Lecture 2

Divisibility

An integer $a$ divides an integer $b$ if and only if there exists an integer $s$ such that $as=b$. Written as $a\mid b$ (a divides b). In logical notation, this is written as $a,b\in \mathbb{Z},a\mid b$ if $\exists s\in \mathbb{Z}$ such that $as=b$. For example, $7\mid 42$ because $\exists 6\in \mathbb{Z}$. Conversely, $10\nmid 42$ because $\nexists s\in \mathbb{Z}$ such that $10s=42$.

Direct Proofs

Demonstrating a one statement’s truth implies another statement’s truth. For example, for $p\to q$ and $q\to r$ and $r\to s$ and $s\to t$, it logically follows that $p\to t$.

A Simple Direct Proof

An integer $a$ is even if and only if $2\mid a$. A non-even integer is said to be odd. We can prove directly that $2\mid n$ implies $2\mid n^2$, that is a square of an even number is also even.

Let $n^2=(2s{)}^2=2(2s^2)$. This gives us that $\exists (2s^2)\in \mathbb{Z}$ such that $2(2s^2)=n^2$ and hence $2\mid n^2$. The same reasoning can be provided that $(2\mid n)\to (2\mid n^3)$ or that $(3\mid n)\to (3\mid n^2)$.

Example of a Direct Proof not working

“Show that $n$ is an integer, such that $n^2$ is even, then $n$ is also even”, ie we want to prove $\forall n\in \mathbb{Z},(2\mid n^2)\to (2\mid n)$. A similar argument tells us that $(2\mid n^2)$ so $\exists s\in \mathbb{Z}$ such that $2s=n^2$. Although this shows that $n=\sqrt{2}\sqrt{s}$, this doesn’t lead to a statement in the form of $n=2t$ for some integer $t$.

Contrapositives

Where: if every time $p$ is true, $q$ is also true, then every time $q$ is false, $p$ must also be false. For example, the statement “I have no brothers and sisters” implies “I have no brothers”. The contrapositive of this is that “I have a brother” implies “I have a brother or sister”. Contrapositives do not change the truth of an original statement. If true, the contrapositive is true and if false, the contrapositive is false.

Proof by Contrapositive

Relies on negating both statements and reversing the direction of implication. When seeking to prove that if we have an integer $n$ such that $n^2$ is even, then $n$ must also be even. Here, we also then seek to prove that if $n$ is not even, $n^2$ is certainly not even. Let $n\in \mathbb{Z}$ be an odd number. This tells us that $\exists s\in \mathbb{Z}$ such that $(2s+1)=n$. We have $n^2=(2s+1{)}^2=4s^2+4s+1=2(2s^2+2s)+1$, so n^2 is therefore 1 larger than an even number so is odd. We have shown that $\neg (2\mid n)\to \neg (2\mid n^2)$ and hence that $(2\mid n^2)\to (2\mid n)$. If $n^2$ is even and $n$ is an integer, $n$ is even.

Proof by Contradiction

This proofs makes an assumption which is the negation of what we wish to prove, and then constructing a logical argument that leads to a statement which is clearly false. Formally, this consists of starting with a statement $p$ (assumed to be true) and showing that $p\to \neg p$. Since no statement can be true and not true, the initial assumption of $p$ being true must be incorrect.

Example Proofs

Prime Numbers

A prime number is an integer greater than 1 which is divisible only by 1 and itself. That is, $n\in \mathbb{Z}$ is prime if and only if $(s\nmid n)\forall s\in {\mathbb{Z}}^{+}1,n$. Rather than proving that $(n\text{ is prime})\wedge (n>2)\to (2\nmid n)$ , we can instead show that $(2\mid n)\to \neg ((n\text{ is prime})\wedge (n>2))$. This is trivial to prove, since if 2 divides $n$, $\exists s\in \mathbb{Z}$ such that $2s=n$. If this is the case, then $n$ is not prime since it has 2 and $s$ as divisors. When writing out all primes, we observe that they appear to be less and less frequent. The question is if there are a finite or infinite number of prime numbers. We can show that there are infinitely many by first assuming that there are not infinitely many, leading to a contradiction. If we assume that there are a finite number of primes, there is a list of distinct primes $p_1,p_2,...,p_n$ (any number not in this list is not prime). Consider the number $N=(p_1,p_2,...,p_n)+1$. It is easy to see that N is not divisible by any of the primes, but is not itself a prime. This is a contradiction, since N should’ve been in the original list, hence our initial assumption that there are a finite number of primes must be false.

Irrational Numbers

To prove that $\sqrt{2}$ is irrational, we can prove by contradiction. Assume that $\sqrt{2}$ is rational, and that its simplest form is $\sqrt{2}=\frac{a}{b}$ for the integers $a$ and $b$. This means that $a^2=2b^2$, which tells us that $a^2$ is even, which implies that $a$ is even. If $a$ is even, $\forall c\in \mathbb{Z}$ such that $a=2c$. This gives $a^2=(2c{)}^2=2b^2$, hence $2c^2=b^2$. $\sqrt{2}=\frac{b}{c}$, which contradicts the original assumption that $\sqrt{2}=\frac{a}{b}$ is the simplest form.

Rational Numbers

We can directly prove that between any two distinct rational numbers, there exists another rational number. Let our two distinct rational numbers be $a$ and $b$, such that $a<b$. These can be expressed in their simplest forms $a=\frac{p}{q},b=\frac{r}{s}$. The value $\frac{a+b}{2}=\frac{p}{2q}+\frac{r}{2s}=\frac{ps+qr}{2qs}$ is midway between $a$ and $b$ and is also rational (since $p,q,r,s$ are integers).