Square Roots and Irrational Numbers

Squaring tells you an area from a side length. A square root works backward: it finds a number that produces the given result when multiplied by itself.

One radical sign, one value

Both $7$ and $-7$ are square roots of $49$. Check them: $7\times 7=49$ and $(-7)\times (-7)=49$. Every positive number has two square roots, with equal size and opposite signs. Zero has just one: $0$.

The sign $\sqrt{\phantom{p}}$ asks for the principal square root, the one that is not negative. Therefore $\sqrt{49}=7$. To request the negative root, put a minus sign outside: $-\sqrt{49}=-7$. To name both, write $\pm \sqrt{49}=\pm 7$. Read $\pm$ as “plus or minus.”

These two tasks have different answers:

Use squares you know

A perfect square is the square of a whole number. The squares from $0^2$ through ${15}^2$ are $0,1,4,9,16,25,36,49,64,81,100,121,144,169,196,225$. Read this relationship backward to find roots: since ${12}^2=144$, $\sqrt{144}=12$.

Taking a square root undoes squaring, not doubling. For example, $\sqrt{16}=4$ because $4^2=16$. Halving $16$ gives $8$, but $8^2=64$, so that check rules it out.

For a fraction whose numerator and denominator are perfect squares, take their roots separately:

Check by multiplication: $\frac{3}{4}\times \frac{3}{4}=\frac{9}{16}$. Roots greater than one work too: $\sqrt{25\mathrm{/}4}=5\mathrm{/}2$.

Keep the decimal point in the right place

Since $0.5\times 0.5=0.25$, $\sqrt{0.25}=0.5$. Similarly, $\sqrt{1.44}=1.2$ because ${1.2}^2=1.44$. Squaring a decimal with one decimal place gives a product written with two places before removing trailing zeros. A noninteger terminating decimal squared has an even number of decimal places after trailing zeros are removed; an integer square has zero decimal places.

Do not just take a root of the digits and guess the point. ${0.3}^2=0.09$, so $\sqrt{0.09}=0.3$. It does not follow that $\sqrt{0.9}=0.3$; in fact, $\sqrt{0.9}\approx 0.949$.

Watch signs and the radical bar

Squaring a real number always gives a positive number or zero. Therefore a negative number has no real square root. $\sqrt{-9}$ has no real value, while $-\sqrt{9}=-3$. The position of the minus sign matters.

For the same reason, $\sqrt{(-6{)}^2}=\sqrt{36}=6$. The principal root is nonnegative, even when the original number was negative. In the other direction, $(\sqrt{5}{)}^2=5$.

Calculate everything under one radical bar first:

Separate signs give a different calculation: $\sqrt{9}+\sqrt{16}=3+4=7$. A square root does not generally split across addition.

Find a side from an area

A square with area $81{\text{ cm}}^2$ has side $\sqrt{81}=9$ cm. For an area of $20{\text{ cm}}^2$, the exact side is $\sqrt{20}$ cm. Since $4^2<20<5^2$, the side lies between $4$ and $5$ cm. A length uses the positive root.

A cube root asks which number was used as a factor three times. Square-root equations and cube-root equations look similar, but their real solutions behave differently.

Read and check cube roots

The small $3$ in $\sqrt[3]{64}$ tells you to undo cubing. Since $4^3=4\times 4\times 4=64$, $\sqrt[3]{64}=4$. Dividing $64$ by $3$ does not undo cubing. To check a cube root, cube your answer.

The perfect cubes from $0^3$ through ${10}^3$ are $0,1,8,27,64,125,216,343,512,729,1000$. They help you find exact cube roots and locate other cube roots between whole numbers.

A negative number has a negative cube root. For example, $(-3{)}^3=-27$, so $\sqrt[3]{-27}=-3$. Two negative factors first give a positive product; the third makes it negative. Every real number has exactly one real cube root, including $\sqrt[3]{0}=0$.

Fractions and decimals follow the same check:

Also, $\sqrt[3]{0.064}=0.4$ because $0.4\times 0.4\times 0.4=0.064$. For $0.001$, the cube root is $0.1$.

Solve a squared equation

An equation asks for every value that makes it true. For $x^2=81$, both signs work:

You can write the pair as $x=\pm 9$. This is different from evaluating $\sqrt{81}$, whose answer is just $9$.

More generally, $x^2=p$ has two real solutions when $p>0$: $x=\pm \sqrt{p}$. When $p=0$, the two signs name the same value, so there is one solution, $x=0$. When $p<0$, there is no real solution because a real square cannot be negative.

