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Posted 15 April 2008

# DENSITY OF THE REALS

The false picture of the reals as a “row of points” suggests that to the right of each real number is the “next” real number  That is wrong:  The points on the real line are dense, meaning that the real line has no gaps.   More precisely:

Between any two distinct points on the real line

is an infinite number of other points.

The proof is just like that for the density of the rationals:  If r and s are two different real numbers with , then  is between them:

(WA)

By repeating this process you can produce as many points between r and s as you want.   More generally, if a and b are any positive real numbers, then

which gives an infinite number of points between r and s directly   See detailed analysis of this proof.

## Density of the rationals and irrationals in the reals

A subset S of the real numbers is dense in the reals if for every real number r you can find numbers  that are as close as you want to r.

#### Theorem

The subsets of rational numbers and of irrational numbers are both dense in the reals.

#### Proof

¨  If r is any real number you can find a rational number as close as you want to it by truncating the decimal representation of r.  For example, , so  is a rational number that is within a quadrillionth of .

¨   is irrational, so  is also irrational for any integer n.  So if r is any rational number,  is irrational and you make it as close as you want to r by making n big enough.  (You could use any irrational number in place of .)  If r is an irrational number,  might not be irrational, but that does not matter because r itself is as close to r as you want!

## More about density of the reals:

How do you go about understanding this inequality?

(WA)  for

You can play with it and you can prove it.  You really need to do both to see what is happening here.

#### Play with it

For example, you can calculate some values of  for, say, :

 a b (ar+bs)/(a+b) 1 1 2.45 2 3 2.48 5 2 2.38571 2 22 2.575

If you stare at this formula and these calculations for awhile, you discover:

¨  You don’t have to worry about the denominator a + b being zero because both a and b are positive.  (You should always consider the possibility that a denominator might be 0.)

¨  When a = b =1, the inequality (WA) becomes

¨

which says: “The average of r and s is between r and s.”

¨  The calculations show that a and b are like weights.  When a is bigger the result is closer to r and when b is bigger the result is closer to s.

This sort of random pondering and calculating instances is the sort of thing all good mathematicans do to understand a situation.

#### Prove it

You can also prove (WA), using these facts about the “<” ordering on the reals:

¨  (T) If  x < y  and y < z, then x < z.

¨  (A) If x < y  and z is any real number, then z + x < z + y.

¨  (M) If x < y  and z is any positive real number, then zx < zy.

Proof of (WA):

Note the use of pattern recognition in understanding this proof.  For example, to prove the first “<”,

¨  We have br < bs by (M), with  x = r, y = s, z = b.

¨  Then ar + br < ar + bs  by (A) with  x = br, y = bs, z = ar.

¨  Then  by (M) with x = ar + br, y = ar + bs,

¨  Try proving the second “<” yourself.  (WA) then follows from (T).