How can be proven that any number X is greater,lesser or equal to any other number Y?

Question

I have looked for it on the internet, really, but all I have found are particular cases like 1 > 0, or such. Is there an algebraic proof for proving that x > y or, x = y, or x < y? I thought of using Euclid's Fundamental theorem of Arithmetic as a tool, and comparing the primes that produce a compound number, but I still would have to prove that those primes (and primes in general) are ones greater (or lesser) than others.

Answer 1

You have to start somewhere. Say we have the integers and we know how to add, subtract, and multiply. This distinguishes the multiplicative identity element $1$. We say that $1$ is "positive," and recursively define positive numbers by declaring that if $p$ is positive, then so is $p+1$. This gives us the usual set of positive numbers.

Define $a<b$ if $b-a$ is positive. Then if $a\neq b$, then either $a-b$ is positive or $-(a-b)=b-a$ is positive, so either $a<b$ or $b<a$, which is what we wanted to show.

What about $52$ and $48$? Well, $52-48=4\ldots$

Answer 2

Different text may do these differently but they are all equivalent.

x < y;x=y; or y < x are mutually exclusive and exhaustive by definition.

Addition and multiplication (and the inverses subtraction and division) are given by fiat so that a+0=a 1xa =a, a+(-a)=0. If a ne 0 then ax1/a =1 and a (b+c)=ab+bc. These are axioms.

From these we can prove basic this like (-(-a))=a and (-a)(-b)=ab and 0xa=0 and so on.

We are then given two essential axioms of order.

x+y < x+z if y < z

And xy > 0 if x >0 and y > 0.

From there we can prove everything.

We can prove if x > 0 then - x < 0

If x >0 and y < z then xy < xz.

That $x^2 > 0$ is $x \ne 0$ and in particular $1^2 = 1 > 0$.

That if 0 < x < y then 0 < 1/y < 1/x

So that's how we can compare any two rational numbers.

But what of irrational. Well when we define the reals it is as the limits of sequences of rationals and the laws of order are extended.

Answer 3

It seems that you're asking how to prove that $\le$ is a total ordering. The answer depends on the definition of $\le$. If I define $x\prec y$ iff $x$ perfectly divides $y$ then $2 \prec 4$ but $3\not\prec 4$ and $4\not\prec 3$ and $3\not= 4$.

Suppose I define '$x\le y$' based upon the sign of $(y-x)$.

$(y-x)$ zero or positive : $x\le y$

$(y-x)$ zero or negative : $y\le x$

Since the difference of two numbers must be at least one of these cases, the ordering has totality. To be a total order I'd also have to show

(antisymmetry) if $a\leq b$ and $b\leq a$ then $a=b$ and

(transitivity) if $a\leq b$ and $b\leq c$ then $a\leq c$