot 9.$$ In fact, a fraction stays the same when its numerator and denominator are multiplied by the same natural number.

Is
$$
x \geq 0 \iff x^2 \geq 0
$$
a true statement? Give me 3 carefully crafted exercises training me in distinguishing $\implies$ and $\iff$. Only use basic mathematics involving numbers and arithmetic operations. After each exercise stop, ask for the answer and give valueable feedback and guidance.

An integer is called even if it is divisible by $2$. So the even integers are $$\{\dots, -4, -2, 0, 2, 4, \dots\}.$$ An integer is called odd if it is not even. So the odd integers are $$\{\dots, -5, -3, -1, 1, 3, \dots\}.$$ Consider the proposition: $$\forall n\in \ZZ: p(n)\implies p(n^2), \tag{1.15}$$ where $p(n) = (n\text{ is odd})$ i.e., the square of an odd integer is odd. This seems true for a first selection of examples: $3^2=9, 5^2=25, \dots$.What does it mean exactly for a number to be odd? This means that it is not divisible by $2$ or that there exists another integer $a$, such that $n = 2 a + 1$. So $$p(n) = \exists a\in \ZZ: n = 2 a + 1.$$ Therefore we need to show that $$\left(\exists a\in \ZZ: n = 2 a + 1\right) \implies \left(\exists b\in \ZZ: n^2 = 2 b + 1\right).$$ Notice that I had to change $a$ into $b$ in the second proposition above. The two variables are not the same: $a$ is associated with $n$ and $b$ is associated with $n^2$.Let us assume that $n = 2 a + 1$. Now we need to argue that $n^2 = 2 b + 1$ for some $b\in \ZZ$. You stare at this for a while and notice that we should use the assumption $n=2 a + 1$ in computing $n^2$: $$n^2 = (2 a + 1)^2 = (2 a)^2 + 2 (2 a) + 1^2 = 4 a^2 + 4 a + 1 = 2(2 a^2 + 2 a) + 1.$$ Thus, using our assumption we may conclude that if $n = 2 a + 1$, then $$n^2 = 2 b + 1,$$ where $b=2a^2+ 2 a$. This completes the proof.

An irrational number is a (real) number that is not rational. It is a startling fact that such numbers exist, but they do! The square root $\sqrt{2}$ of two is an example. We will prove that there exists two irrational numbers $\alpha, \beta$, such that $\alpha^\beta$ is rational.Consider the proposition $p$ given by $$\gamma = \sqrt{2}^{\sqrt{2}}\quad\text{is rational.}$$ Either $p$ is true or false. If $p$ is true we are done putting $\alpha = \beta = \sqrt{2}$. If not, then $p$ must be false and $\gamma$ is irrational. But then $$\gamma^{\sqrt{2}} = \left(\sqrt{2}^{\sqrt{2}}\right)^{\sqrt{2}} = (\sqrt{2})^{\sqrt{2}\cdot \sqrt{2}} = \sqrt{2}^2 = 2$$ and we are done putting $\alpha = \gamma$ and $\beta = \sqrt{2}$.So which one is it? Is $$\sqrt{2}^{\sqrt{2}}$$ rational or irrational? This is really advanced mathematics based on the Gelfond-Schneider theorem.

