Recently I've read the pure mathematics textbook about inequalities, and here I want to discuss the definition about it, it's a philosophic issue more than some kind of mathematical calculation.

How to define inequality?

This is two definition from the book:

1) If x>y, then x-y > 0.

2) If x>y, then x/y > 1.

For me, these definition is not good enough as the nature of inequality wasn't shown.

Back to the numberline with integers, the element forming the integer set is based on a order of magnitude of numbers. Without order, numbers aren't showing their magnitude lost their meaning. On a number line, numbers are arranged in order and from small to big. The problem comes: the integer set contains infinitely many integers, so how do we arrange them in order? Wihtout any axioms we don't know their magnitude, so a base point is set, zero.

Zero is the symmetrical point and the ordering is based on the difference of the point from the zero. Another problem come as one number X are on the left of zero and another number Y is on the right of zero, |OX|=|OY| (lengths are directly comparable before it is quantitzed, they don't need the ordering of numbers), but since they are different points so they are unequal, how to distinguish X and Y? Then there's the need to invent the term "difference".

Difference implies unequal, then the unequal sign works in two way. If x>y, then y is smaller than x, otherwise x > y > x cause the contradiction of self unequal to self. By symmetrical means we will need two signs to interpret these relation.

Positive and negative.

When one is "more positive than" another, the "anthoer" is "more negative" to the original numbers. (These aren't limited to +ve vs +ve numbers because we consider the whole line).

Back to our main concern, x > y.

x is more positive than y.

What's the difference? x-y, the "more positive part".

Therefore "x-y" is positive.

That's the basic definition of inequalities rather than saying it "x-y > 0".

The difference "x-y" let it to be a.

Then x-y = a, x = a + y. y need to be "a more positive" to reach x.

On the numberline, OY + OA = OY + YA = OX, then y + a = x.

Now we can explain x/y > 1 since

"x can be divided into more than 1 portion of y, so x = y + a."

x/y = (y+a)/y = 1 + y/a > 1.

So we observe inequality by difference, not by dividing, this is the fundamental concept of inequalities. Of course in proving inequalites we don't need to be such strict because the properties are widely available from the basic definition already.

For the unequal sign, we know there's a difference but we don't know which why the inequality it is.

for x unequal to y,

case 1: x>y, then x = a + y, where a is positive

case 2: x < y, then x + a = y, or we say x = y + b, where b is negative.

combining gives "x unequal to y" is equivalent to "x = y + a" where a is non-zero. (+ve or -ve).

They looks easy, but worth thinking, that's what I've learnt.

e.g. Show the triangular inequality in terms of difference.

Consider triangle ABC, we are going to show AB+BC > CA.

In geometrical visualization we copy segment AB and BC on CA, and if it overlaps the inequlity is valid.

Step 1: draw a circle, center A and radius AB, intersect AC on D.

Step 2: draw a circle, center C and radius BC, intersect AC on E.

There'll be two intersection point, B and another point B'. It's easy to show that triangle ABC is congruent to triangle AB'C.

Now we consider the following logic:

1) two circle not sharing common tangents gives two intersection points.

2) the area enclosed by two intersection point is overlapped.

Since B' must be out of triangle ABC (otherwise it mustn't congruent to each other), therefore AD and BE overlapped, i.e. |DE| > 0. Proof complete.

Exercise:

1) Show that a^2 is non-negaitve in terms of difference.

2) Show that AM-GM inequality in terms of difference.

Subscribe to:
Post Comments (Atom)

## No comments:

## Post a comment