Irrational Numbers

Irrational Number

An irrational number is an element

x \in ℝ ∖ ℚ

irrational to the power of irrational can be rational

Prove that there exists two irrational numbers

a, b

such that

a^{b}

is rational

Consider

a = \sqrt{2}

and

b = \sqrt{2}

, then there are two cases, either

{\sqrt{2}}^{\sqrt{2}}

is rational, and the proof is over, or

{\sqrt{2}}^{\sqrt{2}}

is irrational and in that case consider

a = {\sqrt{2}}^{\sqrt{2}}

and

b = \sqrt{2}

, then

a^{b} = {({\sqrt{2}}^{\sqrt{2}})}^{\sqrt{2}} = {\sqrt{2}}^{\sqrt{2} \cdot \sqrt{2}} = {\sqrt{2}}^{2} = 2

, so in any case we've found an

a, b

that are irrational such that

a^{b}

is rational.

Square Root of a Non Perfect Square is Irrational

Suppose that

n \in ℕ_{1}

is not a perfect square, then

\sqrt{n}

is irrational.

Suppose for the sake of contradiction that

\sqrt{n}

was rational so that

\sqrt{n} = \frac{p}{q}

therefore

q^{2} n = p^{2}

, since

n

is not a perfect square then we know that

n = \prod_{i \in ℕ_{1}} p_{i}^{α_{i}}

where there is some

j \in ℕ_{1}

such that

α_{j}

is odd, now we also know that

q^{2}

is equal to

\prod_{i \in ℕ_{1}} p_{i}^{2 β_{i}}

since odd plus even is still odd then we know that in the prime factorization of

q^{2} n

the power on the

j

-th prime is still odd, thus

q^{2} n

is not a perfect square, but on the other hand

p^{2}

is, which is a contradiction, therefore

\sqrt{n}

must be irrational.

There is a Rational Between any Two Reals

For any

x, y \in ℝ

such that

x < y

there exists a

p \in ℚ

such that

x < p < y

Since $x < y$ we have $0 < y - x$ therefore by the archimediean property we have an $n \in ℕ_{1}$ such that $n (y - x) > 1$

Consider the set $X = {z \in ℤ : z > n x}$ , this set is not empty because $n x + 1 \in X$ , additionally it is bounded below, and so as a subset of $ℤ$ it has a least element say $m$ such that $m > n x$ . Since it is the least element we know that $m - 1 \leq n x$ , which implies that $m \leq n x + 1$ .

Recall that our $n$ satisfies $n (y - x) > 1$ therefore $m \leq n x + 1 \leq n x + (n (y - x)) = n y$ , thus we have $n x < m < n y$ and since $n \neq 0$ we have $x < \frac{m}{n} < y$ as needed.

There is an Irrational Number between any two Real Numbers

Suppose that

x, y \in ℝ

such that

x < y

then there is a

p \in ℝ ∖ ℚ

such that

x < p < y

x < y

then

\sqrt{2} + x < \sqrt{2} + y

, and thus there is a rational number

r \in ℚ

such that

\sqrt{2} + x < r < \sqrt{2} + y

and therefore

x < r - \sqrt{2} < y

, but

r - \sqrt{2}

is irrational, because the sum of a rational and irrational is irrational, then we are done.