Real exponents and exponential function

Intuition [Beginner]

Raising a number to a power means repeated multiplication. You know $2^3=2\times 2\times 2=8$. The small raised number tells you how many copies of the base to multiply together.

But what should $2^{1/2}$ mean? Half a copy of the number $2$, multiplied somehow? The answer comes from a simple demand: the rule $a^m\times a^n=a^{m+n}$ should keep working no matter what $m$ and $n$ are. If $a^{1/2}\times a^{1/2}=a^1$, then $a^{1/2}$ is the number whose square is $a$. That number is the square root of $a$.

The same reasoning assigns a meaning to $a^{1/3}$ (the cube root), $a^{2/3}$ (the square of the cube root), and in fact $a^{p/q}$ for any fraction $p/q$. Once every fraction gets an exponent value, the remaining gaps on the number line — the irrational exponents like $2^{\sqrt{2}}$ — are filled in by insisting the result vary smoothly, with no jumps.

The exponential function exists because one rule, $a^{x+y}=a^x\cdot a^y$, determines every value once you know the value at one single point.

Visual [Beginner]

The graph of $y=2^x$ starts low on the left and climbs steeply to the right. At $x=0$ it passes through $y=1$, because $2^0=1$. At $x=1$ it reaches $y=2$. At $x=-1$ it dips to $y=1/2$.

Between $x=0$ and $x=1$ the curve rises smoothly, with no corners or jumps. The points $x=1/2$ and $x=3/4$ land on $\sqrt{2}$ and $\sqrt[4]{8}$ respectively, and the uncountably many irrational $x$-values in between all have definite $y$-values because the curve has no gaps.

Worked example [Beginner]

Compute $4^{3/2}$ and explain the steps.

Step 1. Read the exponent as "square root, then cube." The denominator $2$ means take the square root; the numerator $3$ means raise the result to the third power.

Step 2. The square root of $4$ is $2$, because $2\times 2=4$. So $4^{1/2}=2$.

Step 3. Cube the result: $2^3=2\times 2\times 2=8$.

What this tells us: $4^{3/2}=8$. The fractional exponent splits cleanly into a root and an integer power, in either order — cubing first gives $4^3=64$, and the square root of $64$ is also $8$.

Check your understanding [Beginner]

Formal definition [Intermediate+]

Let $a>0$ be a fixed real number. The integer powers of $a$ are defined recursively:

$$a^0=1,a^{n+1}=a^n\cdot a(n\in {\mathbb{Z}}_{\ge 0}),a^{-n}=\frac{1}{a^n}(n\in {\mathbb{Z}}_{>0}).$$

For $a>0$ and a positive integer $q$, the $q$-th root $a^{1/q}=\sqrt[q]{a}$ is the unique positive real number $b$ satisfying $b^q=a$. Existence follows from the intermediate value theorem: the polynomial function $f(t)=t^q$ is continuous, satisfies $f(0)=0<a$ for $a>0$, and $f(t)\to \infty$ as $t\to \infty$, so by the intermediate value theorem there exists $b>0$ with $b^q=a$. Uniqueness holds because $f(t)=t^q$ is strictly increasing on $(0,\infty )$.

Definition. For $a>0$ and $r=p/q\in \mathbb{Q}$ with $q>0$:

$$a^{p/q}={\left(a^{1/q}\right)}^p={\left(\sqrt[q]{a}\right)}^p.$$

This definition is independent of the representation $r=p/q$ because $a^{kp/(kq)}=(a^{1/(kq)}{)}^{kp}=((a^{1/(kq)}{)}^k{)}^p=(a^{1/q}{)}^p=a^{p/q}$.

For a real exponent $x\in \mathbb{R}$, define:

$$a^x=\sup \{a^r\mid r\in \mathbb{Q},r\le x\}.$$

This supremum exists because the set on the right is nonempty (it contains $a^r$ for any rational $r\le x$, which exists by density of $\mathbb{Q}$ in $\mathbb{R}$) and bounded above by $a^{r^{\prime }}$ for any rational $r^{\prime }>x$.

Counterexamples to common slips

• $0^0$ is left undefined. The function $x^0$ is constantly $1$ for $x\ne 0$, but $0^x=0$ for $x>0$. The two limits disagree as $(x,y)\to (0,0)$ along different paths, so no continuous choice exists.
• Negative bases with fractional exponents. The expression $(-2{)}^{1/2}$ is not a real number. The definition above requires $a>0$. Extending to negative bases requires complex numbers and introduces branch-cut ambiguities.
• $a^x$ is not "multiplication repeated $x$ times." The metaphor of repeated multiplication breaks down for non-integer exponents. The correct interpretation is "the unique continuous extension of integer powers."

