Radicals

Definition of radicals

Radicals emerge from the problem of solving equations of the form $x^{n} = a$, where $n \in \mathbb{N}$, $n \geq 2$, and $a \in \mathbb{R}$. In this context, the $n$-th root of a number is defined as a value whose $n$-th power yields the original number.

When $a \geq 0$, the principal real $n$-th root of $a$ is the unique non-negative real number $b$ satisfying $b^{n} = a$. This value is denoted by $\sqrt[n]{a}$. The defining property is that $\sqrt[n]{a} = b$ if and only if $b^{n} = a$ and $b \geq 0$. The condition $b \geq 0$ guarantees uniqueness in the real case when $n$ is even.

The value $a$ is referred to as the radicand, and the integer $n$ is known as the index of the root. This notation specifies both the root extraction operation and its degree.

The properties of the roots depend on whether the index is even or odd.

• For even $n$, the equation $x^{n} = a$ has a real solution only if $a \geq 0$, in which case the principal root $\sqrt[n]{a}$ is the non-negative solution.
• For odd $n$, the equation $x^{n} = a$ has exactly one real solution for every real number $a$, so the function $a \rightarrowtail \sqrt[n]{a}$ is defined for all $a \in \mathbb{R}$.

For example, since $2^{3} = 8$, it follows that $\sqrt[3]{8} = 2$, as 2 is the unique real number whose cube equals 8. More generally, extracting an $n$-th root is the inverse operation of raising a number to the $n$-th power. Solving the equation $x^{n} = a$ is therefore equivalent to applying the $n$-th root to $a$.

Radicals such as $\sqrt{2}$, $\sqrt{3}$, and $\sqrt{5}$ are classified as irrational numbers because they cannot be expressed as exact fractions of two integers.

Their decimal expansions are infinite and non-repeating, lacking any predictable pattern. No rational number squared yields 2, 3, or 5. These values nonetheless occupy precise positions on the real number line, interspersed among the rational numbers, with no gaps.

Formally, if $a \in \mathbb{N}$ is not a perfect square, then $\sqrt{a} \notin \mathbb{Q}$.

Square roots are not always irrational. The square root of a perfect square, such as $4$ or $9$, is rational. In contrast, the square root of a non-perfect square, such as $2$ or $5$, is irrational because it cannot be expressed as a fraction.

Why is $\sqrt{2}$ irrational?

To prove that $\sqrt{2}$ is not a rational number, consider a proof by contradiction. Assume that $\sqrt{2}$ is rational. Then it can be expressed as a fraction of two integers in lowest terms, where $a , b \in \mathbb{Z}$, $b \neq 0$, and $gcd ( a , b ) = 1$:

$gcd ( a , b )$ denotes the greatest common divisor of $a$ and $b$, which is the largest positive integer dividing both numbers. The condition $gcd ( a , b ) = 1$ indicates that $a$ and $b$ are coprime, meaning they share no common divisors other than 1. Thus, the fraction $a / b$ is already in lowest terms.

Squaring both sides: \(2 = \frac{a^{2}}{b^{2}} \rightarrow a^{2} = 2 b^{2}\)

This implies that $a^{2}$ is even, which means $a$ must also be even. So we can write $a = 2 k$ for some integer $k$. Substituting back: \(( 2 k )^{2} = 2 b^{2} \rightarrow 4 k^{2} = 2 b^{2} \rightarrow b^{2} = 2 k^{2}\)

This means $b^{2}$ is also even, so $b$ must be even too. But if both $a$ and $b$ are even, they share a common factor which is $2$ which contradicts our initial assumption that $\frac{a}{b}$ is in lowest terms. This means $\sqrt{2}$ is irrational.

Powers with rational exponents

The connection between radicals and powers becomes explicit when the exponent is a rational number. For $a \in \mathbb{R}^{+}$ and $n \in \mathbb{N}$ with $n \geq 1$, the $n$-th root of $a$ is equivalently written as:

More generally, for $m \in \mathbb{Z}$, a radical whose radicand is raised to an integer power corresponds to a power with rational exponent $\frac{m}{n}$:

This representation is not merely a notational convenience. Because radicals are powers with rational exponents, all standard rules of exponentiation extend to them without modification. For $a , b \in \mathbb{R}^{+}$ and $\frac{m}{n} , \frac{p}{q} \in \mathbb{Q}$:

For example, $\sqrt{a^{3}} = a^{\frac{3}{2}}$, $\sqrt[3]{a^{2}} = a^{\frac{2}{3}}$, and $\sqrt[4]{a} = a^{\frac{1}{4}}$.

Properties

The following identities govern the manipulation of radicals. Each property is stated with the conditions on the radicand and index required for the expression to remain well-defined in the real numbers; these conditions depend in particular on whether the index is even or odd.

For $a \geq 0$, $n \in \mathbb{N}$ with $n \geq 1$, and $m \in \mathbb{Z}$, every radical can be written as a power with rational exponent:

If $n$ is even, the condition $a \geq 0$ is necessary to remain in the real numbers.

