Real Analysis: The Real Numbers

Problems

1. Natural Number Exponents - Algebra: Let $x \in \mathbb{R}$ and $n \in \mathbb{N}$. Show the following:

   1. $x^n \cdot x^m = x^{n + m}$

   2. $\left(x^n\right)^m = x^{nm}$

   1. Proof by induction. Base case: $x^n \cdot x^0 = x^n \cdot 1 = x^n = x^{n + 0}$. Inductive Step: Assume $x^nx^m=x^{n+m}$. Then $x^nx^{m+1} = x^nx^mx = x^{n+m}x = x^{n+m+1}$. The result follows by induction.

   2. Proof by induction. Base case: $(x^n)^0 = 1 = x^{0} = x^{n \cdot 0}$. Inductive Step: Assume $(x^n)^m = x^{nm}$. Then $(x^n)^{m+1} = (x^n)^mx^n = x^{nm}x^n = x^{nm+n} = x^{n(m+1)}$. The result follows by induction.

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2. Show that $x^2 \geq 0$ for all $x \in \mathbb{R}$.

   If $x > 0$, then $x^2 = x \cdot x > 0$ by the second ordering axiom. If $x < 0$, then $-x > 0$, and $x^2 = (-1)^2x^2 = ((-1)x)^2 = (-x)^2 > 0$. Finally, if $x = 0$, then $x^2 = 0$.

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3. Natural Number Exponents - Ordering I: Show that for all non-negative $x \in \mathbb{R}$ and all $n, m \in \mathbb{N}^+$ where $n < m$ that

   1. $x^n > 0.$

   2. $0 < x^n < 1$ when $x \in (0, 1)$ and $n > 1.$  

   3. $0 < x^m < x^n < 1$ when $x \in (0, 1).$

   4. $x < x^n$ when $x > 1$ and $n > 1.$

   5. $x^n < x^m$ when $x > 1.$

   All five proofs will be by induction.

   1. Base case: $x > 0$ is given, and $x^0 = 1 > 0$. Inductive step: Assume $x^n > 0$. By definition, $x^{n+1} = x^nx$. Since $x^n > 0$ and $x > 0$, it follows from the ordering axioms on $\mathbb{R}$ that $x^{n}x > 0,$ so therefore $x^{n+1} > 0.$

   2. Assume $x \in (0, 1).$ Base case: Since $0 < x < 1,$ it follows that $0 < x \cdot x < x.$ By definition of natural number exponents, $x^2 = x \cdot x.$ Therefore $0 < x^2 < x < 1.$ Inductive step: Assume $0 < x^n < x < 1.$ Multiplying by $x$ gives us $0 < x^n \cdot x < x \cdot x < x < 1.$ By definition of natural number exponents, this simplifies to $0 < x^{n+1} < x < 1.$ The result follows by induction.

   3. Assume $x \in (0, 1).$ Base case: Since $0 < x < 1,$ it follows from the above that $0 < x^n < x$ for all $n > 1.$ Since $0 < x < 1,$ it also follows that $x^{n+1} = x^n \cdot x < x^n.$ Inductive step: Assume $x^m < x^n$ when $m > n.$ Since $0 < x < 1$, it follows that $x^{m+1} = x^m \cdot x < x^m.$ It follows by transitivtiy that $x^{m+1} < x^n.$ The result follows by induction.

   4. Assume $x > 1$. Base case: Since $x > 1$, it follows that $x \cdot x > x$. By definition of natural number exponents, $x^2 = x \cdot x$. Therefore $x^2 > x.$ Inductive step: Assume $n > 1$ and $x^n > x.$ Since $x > 1$, it follows that $x^{n+1} = x^n \cdot x > x^n.$ It follows by transitivity that $x^{n+1} > x.$ The result follows by induction.

   5. Assume $x > 1.$ Base case: Since $x > 1$, it follows from the above that $x^n > x$ for all $n > 1.$ Since $1 < x,$ it also follows that $x^{n+1} = x^n \cdot x > x^n.$ Inductive step: Assume $x^m > x^n$ where $m > n.$ Since $x > 1$, it follows that $x^{m+1} = x^m \cdot x > x^m.$ It follows by transitivity that $x^{m+1} > x^n.$ The result follows by induction.

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4. Natural Number Exponents - Ordering II: Consider $a, b \in \mathbb{R}$ where $0 < a < b$ and $n \in \mathbb{N}^+$. Show that $a^n < b^n.$

   Proof by induction.

   Base case: It is given that $a^1 = a < b = b^1$.