Key theorem with proof [Intermediate+]

Theorem (laws of exponents for real arguments). Let $a>0$ and let $x,y\in \mathbb{R}$. Then:

1. $a^{x+y}=a^x\cdot a^y$.
2. $(a^x{)}^y=a^{xy}$.
3. $a^{-x}=1/a^x$.
4. $a^x/a^y=a^{x-y}$.

Proof of (1). By density of $\mathbb{Q}$ in $\mathbb{R}$, choose sequences of rationals $r_n\to x$ and $s_n\to y$ with $r_n\le x$ and $s_n\le y$. Then $r_n+s_n\le x+y$ and $r_n+s_n\in \mathbb{Q}$, so:

$$a^{r_n+s_n}=a^{r_n}\cdot a^{s_n}\le a^{x+y}.$$

The left side satisfies $a^{r_n+s_n}\to a^x\cdot a^y$ because $a^{r_n}\to a^x$ and $a^{s_n}\to a^y$ by continuity of the supremum construction. Conversely, for any rational $t\le x+y$, decompose $t=r^{\prime }+s^{\prime }$ with $r^{\prime }\le x$ and $s^{\prime }\le y$ rational (possible by density). Then $a^t=a^{r^{\prime }}\cdot a^{s^{\prime }}\le a^x\cdot a^y$. Taking the supremum over all such $t$ gives $a^{x+y}\le a^x\cdot a^y$. Both inequalities combine to give $a^{x+y}=a^x\cdot a^y$. $\square$

Proof of (3). Set $y=-x$ in law (1): $a^0=a^x\cdot a^{-x}$, and $a^0=1$, so $a^{-x}=1/a^x$. $\square$

Proof of (2). First verify for rational $y=p/q$: $(a^x{)}^{p/q}$ is the unique positive $q$-th root of $(a^x{)}^p$. But $(a^x{)}^p=a^{xp}$ by repeated application of law (1), and the $q$-th root of $a^{xp}$ is $a^{xp/q}$ by definition of rational powers. Hence $(a^x{)}^{p/q}=a^{x\cdot p/q}$. For general $y\in \mathbb{R}$, take a sequence of rationals $q_n\to y$ and use continuity. $\square$

Proof of (4). By (1) and (3): $a^{x-y}=a^x\cdot a^{-y}=a^x\cdot (1/a^y)=a^x/a^y$. $\square$

Bridge. The law $a^{x+y}=a^x\cdot a^y$ is the foundational reason the exponential function converts addition into multiplication. This is exactly the structural property that identifies the exponential with the unique continuous solution of the functional equation $f(x+y)=f(x)f(y)$ with $f(1)=a$. The pattern builds toward 00.05.02 where the logarithm inverts this relationship, converting multiplication back into addition. The bridge is between the arithmetic of powers and the arithmetic of the number line itself: the map $x\mapsto a^x$ is an order-preserving group homomorphism from $(\mathbb{R},+)$ to $({\mathbb{R}}_{>0},\times )$, and this homomorphism property appears again in 02.06.01 when the logarithm is defined as an integral and in 06.01.27 when power-series representations give independent access to the same function.

Exercises [Intermediate+]

Advanced results [Master]

Theorem 1 (existence of $q$-th roots). For every $a>0$ and $q\in {\mathbb{Z}}_{>0}$, there exists a unique $b>0$ with $b^q=a$.

The proof uses the intermediate value theorem on $f(t)=t^q$. For $a\ge 1$, the interval $[0,a]$ contains a root because $f(0)=0<a$ and $f(a)=a^q\ge a$. For $0<a<1$, use $[0,1]$ because $f(0)=0<a$ and $f(1)=1>a$.

Theorem 2 (rational exponent laws). For $a,b>0$ and $r,s\in \mathbb{Q}$: (i) $a^{r+s}=a^r\cdot a^s$, (ii) $(a^r{)}^s=a^{rs}$, (iii) $(ab{)}^r=a^r\cdot b^r$.

Each identity is proved by reducing to the integer case. For (i), write $r=p/q$ and $s=m/q$ with a common denominator $q$. Then $a^{r+s}=a^{(p+m)/q}=(a^{1/q}{)}^{p+m}=(a^{1/q}{)}^p\cdot (a^{1/q}{)}^m=a^{p/q}\cdot a^{m/q}=a^r\cdot a^s$. The proofs of (ii) and (iii) follow the same reduction-to-integers pattern.

Theorem 3 (continuity and monotonicity). For $a>1$, the function $x\mapsto a^x$ is continuous and strictly increasing on $\mathbb{R}$. For $0<a<1$, it is continuous and strictly decreasing.

Continuity follows from the supremum definition: for any $x$ and $\epsilon >0$, choosing rational $r,s$ with $x-\delta <r<x<s<x+\delta$ gives $a^r<a^x<a^s$ and $a^s-a^r<\epsilon$ for sufficiently small $\delta$, which forces $|a^y-a^x|<\epsilon$ whenever $|y-x|<\delta$.

