Logo

A problem arises when we speak of subtraction.

We define for a given a,b∈Na, b \in \mathbb{N}

We define an expression −(a—b)∈Z- (a — b) \in \mathbb{Z}

(Integers)

(a—b)is an integer,a,b∈N(a — b) \text{is an integer}, a , b \in \mathbb{N} if (a—b)=(c—d)(a — b) = (c — d), then a + d = b + c

If (a—b)=(a′—b′)⇒a+b′=a′+b(a — b) = (a' — b')\Rightarrow a + b' = a' + b

(a—b)+(c—d)=(a′—b′)—(b+d)(a—b)×(c—d)=(a′—b′)×(c+d)\begin{aligned} (a — b) + (c — d) &= (a' — b')— (b + d) \\ (a — b) \times (c — d) &= (a\prime — b\prime) \times (c + d) \\ \end{aligned}
Proof. LHS:(a—b)+(c—d)=(a+c)—(b+d)RHS:(a′—b′)+(c—d)=(a′+c)—(b′+d) Adding(c+d)on both sides\begin{aligned} LHS: (a — b) + (c — d) &= (a + c) — (b + d) \\ RHS: (a^\prime — b^\prime) + (c — d) &= (a^\prime + c) — (b^\prime + d) \\ \text{ Adding} &(c+d) \text{on both sides} \end{aligned}

We can represent natural numbers as (n — 0) ∈Z\in \mathbb{Z}, you will see that every operation holds.

(Negation)

If (a — b) is an integer, we define the negation -(a — b), to be the integer (b — a). In particular, if n = n— 0 is a positive natural number, we can define its negation -n = 0 — n

When x∈Zx \in \mathbb{Z},

  1. x=0x = 0

  2. x=nx = n, n is +ve natural number

  3. x=−nx = -n, n is +ve natural number

Rationals

(Rationals)

a//b,a∈Z,b∈Z,b≠0a // b, a \in \mathbb{Z}, b \in \mathbb{Z}, b \ne 0, we define a//ba //b as such, and when he have two such rational numbers that,

a//b=c//d⇔ad=bca // b = c // d \Leftrightarrow ad = bc

We define the operations for addition, multiplication and negation as such,

  1. a//b+c//d:=(ad+cb)//(bd)a // b + c // d := (ad + cb)//(bd)

  2. (a//b)×c//d:=ac//bd(a // b) \times c // d := ac // bd

  3. −(a//b):=(−a)//b-(a // b) := (-a)// b

The output of the above operations should be equal when a//ba // b is replaced by an equivalent rational number a′//b′a' // b'

If x < y, and z < 0, then xz>yzxz > yz
Proof. Say, x<y ⟹ y−x>0x< y \implies y - x > 0
xz>yz, where z=−p,p>0xz−yz>0 ⟹ (x+(−y))z>0(x+(−y))(−p)=(−1)(x+(−y))p=(y−x)p\begin{aligned} xz &> yz, \text{ where } z = -p, p > 0 \\ xz - yz &> 0 \implies (x + (-y))z > 0 (x + (-y))(-p) & = (-1)(x + (-y))p & = (y - x)p \end{aligned}
(Reciprocal)

If x=a//bx = a // b then x−1:=b//ax^{-1} := b // a but 0−10^{-1} is undefined.

(Division)

We define the operation x / y, where x,y∈Qx,y \in \mathbb{Q} such that,

x/y:=x×y−1=(x//1)×(y//1)−1=(x//1)×(1//y)=x//y\begin{aligned} x/y &:= x \times y^{-1} \\ &= (x // 1) \times (y // 1)^{-1} \\ &= (x//1) \times (1 // y) \\ &= x // y \end{aligned}

(Order)

If x>0x > 0, if x=a/bx = a/b where a,b∈Z,a,b>0a,b \in \mathbb{Z}, a,b > 0 or a,b<0a,b <0
and x<0x < 0 if x=−y,y>0,y∈Qx = -y, y > 0, y \in \mathbb{Q}

This makes rational numbers the first ordered field, real numbers fall under this category as well.

(Trichotomy Of Order)

For x∈Qx \in \mathbb{Q},

x=0x∈Q+x∈Q−\begin{aligned} x &= 0 \\ x &\in \mathbb{Q}^+ \\ x &\in \mathbb{Q}^- \\ \end{aligned}

(Equality For Rationals)

Show that the definition of equality for the rational numbers is reflexive, symmetric and transitive.

Proof. For reflexivity,

We define x=a//bx = a // b

Then if x=xx = x,

a//b=a//b,ab=aba//b = a//b, ab = ab

This reduces it down to integer multiplication, which we know is reflexive.

For symmetry,

For two integers,

a//b=c//d⇔ad=cb⇔cb=ad⇔c//d=a//ba//b = c//d \Leftrightarrow ad = cb \Leftrightarrow cb = ad \Leftrightarrow c //d = a // b

Thus symmetry is proved

For transitivity,

When,

a//b=c//d;c//d=e//f⇔a//b=e//fa // b = c // d; c//d = e//f \Leftrightarrow a // b = e // f ⇔ad=cb⇔cf=ed\Leftrightarrow ad = cb \Leftrightarrow cf = ed

adf=cbf=cfbafd=edb=ebdaf=eb By cancellation law, which is defined for integersa//b=e//f\begin{aligned} adf = cbf &= cfb \\ afd &= edb \\ &= ebd \\ af &= eb \text{ By cancellation law, which is defined for integers} \\ a // b &= e// f \end{aligned}

CC BY-SA 4.0 Adithya Nair.
Last modified: March 04, 2025.
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