   Inductive step: Assume $a^n < b^n$. Since $a > 0$, we can multiply both sides by $a$ to get $a^{n+1} < ab^n$. Next, because $a < b$, it follows that $1 < \dfrac{b}{a}$. Therefore $ab^n < \dfrac{b}{a}(ab^n) = b^{n+1}$. Thus by transitivity we see that $a^{n+1} < b^{n+1}.$ The result follows by induction.

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5. Let $x \in \mathbb{R}^+$ and $n \in \mathbb{N}^+.$ Show that $x^n = 1$ if and only if $x = 1$.

   Assume $x = 1$. Proof by induction. Base case: $x^1 = x = 1.$ Inductive step: Assume $x^{n} = 1.$ Then $x^{n+1} = x^nx = 1 \cdot 1 = 1.$ The result follows by induction.

   Conversely, assume $x^n = 1$. By trichotomy, either $x < 1$, $x = 1$, or $x > 1.$ If $x < 1,$ then $x^n < 1,$ and if $x > 1,$ then $x^n > 1.$ Therefore $x = 1.$

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6. Integer Exponents - Algebra: Let $x \in \mathbb{R} - \{0\}$ and $n, m \in \mathbb{Z}.$ Show the following:

   1. Equivalent definition: $x^{-n} = (x^{-1})^n.$

   2. $x^nx^m = x^{n+m}$

   3. $\left(x^n\right)^m = x^{nm}$

   1. By commutativity, it follows that $x^n \cdot (x^{-1})^n = (x \cdot x^{-1})^n.$ The righthand side then simplifies to $1^n = 1.$ It then follows by the definition of multiplicative inverse that $(x^{-1})^n = (x^n)^{-1},$ and thus $(x^{-1})^n = x^{-n}.$

   2. The result for when $n$ and $m$ are both non-negative is already proved. If $n$ is nonnegative and $m$ is negative, then

      $ x^nx^m = x^n(x^{-m})^{-1} \\ x^nx^m = x^n(x^{-m + n - n})^{-1} \\ x^nx^m = x^n(x^nx^{-m - n})^{-1} \\ x^nx^m = x^n(x^{n})^{-1}(x^{-m - n})^{-1} \\ x^nx^m = (x^{-m - n})^{-1} \\ x^nx^m = x^{m + n} $

      The case when $n$ is negative and $m$ is nonnegative is proved identically. If both $n$ and $m$ are negative, then $x^nx^m = (x^{-n})^{-1}(x^{-m})^{-1} = (x^{-n}x^{-m})^{-1} = (x^{-n-m})^{-1} = x^{n + m}$.

   3. The result for when $n$ and $m$ are both nonnegative is already proved. If $n$ is positive and $m$ is negative, then $(x^n)^m = \dfrac{1}{(x^n)^{-m}} = \dfrac{1}{x^{-nm}} = x^{nm}$. The case for when $m$ is positive and $n$ is negative is proved identically. If both $n$ and $m$ are negative, then $(x^n)^m = \dfrac{1}{\frac{1}{(x^{-n})^{-m}}} = \dfrac{1}{\frac{1}{x^{nm}}} = x^{nm}$.

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7. Integer Exponents - Ordering: Let $x \in \mathbb{R}^+,$ and let $n, m \in \mathbb{N}^+$ where $n < m.$ Show the following:

   1. $x^{-n} > 0.$

   2. $1 < x^{-n}$ when $x \in (0, 1).$

   3. $1 < x^{-n} < x^{-m}$ when $x \in (0, 1).$

   4. $0 < x^{-n} < x$ when $x > 1.$

   5. $0 < x^{-m} < x^{-n} < 1$ when $x > 1.$

   Hint: Do not use induction. Make use of the ordering properties for multiplicative inverses.

   All proofs will make use of the fact that $(x^n)^{-1} = x^{-n}$.

   1. Let $x > 0.$ Then $x^n > 0$ for all $n \in \mathbb{N}^+.$ By trichotomy, either $x^{-n} < 0,$ $x^{-n} = 0,$ or $x^n > 0.$ It cannot be that $x^{-n} = 0$, as otherwise $x^n \cdot 0 = 1.$ It also cannot be the case that $x^{-n} < 0$, as otherwise $1 = x^n \cdot x^{-n} < 0.$ Therefore $x^{-n} > 0.$

