$\sqrt{16} = 4$ — exact, rational, integer. $16$ is a perfect square because $4 \cdot 4 = 16$, so the answer comes out clean.
Result: $\sqrt{16} = 4$ (principal square root); both $4$ and $-4$ satisfy $x^2 = 16$.
Notation: Integer; radical form $\sqrt{16}$ simplifies to $4$.
Method shown: Prime factorization (Quick), repeated subtraction (Standard, with a Wrong-Path-First detour on the four-prime-factors question), long division (Stretch).
Approximate value: No approximation needed — the answer is exact.
Exact form: $4$.
Quick Reference Table — Small Perfect Squares
$n$ | $\sqrt{n}$ (exact) | Perfect square? | Rational / irrational |
|---|---|---|---|
$9$ | $3$ | Yes | Rational |
$12$ | $2\sqrt{3} \approx 3.4641$ | No | Irrational |
$15$ | $\sqrt{15} \approx 3.8730$ | No | Irrational |
$16$ | $\boldsymbol{4}$ | Yes | Rational |
$18$ | $3\sqrt{2} \approx 4.2426$ | No | Irrational |
$20$ | $2\sqrt{5} \approx 4.4721$ | No | Irrational |
$24$ | $2\sqrt{6} \approx 4.8990$ | No | Irrational |
$25$ | $5$ | Yes | Rational |
$36$ | $6$ | Yes | Rational |
$\sqrt{16}$ sits between $\sqrt{9} = 3$ and $\sqrt{25} = 5$ — the fourth-smallest non-trivial perfect square after $1, 4, 9$.
Where √16 appears
$\sqrt{16} = 4$ is the side length of any square with area $16$ — a $4 \times 4$ grid, a flower bed laid out in a $4 \times 4$ pattern, a chessboard quadrant. The number $16$ itself appears all over computing because $2^4 = 16$ — a hexadecimal digit takes values $0$–$15$, exactly $16$ possibilities, and $16$-bit integers run from $-32{,}768$ to $32{,}767$. In music, a $\tfrac{16}{16}$ time signature carries $16$ sixteenth-notes per measure, and $\sqrt{16} = 4$ is the count of quarter-note beats that equal one measure.
What "square root of 16" means
The square root of a non-negative number $n$ is the value $x$ such that $x^2 = n$. For $\sqrt{16}$, the positive $x$ with $x^2 = 16$ — which is $4$, because $4 \cdot 4 = 16$.
The $\sqrt{\phantom{x}}$ symbol denotes the principal (non-negative) square root. The equation $x^2 = 16$ has two solutions, $x = 4$ and $x = -4$; the expression $\sqrt{16}$ refers only to the positive one. Same arithmetic, different question.
Is √16 rational or irrational?
$\sqrt{16} = 4$ is rational. A number is a perfect square if and only if every prime in its factorisation appears to an even power. $16 = 2^{4}$ — the $2$ has exponent $4$, which is even, so $16$ is a perfect square and its square root is the integer $4$.
Every integer is rational ($4 = 4/1$), so $\sqrt{16}$ is rational. The decimal expansion of $\sqrt{16}$ is $4.000\ldots$ — terminating immediately at the integer.
How to find √16 — three methods
Method 1 — Prime factorization (Quick)
Factor $16$ into primes, then pair them.
$$16 = 2 \cdot 2 \cdot 2 \cdot 2 = 2^{4}$$
There are four $2$s, forming two pairs. Each pair leaves the radical as a single $2$:
$$\sqrt{16} = \sqrt{2^{4}} = \sqrt{2^{2} \cdot 2^{2}} = 2 \cdot 2 = 4$$
Final answer: $\sqrt{16} = 4$.
Method 2 — Repeated subtraction (Standard, with a Wrong-Path-First on the four-prime question)
A common student instinct: since $16 = 2 \cdot 2 \cdot 2 \cdot 2$ has four prime factors, the answer must somehow involve $4$. Let us check that intuition — it lands on the right answer, but for the wrong reason.
The wrong reason: "$16$ has four prime factors, so $\sqrt{16} = 4$." The right answer is $4$, but this logic fails on $\sqrt{36}$: $36 = 2 \cdot 2 \cdot 3 \cdot 3$ also has four prime factors — yet $\sqrt{36} = 6$, not $4$. The number of prime factors is a coincidence here, not the cause.