Isolate the power first

For $2x^2=50$, first divide both sides by $2$:

Then find both roots: $x=5$ or $x=-5$. Check in the original equation: $2(5{)}^2=50$ and $2(-5{)}^2=50$. Taking a root before removing the coefficient would solve a different equation.

For $x^3+4=12$, subtract $4$ first:

There is no second negative answer: $(-2{)}^3=-8$. A cubed equation has one real solution for any real value on its right side.

Keep roots exact when necessary

If $x^2=10$, neither solution is a whole number. Write $x=\pm \sqrt{10}$ exactly, or approximately $\pm 3.16$ if a rounded answer is requested. For $x^3=20$, write $x=\sqrt[3]{20}$. Since $2^3=8$ and $3^3=27$, this solution lies between $2$ and $3$, about $2.71$.

Interpret a length

A square field with area $49{\text{ m}}^2$ leads to $s^2=49$. Algebra gives $s=7$ or $s=-7$, but the field’s side is $7$ m because lengths are positive. A cube with volume $216{\text{ cm}}^3$ has edge $\sqrt[3]{216}=6$ cm. The units change from area or volume to units of length.

A number can look different without changing what it is. A fraction, a decimal, and an integer can all name a rational number.

What makes a number rational?

A rational number can be written as $a\mathrm{/}b$, where $a$ and $b$ are integers and $b\mathrm{\ne }0$. The integers may be negative. For example, $-1.25=-5\mathrm{/}4$, $7=7\mathrm{/}1$, and $0=0\mathrm{/}1$. The word comes from ratio.

The number need not already be written as a fraction. $0.6$ is rational because $0.6=3\mathrm{/}5$. To write any fraction as a decimal, divide the numerator by the denominator.

For $3\mathrm{/}8$, division gives $0.375$ and then stops. This is a terminating decimal. For $2\mathrm{/}3$, division gives $0.666\dots$ and continues forever with the same digit. This is a repeating decimal, also called a recurring decimal.

Show exactly what repeats

A bar covers one copy of the repeating block:

In the last row, only the $6$ repeats. Moving the bar changes the value. Some countries mark a repeat with dots over the first and last repeating digits instead.

Why fractions end or repeat

Think about long division by $7$. After each step, the remainder must be one of $0,1,2,3,4,5,6$. A remainder of zero ends the division. Otherwise there are only six possible remainders, so one must return if division continues.

When the same remainder returns, the same division step follows. That produces the same next digit and next remainder. The cycle then continues forever. For $1\mathrm{/}7$, the remainders are $3,2,6,4,5,1$ before returning to $3$, giving $1\mathrm{/}7=0.\overline{142857}$. A finite choice of remainders explains why every fraction ends or eventually repeats.

Predict whether the decimal terminates

First simplify the fraction. Then write its denominator as prime factors. Its decimal terminates exactly when the denominator has no prime factors except $2$ and $5$. A denominator of $1$ also passes: it has no prime factors and gives an integer.

Why those primes? A terminating decimal is a fraction over $10$, $100$, $1000$, or another power of ten. Since $10=2\times 5$, these denominators contain only twos and fives. A denominator made from twos and fives can be scaled to a power of ten.

For example, $7\mathrm{/}20=35\mathrm{/}100=0.35$ terminates. But $7\mathrm{/}12$ is already simplified, and $12=2^2\times 3$ contains a $3$, so its decimal repeats: $0.58\bar{3}$.

Simplifying first matters. Although $12$ contains a factor of $3$, $3\mathrm{/}12=1\mathrm{/}4=0.25$. The factor of $3$ cancels and cannot prevent termination.

Read a calculator carefully

A screen might show $2\mathrm{/}3$ as $0.6666666667$. The final $7$ comes from rounding, not from a change in the repeat. The exact value is $0.\bar{6}$; $0.67$ is only an approximation. Write $2\mathrm{/}3\approx 0.67$.

A repeating block may be longer than the display: $1\mathrm{/}17$ repeats a block of sixteen digits. Use the fraction or an exact repeating form to decide its type. Every terminating or repeating decimal can also be turned back into a fraction, so both kinds are rational.

A terminating decimal has a last place, so place value turns it into a fraction. An endless repeating decimal needs a way to remove its repeating tail.

