Thread: Irrational Numbers

1.) Given that I is the set of irration numbers,

a.) Show that if a, b are in the rational numbers, then ab and a + b are elemenets of the rationals as well.

b.) Show that if a are in the rationals, and t is in the irrationals, then a + t is in the irrationals and at is in the irrationals as long as a does not equal 0.

c.) Part a says that the rationals is closed under addition and multilication. Are the irrationals closed under addition/multiplication? Given 2 irrationals (s and t), what are we able to say about s + t and st?

2.) Using the above, and using that the rationals are dense in the reals a - sqrt(2) and b - sqrt(2), prove that given any 2 real numbers where a < b, there exists an irrational number t that satisfies a < t < b.

I will help with #2.
Given .
There is an rational number such that, .
But and is irrational.

Originally Posted by Plato

I will help with #2.
Given .
There is an rational number such that, .
But and is irrational.

Thanks Plato. And I figured out #1 too.

Originally Posted by seerTneerGevoLI

1.) Given that I is the set of irration numbers,

a.) Show that if a, b are in the rational numbers, then ab and a + b are elemenets of the rationals as well.

a=p/q and b=n/m so their sum is (pm+nq)/qm which is a rational number.
Similarly ab.

b.) Show that if a are in the rationals, and t is in the irrationals, then a + t is in the irrationals and at is in the irrationals as long as a does not equal 0.

If a+t is rational then (a+t)-a is rational so t is rational a contradiction.
If at is rational and a!=0 then at/a=t is rational a contradiction.

c.) Part a says that the rationals is closed under addition and multilication. Are the irrationals closed under addition/multiplication? Given 2 irrationals (s and t), what are we able to say about s + t and st?

No the irrationals are not. Try sqrt(2) and sqrt(8).

2.) Using the above, and using that the rationals are dense in the reals a - sqrt(2) and b - sqrt(2), prove that given any 2 real numbers where a < b, there exists an irrational number t that satisfies a < t < b.

This should be easy now.