For $n \in \mathbb{N}$ with $n \geq 2$, the $n$-th root distributes over multiplication and division:

If $n$ is even, the product rule requires $a \geq 0$ and $b \geq 0$, and the quotient rule requires $a \geq 0$ and $b > 0$, in order to remain in the real numbers. If $n$ is odd, both identities hold for all admissible real values: any $a , b \in \mathbb{R}$ for the product, and any $a \in \mathbb{R}$, $b \in \mathbb{R} \backslash 0$ for the quotient.

For $a \geq 0$, $n \in \mathbb{N}$ with $n \geq 1$, and $m \in \mathbb{Z}$, raising a radical to an integer power is equivalent to raising the radicand to that power and then extracting the root:

If $n$ is even, the condition $a \geq 0$ is required to remain in the real numbers.

Let $k \in \mathbb{N}$ with $k \geq 1$. Multiplying both the index of the root and the exponent of the radicand by the same positive integer $k$ does not alter the value of the radical:

This identity allows the index of a radical to be reduced to its lowest form. For example, $\sqrt[4]{a^{2}} = \sqrt[2]{a} = \sqrt{a}$, obtained by dividing both exponents by 2. If $n$ is even, the condition $a \geq 0$ applies.

For $a \geq 0$ and $m , n \in \mathbb{N}$ with $m , n \geq 1$, a nested radical reduces to a single radical whose index is the product of the two indices:

If either $m$ or $n$ is even, the condition $a \geq 0$ is necessary to remain in the real numbers. If both $m$ and $n$ are odd, the identity holds for all $a \in \mathbb{R}$.

A radical $\sqrt[n]{a^{m}}$ can be simplified when $m \geq n$ by decomposing the exponent as $m = n q + r$, where $q$ is the quotient and $0 \leq r < n$ is the remainder of the division of $m$ by $n$. The factor $a^{q}$ can then be extracted from the radical:

For example, $\sqrt{a^{5}} = \sqrt{a^{4} \cdot a} = a^{2} \sqrt{a}$, since $5 = 2 \cdot 2 + 1$. Similarly, $\sqrt[3]{a^{7}} = a^{2} \sqrt[3]{a}$, since $7 = 3 \cdot 2 + 1$. The condition $a \geq 0$ applies when the index is even.

Two radicals are said to be like if they share the same index and the same radicand. Like radicals can be added and subtracted by combining their coefficients, in the same way as like terms in a polynomial:

For example: \(3 \sqrt{2} + 5 \sqrt{2} = 8 \sqrt{2}\) \(7 \sqrt[3]{5} - 2 \sqrt[3]{5} = 5 \sqrt[3]{5}\)

Radicals with different indices or different radicands are not like radicals and cannot be combined in this way. In some cases, simplification may reveal that two radicals are in fact like. For example:

since:

Example 1

Simplify the following expression and write the result in radical form:

By converting each radical to a power with rational exponent:

Applying the quotient rule for powers:

Therefore we obtain:

Rationalizing the denominator

An expression containing a radical in the denominator is generally rewritten in an equivalent form in which the denominator is free of radicals. This process is called rationalizing the denominator and relies on multiplying numerator and denominator by a suitably chosen expression without altering the value of the fraction. When the denominator is a single radical $\sqrt[n]{a^{m}}$, the goal is to complete the exponent of $a$ inside the radical to a multiple of $n$. Multiplying numerator and denominator by $\sqrt[n]{a^{n - m}}$ yields an integer in the denominator:

The most common case is $n = 2$ and $m = 1$, where the denominator is a square root:

When the denominator has the form $\sqrt{a} + \sqrt{b}$ or $\sqrt{a} - \sqrt{b}$, multiplying by the conjugate expression eliminates the radicals by applying the difference of squares identity $( x + y ) ( x - y ) = x^{2} - y^{2}$:

For example:

Example 2

Rationalization can also be applied to the numerator when this simplifies an expression. Consider the following limit, which arises naturally in the definition of the derivative:

Direct substitution of $h = 0$ yields the indeterminate form $\frac{0}{0}$. To resolve this, multiply numerator and denominator by the conjugate of the numerator:

Taking the limit as $h \rightarrow 0$:

This result is the derivative of $\sqrt{x}$, obtained here without invoking the general power rule.

Geometric construction of the segment $\sqrt{a}$

The square root $\sqrt{a}$ is more than a number: it can be constructed as a segment using only a compass and straightedge. This method transforms an algebraic idea into a geometric form, revealing the deep connection between numbers and shapes. Given a segment of length $a$, follow these steps:

• Draw a segment $A B$ of length $a$.
• Extend the segment to the left by 1 unit. Let point $C$ be such that $C A = 1$. Now $C B = a + 1$.
• Draw a semicircle with diameter $C B$.
• From point $A$, draw a perpendicular to $C B$, intersecting the semicircle at point $D$.
• Segment $A D$ has length $\sqrt{a}$.

In fact, in the right triangle $\triangle D A B$, the segment $A D$ is the height from point $A$ to the hypotenuse $C B$. According to Euclid’s theorem on right triangles, the height is the geometric mean between the two segments into which it divides the hypotenuse. That is:

Multiplying both sides by $A D$, we get:

Since $A C = 1$ and $A B = a$, we find:

This completes the construction: the segment $A D$ has length exactly equal to $\sqrt{a}$.

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