Now run repeated subtraction. Subtract consecutive odd numbers from $16$: $16 - 1 = 15$, $15 - 3 = 12$, $12 - 5 = 7$, $7 - 7 = 0$. Four subtractions, remainder zero — so $\sqrt{16} = 4$. The method counts how many odd numbers (starting from $1$) sum to $16$, and that count is the square root for any perfect square. This is the correct reason: $1 + 3 + 5 + 7 = 16$, four terms, so $\sqrt{16} = 4$.
Final answer: $\sqrt{16} = 4$.
The silent-understander archetype reaches the answer by intuition but cannot articulate why — repeated subtraction is the method that turns intuition into a verifiable count.
Method 3 — Long division (Stretch)
Long division on a perfect square terminates immediately. Run it on $16.00$.
Step 1. Largest integer with square $\leq 16$ is $4$ ($4^2 = 16$). Subtract: $16 - 16 = 0$. Bring down $00$: $0$.
Step 2. Double $4$: $8$. Find $d$ with $(80 + d) \cdot d \leq 0$. $d = 0$. Subtract: $0 - 0 = 0$.
Every further step gives $d = 0$, so $\sqrt{16} = 4.000\ldots = 4$.
Final answer: $\sqrt{16} = 4$.
Long division is overkill on $\sqrt{16}$ — but it confirms the answer and shows how the same algorithm that struggles to four decimal places on $\sqrt{20}$ terminates immediately here.
Three Errors That Cost The Most Marks on √16
1. Confusing the square root expression with the quadratic equation
Where it slips in: A student sees $\sqrt{16}$ and writes $\pm 4$, transferring the rule from $x^2 = 16 \Rightarrow x = \pm 4$.
Don't do this: $\sqrt{16} = \pm 4$ — written as the final answer to a square-root expression.
The correct way: $\sqrt{16} = 4$ (the principal, non-negative root). The $\pm$ shows up only when solving $x^2 = 16$, because taking the square root of both sides introduces $\pm$ explicitly: $x = \pm \sqrt{16} = \pm 4$. The expression alone is single-valued.
2. Reading $\sqrt{16}$ as $16 \div 2$ or $16 \div 4$
Where it slips in: A student rushes and answers $8$ ($16 \div 2$) or $4$ ($16 \div 4$, right answer wrong method).
Don't do this: $\sqrt{16} = 16 \div 2 = 8$.
The correct way: $\sqrt{16}$ is the number whose square equals $16$. $8^2 = 64$, not $16$ — so $8$ is wrong. The defining check is squaring back: $4 \cdot 4 = 16$ ✓; $8 \cdot 8 = 64$ ✗.
3. Writing $\sqrt{16}$ when $4$ is expected (or vice versa).
Where it slips in: A student leaves the answer as $\sqrt{16}$ in a multiple-step problem, then later squares it expecting $16$ — but the marker has already deducted marks for not simplifying.
Don't do this: $\sqrt{16}$ — left unresolved in the final answer.
The correct way: $\sqrt{16} = 4$. For a perfect square, always resolve to the integer. The radical form $\sqrt{16}$ is only acceptable as an intermediate expression while computing; the final answer must show $4$. Read the question — "evaluate" and "simplify" both want the integer for a perfect-square radicand.
The Short Version
The square root of 16 is $4$ — the principal, non-negative root.
$\sqrt{16}$ is rational because $16$ is a perfect square ($16 = 2^4$, every prime to an even power).
The expression $\sqrt{16}$ has a single value ($4$); the equation $x^2 = 16$ has two solutions ($\pm 4$).
All three methods reach the same answer — prime factorization in one line, repeated subtraction in four steps, long division terminating immediately.
Always square back: $4 \cdot 4 = 16$ ✓. If the squaring-back check fails, the square-root answer is wrong.
A Practical Next Step
Show that $\sqrt{25} = 5$ using prime factorization, then verify by squaring back.
Use repeated subtraction to find $\sqrt{36}$ — count how many odd numbers sum to $36$.
Solve $x^2 = 16$ and explain in one sentence why the answer has two values while $\sqrt{16}$ has one.
Want a live Bhanzu trainer to walk through more perfect-square problems? Book a free demo class — online globally.
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