Square Root of 10 - Value, Calculation, Examples

The square root of 10 is approximately $3.162$ — more precisely $\sqrt{10} \approx 3.16227766\ldots$, an irrational number whose decimal expansion never terminates or repeats.

Quick Answer:

• Result: $\sqrt{10} \approx 3.16227766$

• Notation: $\sqrt{10}$ (exact, irrational), or $3.162$ to three decimal places

• Method shown: Long division

• Exact form: $\sqrt{10}$ — cannot be simplified further (10 has no perfect-square factors greater than 1)

• Irrational? Yes — non-terminating, non-repeating decimal

Quick Reference Table

Where $\sqrt{10}$ Appears

$\sqrt{10}$ shows up as the diagonal of a $1 \times 3$ rectangle — the Pythagorean theorem gives $\sqrt{1^2 + 3^2} = \sqrt{10}$. It also appears in engineering as the length of a unit step in base-10 logarithmic scales — a power ratio of 10 dB corresponds to a voltage ratio of $\sqrt{10}$. And anywhere a problem involves "factors of $\sqrt{10}$ steps," the irrational $3.162\ldots$ is at the heart of it.

What Is a Square Root?

The square root of a number $n$ is a value $r$ such that $r^2 = n$. The square root of 10 is the number that, multiplied by itself, gives 10.

There is no integer that does this — $3^2 = 9$ (too small) and $4^2 = 16$ (too big). So $\sqrt{10}$ lies between 3 and 4, somewhere closer to 3.

Because 10 is not a perfect square, $\sqrt{10}$ is an irrational number — its decimal goes on forever without ending or repeating. The first 12 digits are:

$$\sqrt{10} = 3.16227766017\ldots$$

Is the Square Root of 10 Rational or Irrational?

$\sqrt{10}$ is irrational — meaning it cannot be written as a fraction $\frac{p}{q}$ of two integers, and its decimal expansion neither terminates nor repeats.

Quick reasoning. A whole number $n$ has a rational square root only if $n$ is a perfect square (a number like 4, 9, 16, 25, 36...). 10 sits between $3^2 = 9$ and $4^2 = 16$ — it isn't a perfect square. So $\sqrt{10}$ is irrational.

Formal proof (by contradiction). Suppose $\sqrt{10}$ were rational. Then we could write $\sqrt{10} = \frac{p}{q}$ in lowest terms (no common factor). Squaring both sides:

$$10 = \frac{p^2}{q^2} \quad\Rightarrow\quad p^2 = 10 q^2 = 2 \cdot 5 \cdot q^2$$

So $p^2$ is divisible by 2, which forces $p$ to be divisible by 2 (any prime that divides a square divides the original number). Write $p = 2k$. Then $4k^2 = 10 q^2$, so $2k^2 = 5 q^2$. The right side is now divisible by 2, so $5q^2$ is even, which means $q^2$ is even, which means $q$ is divisible by 2.

But that makes both $p$ and $q$ divisible by 2 — contradicting our "lowest terms" assumption. Hence no such $p/q$ exists, and $\sqrt{10}$ is irrational. $\blacksquare$

The same argument works for $\sqrt{2}$, $\sqrt{3}$, $\sqrt{5}$, $\sqrt{6}$, $\sqrt{7}$, $\sqrt{8}$, $\sqrt{10}$, $\sqrt{11}$, ... — any non-perfect-square whole number has an irrational square root. (The original proof of $\sqrt{2}$ irrationality is attributed to the Pythagorean school around 500 BCE; legend has it the discoverer, Hippasus, was thrown overboard for revealing the result.)

What this means practically. Any decimal you write for $\sqrt{10}$ — even with thousands of digits — is an approximation. The exact value lives only in the symbol $\sqrt{10}$ itself. In algebra, keep the radical form; in numerical work, round to the precision your context requires.

How to Find $\sqrt{10}$ — Long Division Method

Step 1: Pair the digits from the decimal point. $10.\overline{00}\overline{00}\overline{00}\ldots$

Step 2: Find the greatest number whose square is $\leq 10$. That's $3$ ($3^2 = 9$). Write $3$ as the first digit of the quotient.

Step 3: Subtract: $10 - 9 = 1$. Bring down the next pair of zeros: $100$.

Step 4: Double the current quotient: $3 \times 2 = 6$. Write $6_$ and find a digit $d$ such that $(60 + d) \times d \leq 100$. Try $d = 1$: $61 \times 1 = 61$. Try $d = 2$: $62 \times 2 = 124$ — too big. So $d = 1$. Quotient becomes $3.1$.

Step 5: Subtract $61$ from $100$: remainder $39$. Bring down next pair: $3900$.

Step 6: Double the current quotient (ignoring decimal point): $31 \times 2 = 62$. Find $d$ such that $(620 + d) \times d \leq 3900$. Try $d = 6$: $626 \times 6 = 3756$. Try $d = 7$: $627 \times 7 = 4389$ — too big. So $d = 6$. Quotient becomes $3.16$.

Step 7: Continue. Remainder $3900 - 3756 = 144$; bring down next pair: $14400$. Doubled quotient $316 \times 2 = 632$. Try $d = 2$: $6322 \times 2 = 12644$. Try $d = 3$: $6323 \times 3 = 18969$ — too big. So $d = 2$. Quotient becomes $3.162$.

After three decimal places: $\sqrt{10} \approx 3.162$. The process continues indefinitely — that's what makes the number irrational.

Common Mistakes With Square-Root-of-10

Mistake 1: Reporting $\sqrt{10} = \pm 3.162$ in the wrong context

Where it slips in: Solving $x^2 = 10$.

Don't do this: Writing $\sqrt{10} = \pm 3.162$ in a context where only the positive root is wanted.

The correct way: The principal square root $\sqrt{10}$ is positive — about $3.162$. The equation $x^2 = 10$ has two solutions: $x = \sqrt{10}$ and $x = -\sqrt{10}$, written together as $x = \pm \sqrt{10}$. The symbol $\sqrt{10}$ alone refers to the positive root only.

Mistake 2: Trying to simplify $\sqrt{10}$

Where it slips in: Students assume every square root can be simplified.

Don't do this: Writing $\sqrt{10} = \sqrt{2} \cdot \sqrt{5}$ as a "simpler" form.

The correct way: $\sqrt{10}$ is already in simplest radical form. To simplify $\sqrt{n}$, you factor out a perfect square — and 10 has no perfect-square factor greater than 1 (its factors are $1, 2, 5, 10$). So $\sqrt{10}$ stays as $\sqrt{10}$.

Mistake 3: Rounding too early

Where it slips in: Multi-step problems where $\sqrt{10}$ appears mid-calculation.

Don't do this: Replace $\sqrt{10}$ with $3.162$ at the very start and carry that rounded value through every step.

The correct way: Keep $\sqrt{10}$ as $\sqrt{10}$ symbolically until the final step. Decimal approximation introduces error that compounds across multiplications and divisions.