   2. Assume $x \in (0, 1).$ Then $0 < x^n < 1$ for all $n \in \mathbb{N}^+.$ Therefore $x^{-n} > 1.$

   3. Assume $x \in (0, 1).$ Then $0 < x^m < x^n < 1$ for $m > n.$ Therefore $1 < x^{-n} < x^{-m}.$

   4. Assume $x > 1.$ Then $x < x^n$ and $1 < x^n$ by transitivity. Therefore $0 < x^{-n} < 1$, and by transitivity again $0 < x^{-n} < x.$

   5. Assume $x > 1.$ Then $1 < x^n < x^m.$ Therefore $0 < x^{-m} < x^{-n} < 1.$

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8. Rational Exponents I: For a positive real number $x$ and positive natural number $n$, define $x^{\frac{1}{n}}$ to be the positive real number $y$ such that $y^n = x$. The expression $x^{\frac{1}{n}}$ can also be written as $\sqrt[n]{x}.$

   Prove that there exists exactly one real number $y$ such that $y = x^{\frac{1}{n}}.$

   Hint: Use the identity $a^n - b^n = (b - a)(b^{n-1} + b^{n-2}a + \ldots + a^{n-1})$ and consider what happens when $a < b$.

   First we show that one $y$ exists, then show that there is only one.

   Let $T = \{ t > 0 : t^n < x \}$. Consider the number $t = \dfrac{x}{1+x}$. Note that $0 \leq t^n < 1$. It follows that $t^n \leq t < x$. Therefore $T$ is not empty. Consider the number $b = x + 1$. Then $x < b \leq b^n$, and so $b$ is an upper bound of $T$. Thus it follows that $T$ has a least upper bound, call it $y$.

   Next we show that $y^n = x$ by eliminating the possibilities that $y^n < x$ and $y^n > x$.

   Assume $y^n < x$. We would like to show that there exists some $t^n$ between $y^n$ and $x$, as this will lead to a contradiction, specifically with the fact that $y = \text{sup}(T)$.

   If $y^n < x$, then $x - y^n > 0$. Note the identity $b^n - a^n = (b-a)(b^{n-1} + b^{n-2}a + \ldots + a^{n-1})$. If $0 < a < b$, we get the inequality $0 < b^n - a^n < (b-a)(b^{n-1} + b^{n-2}b + \ldots + b^{n-1})) = (b-a)(nb^{-1})$. Dividing through by $(nb^{-1})$ on both sides gives $0 < \dfrac{b^n - a^n}{(nb^{-1})} < b-a$. Now pick a $t$ such that $0 < t < 1$ and $t < \dfrac{b^n - a^n}{(nb^{-1})}$. Set $a = y$ and $b = y + h$. Then $(y + h)^n - y^n < ((y+h)-y)n(y + h)^{n-1} = hn(y+h)^{n-1} < hn(y+1)^{n-1} < x - y^n$. Therefore $(y+h)^n < x$, and $y + h \in A$. But this is a contradiction, since $y < y + h$ and $y = \text{sup}(T)$. Therefore $y \nless \sqrt[n]{x}$.

   Conversely, assume $y^n > x$. Let $k = \dfrac{y^n - x}{ny^{n-1}}$. Then $k = \dfrac{y}{n} - \dfrac{x}{ny^{n-1}} < y$. If $t \geq y - k$, then $y^n - t^n \leq y^n - (y-k)^n < kny^{n-1} = y^n - x$. Therefore $t^n > x$, so $t \notin T$. Therefore $y-k$ is an upper bound of $T$. But $y-k < y$, and $y = \text{sup}(T)$, which is a contradiction. Therefore $y \ngtr \sqrt[n]{x}$. Therefore $y = \sqrt[n]{x}$.

   To prove uniqueness, consider some other $y'$ such that $y'^n = x$ and $y' \neq y$. If $y < y'$, then $x = y^n < y'^n$, so $y' \neq \sqrt[n]{x}$ after all, and the same logic holds if $y' < y$.

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9. Rational Exponents II: Show that $(xy)^{\frac{1}{n}} = x^{\frac{1}{n}}y^{\frac{1}{n}}$ for positive $n \in \mathbb{N}$ and positive $x, y \in \mathbb{R}$.

   Let $a = x^{\frac{1}{n}}$ and $b = y^{\frac{1}{n}}$. Then $xy = a^nb^n = (ab)^n$ by commutativity. Since there is only one $n$th root of any positive real number, it follows that $(xy)^{\frac{1}{n}} = ab = x^{\frac{1}{n}}y^{\frac{1}{n}}$.

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10. Rational Exponents III: Show that $(x^{\frac{1}{n}})^n = (x^n)^{\frac{1}{n}}$ for positive $n \in \mathbb{N}$ and positive $x \in \mathbb{R}$.