Start with place value

For $0.375$, three decimal places mean thousandths:

Similarly, $2.4=24\mathrm{/}10=12\mathrm{/}5$. You cannot use “the last decimal place” for $0.777\dots$ because there is no last digit.

Make matching tails

Call the repeating decimal $x$. Multiplying by a power of ten moves the decimal point without changing the order of the digits. Choose a shift equal to one whole repeating block. The digits after the decimal point will then match those of $x$.

For one repeating digit, use $10x$. For two digits, use $100x$. For three, use $1000x$. Subtracting the original equation removes the identical tails and leaves an equation with integers.

One repeating digit

Let $x=0.\bar{7}=0.777\dots$. Shift one place:

Subtract the original value:

so $9x=7$. Divide by $9$ to get $x=7\mathrm{/}9$. This is exact: the tails match all the way through, not just for the digits written on the page.

A longer block

Let $x=0.\overline{36}$. Its repeating block has two digits, so shift two places:

Subtract $x=0.3636\dots$ to obtain $99x=36$. Therefore

Check by division: $4÷11=0.363636\dots$.

A shift of one place would give $10x=3.6363\dots$. Its tail starts with $63$, while the original starts with $36$. Subtraction is still allowed, but it does not remove the repeating tail. Match the block before you subtract.

Use the shortcut with care

When the repeat begins immediately after the decimal point, write the block over the same number of nines:

The denominator is $99$ for two digits because $100x-x=99x$. It is not $100$: $36\mathrm{/}100$ would end at $0.36$. Simplify only after setting up the fraction correctly.

Include the whole-number part

For $x=2.\overline{45}$, shifting two places gives $100x=245.4545\dots$. Subtract the entire original number:

Thus $x=243\mathrm{/}99=27\mathrm{/}11$. Using $245\mathrm{/}99$ would miss the subtraction of the original whole-number part.

Why repeating nines reach the next value

Set $x=0.\bar{9}$. Then $10x=9.999\dots$, so $9x=9$ and $x=1$. Repeating nines are exactly another way to write $1$. A finite string such as $0.999$ is smaller, but the endless string has no final digit leaving a small gap.

The same equality appears later in a decimal: $2.4\bar{9}=2.5$. You can check a converted fraction by division, keeping enough digits to see the full repeating block.

The bar in a repeating decimal tells you both what repeats and where the repeat starts. If some decimal digits occur first, two shifted equations let you remove the tail.

Read the starting point

Compare $0.1\bar{6}$ and $0.\overline{16}$. The first means $0.166666\dots$; the second means $0.161616\dots$. They agree in two places, then differ. Write several digits before choosing a multiplier.

For $0.2\overline{45}=0.2454545\dots$, the $2$ occurs before the repeating block. Writing $245\mathrm{/}999$ would make all three digits repeat. The block-over-nines shortcut cannot be applied to that whole decimal.

Move the beginning, then one block

Call the number $x$. First shift enough places to put every non-repeating decimal digit before the point. Next shift one full repeating block farther. The two new values now have identical decimal tails.

For $x=0.1\bar{6}$, there is one non-repeating digit and one repeating digit:

Subtract the smaller multiple from the larger:

so $90x=15$ and $x=15\mathrm{/}90=1\mathrm{/}6$. The coefficient is $90$ because you subtracted $10x$, not $x$. Dividing $1$ by $6$ checks the result.

A two-digit repeating block

Let $x=0.2\overline{45}$. Moving the initial $2$ across gives

Then move a complete two-digit block farther:

The tails are now the same, so $1000x-10x=243$. This gives $990x=243$ and

Choosing $100x$ and $10x$ would not work for cancellation. Their tails begin $5454\dots$ and $4545\dots$. One extra place moves only half the block; two extra places move all of it.

Keep whole-number parts in the subtraction

For $x=3.1\bar{2}$, use the same shifts:

Subtract to get $90x=281$, hence $x=281\mathrm{/}90$. This is $3\frac{11}{90}$, which is sensible because the original number is just above $3.1$. The whole-number part does not change how many places the repeat needs, but it changes the integers you subtract.

See why a delayed block changes the denominator

The decimal $0.1\bar{6}$ can also be split:

The second part is one tenth of $0.\bar{6}=6\mathrm{/}9$, so it equals $6\mathrm{/}90$. The first part is $1\mathrm{/}10=9\mathrm{/}90$. Their sum is $15\mathrm{/}90=1\mathrm{/}6$, agreeing with subtraction. The initial place accounts for the extra factor of ten.

