Arithmetic Background Problems

--- tags: mth350, class-journal --- # Arithmetic Background Problems ## Basic Axioms and Definitions :::success **Axioms of Integer Arithmetic** 1. (*Closure of the integers under addition and multiplication*) For all integers $a$ and $b$, both $a+b$ and $ab$ are integers. 2. (*Commutative property of addition and multiplication*) For all integers $a$ and $b$, $a+b = b+a$ and $ab = ba$. 3. (*Associative property of addition and multiplication*) For all integers $a,b$ and $c$, $a + (b + c) = (a+b) + c$ and $(ab)c = a(bc)$. 4. (*Distributive property of multiplication over addition*) For all integers $a,b$ and $c$, $a(b+c) = ab + ac$. 5. (*Additive identity*) For every integer $a$, $0 + a = a$. 6. (*Multiplicative identity*) For every integer $a$, $1a = a$. 7. (*Additive inverses*) For every integer $a$, there exists an integer $b$ such that $a+b = 0$. We call $b$ the *additive inverse* of $a$ and denote it as $-a$. **Ordering Axioms** 1. (*Trichotomy*) For all integers $a$ and $b$, exactly one of the following holds: $a<b$, $b < a$, or $a = b$. 2. (*Transitivity*) For all integers $a,b$ and $c$, if $a<b$ and $b < c$ then $a < c$. 3. (*Translation invariance*) For all integers $a,b,$ and $c$, if $a < b$ then $a+c < b + c$. 4. (*Scaling*) For all integers $a,b,$ and $c$, if $a < b$ and $c > 0$, then $ac < bc$. ::: **Definition:** Let $x,y \in \mathbb{Z}$. Define *subtraction* as $x - y = x + (-y)$ where $-y$ is the additive inverse of $y$ from Arithmetic Axiom 7. :::danger Some things the axioms *do not* say and therefore do not constitute valid proof steps unless they are proven separately: * That $-x = (-1)x$. **The negative sign notation is not defined as multiplication by $-1$**; it only denotes the additive inverse of the integer. Also as noted in class, **"division" is not defined, and therefore fractions and negative exponents do not have meaning**. Don't use any of those until we get around to defining what we mean. We'll add to this list if an issue becomes common. ::: ## Problems :::info Most of these proofs are self-contained and do not need or use any of the other problems in the list. You should try your utmost not to use any of the other problem results in your proof! But if it cannot be avoided, i.e. you *must* use the result of another one of the problems in your solution, go ahead, but be clear about which problem you are using. ::: **Problem AB.1:** Let $a,b,c \in \mathbb{Z}$. Then $a(b-c) = ab - ac$. **Problem AB.2:** For all $a \in \mathbb{Z}$, $a \cdot 0 =0$. **Problem AB.3:** For all $a,b \in \mathbb{Z}$, $-(ac) = a(-c)$. **Problem AB.4:** For all $a,b \in \mathbb{Z}$, $-(ac) = (-a)c$. **Problem AB.5:** For all $a,b \in \mathbb{Z}$, $-(a + b) = -a -b$. **Problem AB.6:** For all $a \in \mathbb{Z}$, $-(-a) = a$. **Problem AB.7:** For all $a,b \in \mathbb{Z}$, $(-a)(-b) = ab$. **Problem AB.8:** For all $a \in \mathbb{Z}$, if $0 < a$, then $-a < 0$. **Problem AB.9:** For all $a,b \in \mathbb{Z}$, if $a< b$ and $c < 0$, then $ac > bc$. **Problem AB.10:** For all $a,b \in \mathbb{Z}$: (a) If $a, b > 0$ then $ab > 0$. (b) If $a > 0$ and $b < 0$ then $ab < 0$. (c) If $a,b < 0$ then $ab > 0$. **Problem AB.11:** For all $a,b \in \mathbb{Z}$, if $ab = 0$ then either $a = 0$ or $b = 0$. ## Hints/How-Tos **How to prove that something equals $0$**: Take an arbitrary integer, say $a$, and add it to the thing you want to show is equal to $0$. If the result is $a$, then the thing is $0$. **How to prove that something equals the negative of something else:** For example, suppose you're given $u$ and $v$ and you want to show $u = -v$. Look at $u+v$. If this equals $0$, then by definition (in Arithmetic Axiom 7), $u = -v$. **How to prove that one thing is less than another:** For example suppose you're given $u$ and $v$ and you want to prove $u < v$. There are a few strategies available: - *Use Trichotomy*: By Trichotomy, we know that either $u < v$, $v < u$, or $v = u$. Use this to create three cases and rule out/derive contradictions for the second two cases. - *Use Transitivity*: Find some third item $z$ and show that $u < z$ and $z < v$. Therefore $u < v$ by transitivity. - It's sometimes possible to use Scaling or Translation Invariance as well --- start with some other inequality and add stuff to both sides or multiply on both sides to get $u < v$.