    By definition, $(x^{\frac{1}{n}})^n = x$. Likewise, $(x^n)^{\frac{1}{n}} = y$, where $y^n = x^n$. But this implies $y = x$. Therefore $(x^{\frac{1}{n}})^n = (x^n)^{\frac{1}{n}}$.

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11. Rational Exponents IV: Define for $a, b \in \mathbb{N}$ where $b > 0$ and for positive $x \in \mathbb{R}$ that $x^{\frac{a}{b}} = \left(x^{\frac{1}{b}}\right)^a$. Show that for $c, d, \in \mathbb{N}$ where $\dfrac{a}{b} = \dfrac{c}{d}$ that $x^{\frac{a}{b}} = x^{\frac{c}{d}}$.

    First we show that $x^{\frac{ac}{bc}} = x^{\frac{a}{b}}$:

    By definition, $x^{\frac{ac}{bc}} = \left(x^{ac}\right)^{\frac{1}{bc}}$. Let $y = \left(x^{ac}\right)^{\frac{1}{bc}}$. Then $y^{bc} = x^{ac}$. Therefore $(y^b)^c = (x^a)^c$, so $y^b = x^a$, and thus $y = (x^a)^{\frac{1}{b}} = x^{\frac{a}{b}}$. Therefore $x^{\frac{ac}{bc}} = x^{\frac{a}{b}}$.

    Now we show $x^{\frac{a}{b}} = x^{\frac{c}{d}}$ when $\dfrac{a}{b} = \dfrac{c}{d}$. Note that $ad = bc$. Therefore $ x^{\frac{a}{b}} = x^{\frac{ad}{bd}} = x^{\frac{bc}{bd}} = x^{\frac{c}{d}}$.

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12. Rational Exponents V: Show the following:

    1. $\left(x^{\frac{1}{n}}\right)^{\frac{1}{m}} = x^{\frac{1}{nm}}.$

    2. $\left(x^{\frac{a}{b}}\right)^{\frac{c}{d}} = x^{\frac{ac}{bd}}.$

    3. $x^{\frac{a}{b}}x^{\frac{c}{d}} = x^{\frac{a}{b} + \frac{c}{d}}.$

    1. To show that $\left(x^{\frac{1}{n}}\right)^{\frac{1}{m}} = x^{\frac{1}{nm}},$ first let $a = \left(x^{\frac{1}{n}}\right)^{\frac{1}{m}}$ such that $a^m = x^{\frac{1}{n}}.$ Next, let $b = x^\frac{1}{n}$ such that $b^n = x.$ It follows that $\left(a^m\right)^n = x = a^{nm}.$ Therefore $x^{\frac{1}{nm}} = a = \left(x^{\frac{1}{n}}\right)^{\frac{1}{m}}.$

    2. $ \left(x^{\frac{a}{b}}\right)^{\frac{c}{d}} = \left(\left(x^{\frac{1}{b}}\right)^a\right)^{\frac{c}{d}} \\ \left(x^{\frac{a}{b}}\right)^{\frac{c}{d}} = \left(\left(\left(x^{\frac{1}{b}}\right)^a\right)^c\right)^{\frac{1}{d}} \\ \left(x^{\frac{a}{b}}\right)^{\frac{c}{d}} = \left(\left(x^{\frac{1}{b}}\right)^{ac}\right)^{\frac{1}{d}} \\ \left(x^{\frac{a}{b}}\right)^{\frac{c}{d}} = \left(\left(x^{ac}\right)^{\frac{1}{b}}\right)^{\frac{1}{d}} \\ \left(x^{\frac{a}{b}}\right)^{\frac{c}{d}} = \left(x^{ac}\right)^{\frac{1}{bd}} \\ \left(x^{\frac{a}{b}}\right)^{\frac{c}{d}} = x^{\frac{ac}{bd}} \\ $

    3. $ x^{\frac{a}{b}}x^{\frac{c}{d}} = x^{\frac{ad}{bd}}x^{\frac{bc}{bd}} \\ x^{\frac{a}{b}}x^{\frac{c}{d}} = \left(x^{\frac{1}{bd}}\right)^{ad}\left(x^{\frac{1}{bd}}\right)^{bc} \\ x^{\frac{a}{b}}x^{\frac{c}{d}} = \left(x^{\frac{1}{bd}}\right)^{ad+bc} \\ x^{\frac{a}{b}}x^{\frac{c}{d}} = x^{\frac{ad+bc}{bd}} \\ x^{\frac{a}{b}}x^{\frac{c}{d}} = x^{\frac{a}{b} + \frac{c}{d}}$

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13. For $x, y, \in \mathbb{R}$, does $x^n = y^n$ imply that $x = y$ if $x$ and $y$ are not necessarily non-negative? Prove or give a counterexample.