When checking your work, compare both the first few digits and the location of the repeat. If subtraction leaves an endless decimal, rewrite the shifted values and line up their repeating blocks before continuing.

Fractions describe many numbers, but they do not describe every point on the number line. An irrational number is a real number that cannot be written as a fraction of two integers with a nonzero denominator.

Endless does not always mean irrational

An irrational number’s decimal continues forever without eventually repeating a fixed block. For example, $\sqrt{2}=1.41421356\dots$. A repeating decimal such as $0.333\dots =1\mathrm{/}3$ also continues forever, but it is rational.

A pattern can exist without a repeating block. In $0.101001000100001\dots$, each gap has one more zero than the previous gap. The growing gaps prevent any fixed block from repeating forever, so this number is irrational.

Check a root’s value

For a whole number, its square root is a whole number if it is a perfect square and is irrational otherwise. Thus $\sqrt{36}=6$ is rational, while $\sqrt{8}$ is irrational. The similar cube-root test uses perfect cubes: $\sqrt[3]{27}=3$ is rational, but $\sqrt[3]{2}$ is irrational.

For a positive fraction in simplest form, its square root is rational exactly when both numerator and denominator are perfect squares. For example, $\sqrt{4\mathrm{/}9}=2\mathrm{/}3$, but $\sqrt{1\mathrm{/}2}$ is irrational. Turn decimals into simplified fractions before using this test: $0.04=1\mathrm{/}25$ gives a rational root, while $0.4=2\mathrm{/}5$ gives an irrational root.

A root sign alone does not settle the question. Nor does a fraction bar: $\sqrt{2}\mathrm{/}2$ is irrational because dividing an irrational number by a nonzero rational number leaves it irrational.

Pi is an exact irrational number

The number $\pi$ is a circle’s circumference divided by its diameter. Its decimal begins $3.14159265\dots$ and never ends or repeats. Both $3.14$ and $22\mathrm{/}7$ are rational approximations. In particular, $22\mathrm{/}7=3.\overline{142857}$ is slightly larger than $\pi$, by about $0.0013$.

An irrational number is still a precise value. A unit square has a diagonal of exactly $\sqrt{2}$ units, since the Pythagorean theorem gives diagonal squared $=1^2+1^2=2$. The decimal cannot be finished, but the length is fixed.

Why no fraction equals the square root of two

Suppose $\sqrt{2}=a\mathrm{/}b$ for a fraction already in simplest form. Squaring both sides gives

The right side is even, so $a^2$ is even. An odd number squared is odd, which means $a$ must be even. Write $a=2k$ and substitute:

Now $b^2$, and therefore $b$, must also be even. That would make both parts of $a\mathrm{/}b$ divisible by $2$. But the fraction was in simplest form. This contradiction shows that the proposed fraction cannot exist, so $\sqrt{2}$ is irrational.

Use exact information

A calculator displays only finitely many digits. An irrational value and a rational value close to it can produce the same rounded display. Decide from the exact expression or the stated decimal rule. The display $1.41421356$ by itself names a terminating rational decimal; as an approximation to $\sqrt{2}$, it leaves out the endless non-repeating tail.

A calculation can change the kind of number you have. Simplify any exact parts first, then use the relationship between the numbers to classify the result.

Two rational inputs

Adding, subtracting, multiplying, or dividing rational numbers gives another rational number, provided the divisor is not zero. For example,

Both results can be written as fractions of integers. This is why a calculation using only exact fractions and terminating or repeating decimals stays rational.

Add a rational and an irrational

A rational number added to an irrational number gives an irrational result. Consider $3+\sqrt{2}$. Suppose the sum were rational, and call it $r$. Subtracting $3$ would give $\sqrt{2}=r-3$, a difference of two rational numbers. That would make $\sqrt{2}$ rational, which is impossible.

The same reasoning covers subtraction. Expressions such as $5-\pi$ and $\pi -1$ are irrational. Moving an irrational number by a rational amount cannot make it land on a rational value.

Multiply by a rational number

A nonzero rational number multiplied by an irrational number gives an irrational result. Thus $2\pi$, $-3\sqrt{5}$, and $\sqrt{7}\mathrm{/}4$ are irrational.

Here is the reason for the condition. If $2\pi$ were rational, dividing by $2$ would make $\pi$ rational. But this argument cannot divide by zero. In fact, $0\times \sqrt{2}=0$ is rational. Always check whether the rational factor is zero.