Let us use proof by contradiction to show that the proposition $p:$ ($\sqrt{2}$ is an irrational number) is true. Assuming that $p$ is false, we must have that $\neg p$ is true. But $\neg p$ is the proposition $p_1$ (that $\sqrt{2}$ is a rational number) $$\exists m, n\in \NN: \sqrt{2} = \frac{m}{n}.$$ Here $p_1 \iff p_2$, where $p_2$ is the proposition $$\exists m, n\in \NN: \left(\sqrt{2} = \frac{m}{n}\right) \land ((m \text{ is odd}) \lor (n \text{ is odd})),$$ since we can assume that $2$ is not a common divisor of $m$ and $n$ by Definition 1.54 . However, $$\sqrt{2} = \frac{m}{n} \iff \sqrt{2} n = m \iff 2 n^2 = m^2 \implies m\text{ is even}.$$ The last implication above follows from Exercise 1.81 . If $m$ is even, then $m = 2 k$ for some $k\in \NN$. Therefore $m^2 = 4 k^2$ and $$2 n^2 = 4 k^2 \iff n^2 = 2 k^2$$ so that $n$ is also even. We have proved that $p_2$ implies the proposition $p_3$ given by $$(m \text{ is even}) \land (n \text{ is even}).$$ But notice that $p_3$ is the negation of the second part of $p_2$ i.e., $p_3 \equiv \neg \left((m \text{ is odd}) \lor (n \text{ is odd})\right)$. So if $p_2$ were true, then $p_3$ would be true, since we proved that $p_2 \implies p_3$ is true. But $p_3$ would also be false, being the negation of a part of the true proposition $p_2$. This is impossible. Therefore we must have that $p_2$ is false and therefore that $p_1$ is false. But then according to the law of the excluded middle, we must have that $p = \neg p_1$ is true.

For a real number $r\neq 1$, the extremely useful formula $$1 + r + \cdots + r^n = \frac{1 - r^{n+1}}{1-r} \tag{1.18}$$ holds. Let us prove this formula by induction. For $n=1$ this amounts to the identity $$1 + r = \frac{1-r^2}{1-r},$$ which is true since $1-r^2 = (1+r)(1-r)$. We let $p(n)$ denote the identity in (1.18) . We have seen that $p(1)$ is true. The induction step consists in proving $p(n)\implies p(n+1)$. We can prove this by adding $r^{n+1}$ to the right hand side in (1.18) : $$\frac{1 - r^{n+1}}{1-r} + r^{n+1} = \frac{1 - r^{n+1} + (1-r) r^{n+1}}{1-r} = \frac{1 - r^{n+2}}{1-r}. \tag{1.19}$$ Real life application

def f(n: int) -> int: return(n+1)

Good examples of functions can be found in the cryptographic hash functions. They are examples of complicated functions $f:S \rightarrow T$, where $S$ is infinite and $T$ finite. Here $S$ could be data like plain text files and $T$ could be a $256$ bit number. This is the setup for the widely used sha-256 cryptographic hash function. The whole point of a cryptographic hash function is that it must be humanly impossible to compute $y$ with $f(y) = f(x)$ given $f(x)$ ⁴ A pair $x\neq y$ with $f(x) = f(y)$ is called a collision . In fact, sha-256 is used in the Bitcoin block chain. The precise definition of sha-256 can be found in FIPS PUB 180-4 approved by the Secretary of Commerce.Other interesting functions output a bounded size digital footprint (checksum) of a file (like md5). This is very useful for checking data integrity of downloads over the internet. The md5 hash is a $128$ bit number.Instead of listing $256$ or $128$ bits for the hash value one uses hexadecimal notation with digits in 0, 1, 2, 3, 4, 5, 6, 7, 8, 9 , a, b, c, d, e, f. A pair of hexadecimal digits then represents a byte or $8$ bits. Output from sha-256 and md5 consist of $64$ and $32$ hexadecimal digits respectively. You are welcome to experiment with these two hash functions in the Python cell below.