    No, the implication does not hold. Counterexample: $(-1)^2 = 1$, and $1^2 = 1$, but $-1 \neq 1$.

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14. Theorem: Show that $\sqrt{2}$ is irrational.

    Proof by contradiction. Assume that there is some rational number $\dfrac{a}{b} \in \mathbb{Q}$ such that $\left(\dfrac{a}{b}\right)^2 = 2$. Without loss of generality, assume that $\dfrac{a}{b}$ is in simplest form, i.e. that $a$ is not divisible by $b$. Then $\dfrac{a^2}{b^2} = 2$, so $a^2 = 2b^2$.  This means that $a^2$ is divisible by 2. Since $a^2$ is a square, $a$ must be divisible by $2$. Let $ac = a$. Then $a^2 = 4c^2 = 2b^2$. Dividing both sides through by $2$ gives us $2c = b^2$. The same logic shows that $b$ must also be divisible by $2$. However, this is a contradiction, as it means that $a$ and $b$ have a common factor. Therefore $\sqrt{2} \notin \mathbb{Q}$.

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15. Show that $\sqrt{2} < 2$.

    By trichotomy, either $\sqrt{2} < 2$, $\sqrt{2} = 2$, or $\sqrt{2} > 2$. If $\sqrt{2} = 2$, then $\sqrt{2}^2 = 2^2$, but clearly $2 \neq 4$. If instead $\sqrt{2} > 2$. Then $\sqrt{2}^2 > 2\sqrt{2}$, so $2 > 2\sqrt{2}$. But this implies $0 < \sqrt{2} < 1$, which is a contradiction, as we assumed $\sqrt{2} > 2 > 1$. Therefore $\sqrt{2} < 2$.

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16. Show that $\sqrt{x} < \sqrt{y}$ if and only if $x < y$ for all positive $x, y \in \mathbb{R}$.

    Assume $\sqrt{x} < \sqrt{y}$. By trichotomy, either $x < y$, $x = y$, or $x > y$. If $y = x$, then $\sqrt{x} = \sqrt{y}$, since roots are unique, but this is a contradiction. If instead $y < x$, then $\sqrt{y}\sqrt{y} < \sqrt{x}\sqrt{x}$. But $\sqrt{x}\sqrt{x} < \sqrt{x}\sqrt{y}$, so by transitivity, $\sqrt{y}\sqrt{y} < \sqrt{x}\sqrt{y}$. But this implies $\sqrt{y} < \sqrt{x}$, which is a contradiction. Therefore $x < y$.

    Conversely, let $x < y$. By trichotomy, either $\sqrt{x} < \sqrt{y}$, $\sqrt{x} = \sqrt{y}$, or $\sqrt{x} > \sqrt{y}$. If $\sqrt{x} = \sqrt{y}$, then $x = y$, which is a contradiction. If instead $\sqrt{x} > \sqrt{y}$, then $\sqrt{x}\sqrt{y} > \sqrt{y}\sqrt{y}$. But by our assumption $\sqrt{x}\sqrt{x} < \sqrt{y}\sqrt{y}$, so by transitivity $\sqrt{x}\sqrt{y} > \sqrt{x}\sqrt{x}$, so $\sqrt{y} > \sqrt{x}$, which is a contradiction. Therefore $\sqrt{x} < \sqrt{y}$.

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17. Show that $\sqrt{x + y} \leq \sqrt{x} + \sqrt{y}$.

    Hint: Consider $(\sqrt{x + y})^2 - (\sqrt{x} + \sqrt{y})^2$.

    $ (\sqrt{x + y})^2 - (\sqrt{x} + \sqrt{y})^2 = (x + y) - (x + 2\sqrt{x}\sqrt{y} + y) \\ (\sqrt{x + y})^2 - (\sqrt{x} + \sqrt{y})^2 = -2\sqrt{x}\sqrt{y} \\ (\sqrt{x + y})^2 = (\sqrt{x} + \sqrt{y})^2 - 2\sqrt{x}\sqrt{y} \\ (\sqrt{x + y})^2 \leq (\sqrt{x} + \sqrt{y})^2 \\ \sqrt{x + y} \leq \sqrt{x} + \sqrt{y}$

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