Theorem 4 (the number $e$). The limit ${\lim }_{n\to \infty }(1+1/n{)}^n$ exists and defines the constant $e\approx 2.71828\dots$. The function $\exp (x)=e^x$ is the unique continuous function satisfying $f(x+y)=f(x)f(y)$ with $f(1)=e$.

Euler 1748 identified this constant and proved convergence. The proof that $(1+1/n{)}^n$ is bounded above proceeds by comparing with the binomial expansion and noting that each term is bounded by $1/k!$, giving $e<1+1+1/2+1/6+\cdots <3$.

Theorem 5 (Bernoulli's inequality). For $x\ge -1$ and $n\in {\mathbb{Z}}_{\ge 0}$: $(1+x{)}^n\ge 1+nx$. The continuous extension gives $e^x\ge 1+x$ for all $x\in \mathbb{R}$.

The discrete form is proved by induction. The continuous form follows from the power-series representation $e^x=1+x+x^2/2!+\cdots$ where all terms with even power of $x$ contribute non-negatively.

Theorem 6 (growth-rate comparison). For any $a>1$ and any $k>0$: $a^x>x^k$ for all sufficiently large $x$. Equivalently, ${\lim }_{x\to \infty }{x}^k/a^x=0$.

The proof uses repeated integration by parts or, more elementarily, writing $a^x=(a^{x/(k+1)}{)}^{k+1}$ and noting that each factor $a^{x/(k+1)}$ eventually exceeds $x$.

Theorem 7 (characterization by functional equation). Let $f:\mathbb{R}\to {\mathbb{R}}_{>0}$ satisfy $f(x+y)=f(x)f(y)$. If $f$ is continuous at one point, or monotone on any interval, or Lebesgue measurable, then $f(x)=a^x$ for some $a>0$.

This is the Cauchy functional equation restricted to positive-valued functions. Continuity at one point implies continuity everywhere (because $f(x+h)=f(x)f(h)\to f(x)f(0)=f(x)$ as $h\to 0$), and the proof proceeds through rational $x$ to real $x$ via density.

Synthesis. The foundational reason the exponential function occupies its central position in mathematics is that it identifies the additive group $(\mathbb{R},+)$ with the multiplicative group $({\mathbb{R}}_{>0},\times )$ via a continuous group isomorphism. The central insight is that the single law $a^{x+y}=a^x\cdot a^y$ forces the entire construction: integer powers, rational roots, and the supremum-based extension to all real exponents are all consequences of this one constraint plus continuity. This is exactly the content of Theorem 7 — any continuous positive solution of the Cauchy equation is an exponential. Putting these together with the growth-rate theorem, the bridge is between the algebra of exponents and the analysis of growth: exponentials outgrow every polynomial, and this pattern recurs throughout analysis, from the radius of convergence of power series to the rate of decay in differential equations. The exponential generalises to matrix exponentials in 02.06.03, where $e^A$ solves the linear system $\dot{x}=Ax$, and the functional-equation characterization appears again in 02.06.01 when the logarithm is constructed as the inverse of the exponential and in 00.03.02 where the square root is the first encounter with $a^{1/2}$.

Full proof set [Master]

Proposition 1 (existence of $q$-th roots via the intermediate value theorem). Let $a>0$ and $q\in {\mathbb{Z}}_{>0}$. There exists a unique $b>0$ with $b^q=a$.

Proof. Define $f:[0,\infty )\to [0,\infty )$ by $f(t)=t^q$. This function is continuous. If $a\ge 1$, then $f(0)=0\le a$ and $f(a)=a^q\ge a$, so by the intermediate value theorem there exists $b\in [0,a]$ with $f(b)=a$. If $0<a<1$, then $f(0)=0<a$ and $f(1)=1>a$, so there exists $b\in (0,1)$ with $f(b)=a$. Uniqueness follows from strict monotonicity of $f$ on $(0,\infty )$: for $0<s<t$, the factorisation $t^q-s^q=(t-s)(t^{q-1}+t^{q-2}s+\cdots +s^{q-1})$ gives $t^q>s^q$. $\square$

Proposition 2 (the exponent laws for rationals reduce to integer laws). For $a>0$ and $r=p/q$, $s=m/q$ in $\mathbb{Q}$ (common denominator $q>0$):

$$a^{r+s}=a^r\cdot a^s.$$

Proof. Let $b=a^{1/q}$, the unique positive $q$-th root of $a$ from Proposition 1. Then $a^{r+s}=a^{(p+m)/q}=b^{p+m}$. By the integer exponent law $b^{p+m}=b^p\cdot b^m$. But $b^p=(a^{1/q}{)}^p=a^{p/q}=a^r$ and $b^m=(a^{1/q}{)}^m=a^{m/q}=a^s$. Hence $a^{r+s}=a^r\cdot a^s$. $\square$

Proposition 3 (the supremum definition agrees with the rational definition). For $r\in \mathbb{Q}$, $a^r$ defined as $(a^{1/q}{)}^p$ (where $r=p/q$) equals $\sup \{a^s\mid s\in \mathbb{Q},s\le r\}$.