Live Python

import hashlib print(hashlib.sha256(b"sha-256 is safe so far").hexdigest()) print(hashlib.md5(b"md5 has been broken").hexdigest())

▶︎ Run Explain

d131dd02c5e6eec4693d9a0698aff95c2fcab58712467eab4004583eb8fb7f89 
55ad340609f4b30283e488832571415a085125e8f7cdc99fd91dbdf280373c5b 
d8823e3156348f5bae6dacd436c919c6dd53e2b487da03fd02396306d248cda0 
e99f33420f577ee8ce54b67080a80d1ec69821bcb6a8839396f9652b6ff72a70

d131dd02c5e6eec4693d9a0698aff95c2fcab50712467eab4004583eb8fb7f89 
55ad340609f4b30283e4888325f1415a085125e8f7cdc99fd91dbd7280373c5b 
d8823e3156348f5bae6dacd436c919c6dd53e23487da03fd02396306d248cda0 
e99f33420f577ee8ce54b67080280d1ec69821bcb6a8839396f965ab6ff72a70 

The functions $f_1: \RR \rightarrow \RR$ and $f_2: \{x\in \RR \mid x\geq 0\}\rightarrow \RR$ given by $f_1(x) = x^2$ and $f_2(x) = x^2$ are not the same! Their domains are different.

Suppose that $B = \RR\times \RR = \RR^2$ and $(x, y) \in B$. Then $$\begin{aligned} \pi_1((1, 2)) &= 1\\ \pi_2((1, 2)) &= 2. \end{aligned}$$ Now suppose that $A = \RR^3$ and that $f: A\rightarrow \RR^3$ is given by $$f((x, y, z)) = (x^2 + y, x y , z^3).$$ Then $$\begin{aligned} f_1((1, 2, 3)) &= 3\\ f_2((1, 2, 3)) &= 2\\ f_3((1, 2, 3)) &= 27. \end{aligned}$$

Let $f: S\rightarrow S$, where $S = \{1, 2, 3\}$ be given by the table $$\def\arraystretch{1.5} \begin{array}{c|ccccccc} x & 1 & 2 & 3\\ \hline f(x) & 3 & 1 & 2 \end{array}.$$ Then $f^{-1}$is given by the table $$\def\arraystretch{1.5} \begin{array}{c|ccccccc} x & 1 & 2 & 3\\ \hline f^{-1}(x) & 2 & 3 & 1 \end{array}.$$

Consider the function $f: A\rightarrow B$, where $A = \{1, 2, 3, 4, 5\}$ and $B = \{a, b, c, d\}$ given by $$\def\arraystretch{1.5} \begin{array}{c|ccccccc} x & 1 & 2 & 3 & 4 & 5\\ \hline f(x) & a & b & c & a & d \end{array}$$ For $C = \{a, c\}$, $f^{-1}(C) = \{1, 3, 4\}$ as illustrated below.

Perceptrons model weighted decisions of the kind you make every day. Should you bike to the university tomorrow? Suppose you weigh the chance of rain $x_1$ (in percent) against the time the bike saves you, $x_2$ (in minutes), and decide: bike exactly when $$-x_1 + 4 x_2 - 20 > 0.$$ This is the perceptron $p:\RR^2\rightarrow \RR$ with weights $w_1 = -1$, $w_2 = 4$ and bias $b = -20$. The weights say how much each input matters and in which direction — rain counts against, saved time counts for, and one saved minute outweighs four percentage points of rain. The bias makes you demanding: with nothing speaking for the bike you stay on the bus, since $p(0, 0) = 0$. A $40%$ chance of rain and $20$ saved minutes? Then $-40 + 80 - 20 = 20 > 0$: the neuron fires and you bike. The art is clearly in choosing the weights and the bias. Choosing them automatically from data is called training — the subject of Section 7.9 .

Explain the python code below to me.

def p1(v):
    (x, y) = v
    if -x - y + 3/2 > 0:
        return 1
    else:
        return 0

def p2(v):
    (x, y) = v
    if x + y - 1/2 > 0:
        return 1
    else:
        return 0

def p3(v):
    (x, y) = v
    if x + y - 3/2 > 0:
        return 1
    else:
        return 0

def h(v):
    return (p1(v), p2(v))

def g(v):
    return p3(v)

def f(v):
    return g(h(v))

Explain why a single perceptron cannot compute the logical function XOR, but a neural network with one hidden layer can. I am a first year university student. I know about functions and composition of functions, but please do not use calculus.