Two irrational inputs need a closer look

There is no single answer for the sum or product of two irrational numbers. Cancellation can give a rational value:

Multiplying a root by itself can also do this:

Dividing any nonzero number by itself gives $1$, so $\pi \mathrm{/}\pi =1$. Yet $\pi +\pi =2\pi$ is irrational, since it is a nonzero rational multiple of $\pi$. The fact that both inputs are irrational does not choose the result for you.

Classify the simplified value

Compare these calculations:

For $\sqrt{3}\mathrm{/}\sqrt{3}$, the result is $1$, so it is rational. Treating every expression with a radical as irrational would miss this simplification.

Keep exact values separate from approximations

Replacing $\pi$ with $3.14$ creates an approximate calculation. It does not change the type of $\pi$ itself. The exact expression $2\pi$ is irrational, while the approximate value $6.28$ is a terminating rational decimal. Classify the number the question actually names, and use $\approx$ when showing a rounded replacement.

A root need not have a whole-number value to be located accurately. Known squares and cubes give a starting interval; trial decimals narrow it.

Trap a square root

To estimate $\sqrt{50}$, use $7^2=49$ and $8^2=64$. Since $49<50<64$,

Square roots increase as their nonnegative inputs increase. The number under the radical lies between the squares; the root lies between the numbers that were squared.

To decide which whole number is nearer, test the halfway root, $7.5$. Its square is $56.25$. Since $50<56.25$, $\sqrt{50}$ is closer to $7$. Comparing distances from $50$ to $49$ and $64$ is not the correct halfway test. For instance, $56.4$ is nearer $49$, yet its root is about $7.51$, nearer $8$.

A quick starting estimate takes the fraction of the gap: $7+(50-49)\mathrm{/}(64-49)=7+1\mathrm{/}15\approx 7.07$. Between consecutive perfect squares this estimate lies a little below the true root. Squaring a trial value gives a firmer check.

Narrow the interval with decimals

Try $7.1$. Since ${7.1}^2=50.41>50$, the root is below $7.1$. Try hundredths next:

Therefore $7.07<\sqrt{50}<7.08$. This locates the root between adjacent hundredths, but does not yet choose the nearest one. Test their midpoint:

The root is below $7.075$, so it rounds to $7.07$ to two decimal places. A lower bound and a rounded value are different claims; the midpoint test connects them.

Work with small and large values

For $\sqrt{300}$, ${17}^2=289$ and ${18}^2=324$ give $17<\sqrt{300}<18$. For $\sqrt{0.5}$, use ${0.7}^2=0.49$ and ${0.8}^2=0.64$. Thus $0.7<\sqrt{0.5}<0.8$, and further checks give about $0.707$. Between zero and one, taking a square root increases the value.

The same refinement can continue for as many places as needed. For $\sqrt{2}$, trial squares give $1.4<\sqrt{2}<1.5$, then $1.41<\sqrt{2}<1.42$. An irrational root never reaches a finished decimal expansion.

Use cubes for cube roots

Since $3^3=27<40<64=4^3$, $3<\sqrt[3]{40}<4$. Now check ${3.4}^3=39.304$ and ${3.5}^3=42.875$. This gives $3.4<\sqrt[3]{40}<3.5$; more trials give about $3.42$.

The negative cube root has the opposite sign, so $-4<\sqrt[3]{-40}<-3$. Write the more negative number on the left.

Estimate a whole expression

Estimate roots before combining their values. For example, $2\sqrt{10}\approx 2(3.16)=6.32$. Also, $\sqrt{3}+\sqrt{5}\approx 1.73+2.24=3.97$. These are rough estimates because their inputs were rounded.

For ${\pi }^2$, $3.14<\pi <3.15$ gives $9.8596<{\pi }^2<9.9225$. A calculator refines this to about $9.87$. To check any proposed square-root estimate, square it and compare with the original input.

The same number line holds fractions, roots, and repeating decimals. To compare them, put their values into a form where the difference is clear.

Compare decimal digits

Write enough digits to find the first place where the values differ. Equal rounded values only show that your approximation is too rough to decide.

For example, consider $3.\bar{1}$, $3.14$, $\pi$, $22\mathrm{/}7$, and $\sqrt{10}$. Their decimal forms begin:

Read from the left until digits differ. This gives

Rounding all five to one decimal place would hide several of these differences. Retain more places before deciding the order.

