Square Root of 3 — Value, How to Find It, and Examples

The Answer at a Glance

Quick Answer:

Result: $\sqrt{3} \approx 1.7320508$

Notation: Decimal approximation; exact form is $\sqrt{3}$.

Method shown: Long division by hand, cross-checked with estimation between perfect squares.

Approximate value: $1.7321$ (4 d.p.)

Exact form: $\sqrt{3}$ — cannot be simplified, since $3$ is prime.

Quick Reference Table — Square Roots From 1 to 20

$\sqrt{3}$ sits between $\sqrt{1} = 1$ and $\sqrt{4} = 2$, and a little past the midpoint because $3$ is closer to $4$ than to $1$.

Where √3 Appears

$\sqrt{3}$ is the height of an equilateral triangle whose side length is $2$ — drop a perpendicular and Pythagoras gives $\sqrt{2^2 - 1^2} = \sqrt{3}$. It is also the length of the space diagonal of a unit cube, and the distance across a regular hexagon of side $1$ from one flat edge to the opposite flat edge. It shows up constantly in trigonometry too: $\tan 60^\circ = \sqrt{3}$, which is why $\sqrt{3}$ appears in any 30-60-90 triangle.

What "square root of 3" Means

The square root of a non-negative number $n$ is the value $x$ such that $x^2 = n$. For $\sqrt{3}$, it is the positive $x$ with $x^2 = 3$.

Because $1^2 = 1$ and $2^2 = 4$, the answer must land between $1$ and $2$ — and squaring $1.732$ gives $2.9998\ldots$, which confirms the value.

Is The Square Root of 3 Rational or Irrational?

$\sqrt{3}$ is irrational. A whole number is a perfect square only when every prime in its factorisation appears an even number of times; $3$ is prime, so it carries the single factor $3^1$ — an odd exponent — and cannot be a perfect square.

The classic proof by contradiction makes this airtight: assume $\sqrt{3} = p/q$ in lowest terms, square to get $3q^2 = p^2$, and the parity of the factor $3$ on each side cannot match. NASA's educational archive walks through this same argument step by step. The decimal $1.7320508\ldots$ never terminates and never repeats.

How To Find √3 — Two Methods

Method 1 — Long division (digit by digit)

Write $3$ as $3.000000$ and pair the digits after the decimal point.

Step 1. The largest integer whose square is at most $3$ is $1$ ($1^2 = 1$). Subtract: $3 - 1 = 2$. Bring down $00$ to get $200$.

Step 2. Double the quotient $1$ to get $2$. Find $d$ with $(20 + d)\cdot d \leq 200$. Here $d = 7$ gives $27 \cdot 7 = 189$. Subtract: $200 - 189 = 11$. Bring down $00$ to get $1100$.

Step 3. Double $1.7$ to get $34$. Find $d$ with $(340 + d)\cdot d \leq 1100$. Here $d = 3$ gives $343 \cdot 3 = 1029$. Subtract: $1100 - 1029 = 71$.

Continuing produces $1.7320\ldots$

Final answer: $\sqrt{3} \approx 1.7321$.

Method 2 — Estimation between perfect squares

Since $1^2 = 1$ and $2^2 = 4$, start at $1.7$: $1.7^2 = 2.89$, a touch low. Try $1.73$: $1.73^2 = 2.9929$. Try $1.732$: $1.732^2 = 2.999824$. Each refinement nudges the guess upward toward $3$, landing on $1.732$ for everyday use. This is the same idea as the average (Babylonian) method, just done by inspection.

What Are The Most Common Mistakes With √3?

Mistake 1: Treating √3 as if it can be simplified

Where it slips in: A student sees a radical and reaches for the "pull out a square factor" rule on autopilot.

Don't do this: Writing $\sqrt{3} = \sqrt{1} \cdot \sqrt{3} = \sqrt{3}$ and then "simplifying" further, or splitting $3 = 1 + 2$ under the root.

The correct way: $3$ is prime, so it has no square factor greater than $1$. $\sqrt{3}$ is already in simplest radical form.

Mistake 2: Splitting the root over addition

Where it slips in: When $\sqrt{3}$ appears inside a sum like $\sqrt{1 + 2}$.

Don't do this: $\sqrt{1 + 2} = \sqrt{1} + \sqrt{2} = 1 + 1.414 = 2.414$.

The correct way: $\sqrt{1 + 2} = \sqrt{3} \approx 1.732$. Square roots do not distribute over addition.

Mistake 3: Rounding too early

Where it slips in: Replacing $\sqrt{3}$ with $1.73$ at the start of a longer calculation, then expecting an exact result later.

Don't do this: $1.73^2 = 2.9929 \neq 3$.

The correct way: Carry the exact form $\sqrt{3}$ through the algebra and convert to a decimal only at the final step.

Examples of Square Root of 3

Example 1

Simplify $\sqrt{12}$ using $\sqrt{3}$.

$\sqrt{12} = \sqrt{4 \cdot 3} = 2\sqrt{3} \approx 3.4641$. The factor $4$ comes out as $2$; the $3$ stays under the root.

Example 2 (Wrong path first)

Find the height of an equilateral triangle with side $2$.

Wrong attempt. A student writes the height as half the side, $\frac{2}{2} = 1$ — confusing "half the base" with "the height."

Why it breaks. That would make a triangle with base $2$ and height $1$, whose sides could not all be length $2$; the slant side would be only $\sqrt{1^2 + 1^2} = \sqrt{2} < 2$.

Correct. Drop the altitude to split the base in half: height $= \sqrt{2^2 - 1^2} = \sqrt{3} \approx 1.732$.

Example 3

Evaluate $\tan 60^\circ$.

In a 30-60-90 triangle the sides are in ratio $1 : \sqrt{3} : 2$, so $\tan 60^\circ = \frac{\sqrt{3}}{1} = \sqrt{3} \approx 1.732$.

Example 4

Rationalise $\dfrac{1}{\sqrt{3}}$.

Multiply top and bottom by $\sqrt{3}$: $\dfrac{1}{\sqrt{3}} = \dfrac{\sqrt{3}}{3} \approx 0.5774$.

Example 5

Find the space diagonal of a cube with side $1$.

The diagonal is $\sqrt{1^2 + 1^2 + 1^2} = \sqrt{3} \approx 1.732$ — the same value, now in three dimensions.

Conclusion

• The square root of 3 is approximately $1.732$ — irrational, non-terminating, non-repeating.

• $3$ is prime, so $\sqrt{3}$ cannot be simplified and stays in radical form.

• Long division and estimation both pin the value down by hand.

• Square roots do not distribute over addition: $\sqrt{a+b} \neq \sqrt{a} + \sqrt{b}$.

• $\sqrt{3}$ is the height of an equilateral triangle of side $2$ and equals $\tan 60^\circ$.

A Practical Next Step

1. Find $\sqrt{2}$ to three decimal places by long division, then compare your digits to a calculator.

2. Show that $\sqrt{12}$ simplifies to $2\sqrt{3}$ but $\sqrt{3}$ itself does not simplify.

3. A 30-60-90 triangle has its shortest side equal to $1$. Find the other two sides in exact form.

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