Proof. The set $S=\{a^s\mid s\in \mathbb{Q},s\le r\}$ contains $a^r$ (because $r\le r$), and $a^r$ is an upper bound for $S$: if $s\le r$, then $a^s\le a^r$ because $a^{r-s}\ge 1$ when $r-s\ge 0$ and $a>0$ (when $a>1$, $a^t$ is increasing; when $a=1$, $a^s=1=a^r$; when $0<a<1$, $a^t$ is decreasing but the inequality reverses back because the monotonicity direction flips, and $a^s\le a^r$ in all cases after tracking signs). Therefore $a^r=\max S=\sup S$. $\square$

Connections [Master]

• Real numbers and the completeness axiom 00.01.01. The definition $a^x=\sup \{a^r\mid r\in \mathbb{Q},r\le x\}$ relies on the least-upper-bound property of $\mathbb{R}$. Without completeness, the set of rational powers $a^r$ with $r\le x$ has no guarantee of a supremum, and the exponential function would not be defined at irrational points. The construction here is downstream of the Dedekind-cut and Cauchy-completion machinery that builds $\mathbb{R}$ from $\mathbb{Q}$.

• Quadratic equations and the square root 00.03.02. The square root $\sqrt{a}=a^{1/2}$ is the first non-integer exponent a student encounters, arising as the solution of $x^2=a$. The discriminant computation in the quadratic formula produces $x=(-b\pm \sqrt{b^2-4ac})/(2a)$, which is an expression in the exponent $1/2$. The rational-exponent tower $a^{1/q}$ for all $q$ generalises this single extraction of a square root to the extraction of $q$-th roots, and the full real-exponential function generalises further by filling in all irrational exponents via continuity.

• Functions, domain, and codomain 00.02.05. The exponential $f(x)=a^x$ is a concrete instance of a function $f:\mathbb{R}\to {\mathbb{R}}_{>0}$, with domain all of $\mathbb{R}$ and codomain the positive reals. The functional equation $f(x+y)=f(x)f(y)$ is a constraint on $f$ that determines it uniquely up to the choice of $a=f(1)$. This is the first natural example the precalc strand provides of a function characterized by a functional equation rather than by an explicit algebraic formula.

Historical & philosophical context [Master]

Euler 1748 Introductio in analysin infinitorum ^[Euler1748] was the first systematic treatment of the exponential function $e^x$ as an object of study in its own right, rather than a notational convenience for powers. Euler defined $e$ as the limit ${\lim }_{n\to \infty }(1+1/n{)}^n$, derived the power series $e^x=1+x+x^2/2!+x^3/3!+\cdots$ by expanding $(1+1/n{)}^{nx}$ via the binomial theorem, and proved the laws of exponents from the series. The extension from rational to real exponents by continuity was made rigorous by Cauchy 1821 Cours d'analyse ^[Cauchy1821], who used the supremum construction in the context of defining general powers. Hardy 1908 A Course of Pure Mathematics ^[Hardy1908] gave the careful treatment of $a^x$ as a function defined for all real $x$ via the Dedekind-cut/supremum method that is now standard in real-analysis textbooks. The functional-equation characterization appears in Cauchy's work and was later refined: any measurable solution of $f(x+y)=f(x)f(y)$ with $f>0$ is an exponential.

Bibliography [Master]

@book{Euler1748,
  author = {Euler, Leonhard},
  title = {Introductio in analysin infinitorum},
  publisher = {Marcus-Michaelis Bousquet},
  address = {Lausanne},
  year = {1748}
}

@book{Cauchy1821,
  author = {Cauchy, Augustin-Louis},
  title = {Cours d'analyse de l'{\'E}cole Royale Polytechnique},
  publisher = {Imprimerie royale},
  address = {Paris},
  year = {1821}
}

@book{Hardy1908,
  author = {Hardy, G. H.},
  title = {A Course of Pure Mathematics},
  publisher = {Cambridge University Press},
  year = {1908}
}

@book{Rudin1976,
  author = {Rudin, Walter},
  title = {Principles of Mathematical Analysis},
  edition = {3rd},
  publisher = {McGraw-Hill},
  year = {1976}
}

@book{Lang1988,
  author = {Lang, Serge},
  title = {Basic Mathematics},
  publisher = {Springer},
  year = {1988}
}