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Expert: - 1/27/2009


Question
1. Prove that if x in set Q (rational numbers), y in set Q (rational numbers) and x < y, then there is some z in set Q (rational numbers) such that x < z < y.

2. Prove that if x in set Q (rational numbers), y in set Q (rational numbers) and x < y, then there is some z not in set Q (rational numbers) such that x < z < y.


Answer
1. If x is in the rational numbers,
then x = a/b where a, b are both integers.

If y is in the rational numbers,
then y = c/d where c, d are both integers.

Making them have the same denominator, we get
x = ad/bd and y = bc/bd.

Now we know that x is less than y, so ad < bc.
Now bc - ad might only be 1 (by some chance), so lets multiply them both by 2.

This gives x = 2ad/2bd and y = 2bc/2bd.

That would make (2ad+2bc)/4bd a rational as well, since it is
ad+bc and we know that a, d, b, and c are all integers.

NOw we know that adx < y, which means 2ad/2bd < 2bc/2bd,
which means 2ad < 2bc.

Multiplying the top and bottom of x and y by 2 again, we get
x = 4ad/4bd and y = 4bc/4bd.

Now since we know that 2ad < 2bc, we can change 2ad in x to 2bc,
and that will be greater than x.  The result is (2ad+2bc)/4bd,
which was already mentioned and shown to be rational.

In a similar manner, we could also show that (2ad+2bc)/4bd is less than y.

So what we have is a rational number that is greater than x
and is less than y.  I think that I will call it z, and that
ends the problem.

2. Take root to be short for squareroot.
We know that root(2) is irrational.

Take the number w = (x-y)/root(2).

When compared to x-y, we no that (x-y)/root(2) is less than x-y
since it is known that root(2)>1.

If we add w to x we get z=x+w.  Putting in the value of w gives us
z = x + (x-y)/root(2).

Putting them over the same rational denominator gives
z = (2x + (x-y)root(2))/2.

The denominator is rational, but the numerator is not
since there is an irrational term in it.

It was also shown that z was between x and y.

Not only that, but there is an infinite number of irrationals
between x and y since we can divide by root(3), root(5),
root(6), root(7), root(8), root(10), etc.  Note that the true
squares are skipped so that an irrational is gotten.
We know that {set of all integers} - {set of all squares} still
has infinite size, therefore there is an infinite amount of numbers
that would work.

That was a little more than you needed, but there are even more
irrational numbers than we need as well...

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I can answer any question in general math, arithetic, discret math, algebra, box problems, geometry, filling a tank with water, trigonometry, pre-calculus, linear algebra, complex mathematics, probability, statistics, and most of anything else that relates to math. I can even tell you it takes me over 2,000 steps to go a mile, but is that relevant?

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