Compare positive values by squaring

Squaring keeps the order of nonnegative numbers. To compare $\sqrt{40}$ and $6.3$, compare their squares:

Since $39.69<40$, $6.3<\sqrt{40}$. No decimal expansion of the root is needed.

This also works with positive fractions. Compare $5\mathrm{/}2$ and $\sqrt{6}$:

Therefore $5\mathrm{/}2>\sqrt{6}$. Since ${\pi }^2\approx 9.87<10$, the same idea shows $\pi <\sqrt{10}$.

The input values must both be nonnegative for this method to keep their order. For $-3$ and $2$, squaring gives $9$ and $4$, reversing their original order. Check signs before squaring.

Expand a repeating block carefully

A bar means the digits continue forever. Thus $0.\bar{3}=0.3333\dots$ is greater than $0.33=0.3300\dots$. They first differ in the third decimal place.

Compare $0.\overline{45}$ and $0.4\bar{5}$ by expanding them:

The first two decimal digits match. The third digits are $4$ and $5$, so $0.\overline{45}<0.4\bar{5}$. Writing only two places would miss the difference caused by the bar.

Reverse the order for opposites

On the negative side, values farther left are smaller. Since $\sqrt{5}\approx 2.236>2.2$, their opposites satisfy

A larger distance from zero means a smaller negative number. When ordering a mixed list, separate negatives, zero, and positives first. Within the negatives, compare their distances from zero and reverse that order. This keeps the minus signs attached while you compare the sizes.

An exact expression and a rounded decimal serve different purposes. Keep the exact value while calculating, then choose the precision the answer needs.

Exact does not mean a finished decimal

The expressions $\sqrt{7}$, $\sqrt[3]{20}$, $6\pi$, $2+\sqrt{3}$, and $1\mathrm{/}3$ are exact. Each names one precise number. A rounded version such as $\sqrt{7}\approx 2.65$ is an approximation.

Use $=$ for equality and $\approx$ for approximation. Writing $\sqrt{7}=2.65$ would claim that ${2.65}^2=7$, but its square is $7.0225$.

An instruction such as “give an exact answer” or “give your answer in terms of pi” calls for the expression with $\pi$ or the root left in it. A request for decimal places or significant figures calls for a rounded value.

A circle with a known radius

For a circle of radius $5$ cm, the circumference is

The area is

These are exact values. To one decimal place, they are about $31.4$ cm and $78.5{\text{ cm}}^2$. Keep the units: circumference is a length; area uses square units.

Different replacements for $\pi$ give different results. For the area, $25\times 3.14=78.5$ and $25\times 22\mathrm{/}7\approx 78.57$, while $25\pi \approx 78.5398$. The calculator’s pi key keeps more accuracy than those short replacements.

Carry the exact value through several steps

A square of area $30{\text{ m}}^2$ has side $\sqrt{30}$ m. Its perimeter is therefore

If you round the side to $5.48$ m first, multiplying by four gives $21.92$ m. If you round it to $5.5$ m first, you get $22$ m. The early rounding changes the final answer. Keep $\sqrt{30}$ in the calculation or retain the calculator’s full stored value until the last step.

A cube of volume $100{\text{ cm}}^3$ has exact edge $\sqrt[3]{100}$ cm, about $4.64$ cm. Again, the volume’s cubic units become units of length after taking the cube root.

Combine only matching terms

The expression $3\pi -\pi$ counts three copies of $\pi$ minus one copy, so it equals $2\pi$. Similarly, $2\times 3\pi =6\pi$. But $4\pi +1$ has unlike terms and cannot become $5\pi$. The extra $1$ is not another copy of $\pi$.

A root times itself can give a whole number exactly: $\sqrt{5}\times \sqrt{5}=5$. There is no need to replace each root by a decimal before multiplying.

State the precision

Decimal places count digits after the point. Significant figures count from the first nonzero digit. For $\sqrt{2}$, three decimal places give $1.414$, while three significant figures give $1.41$. Look at the next digit to round, and include the requested number of places, even when a final zero is needed to show that precision.

Number sets let you describe a number’s properties. They overlap in a clear order, so one number can belong to several sets at once.

Start with counting and whole numbers

The natural numbers, written $\mathbb{N}$, are $1,2,3,\dots$ under the convention used here. The whole numbers add zero: $0,1,2,3,\dots$. Some courses also include zero in the natural numbers; check your course’s convention when using that name.

The integers, $\mathbb{Z}$, include whole numbers and their negative opposites:

A value such as $-3$ is an integer but is not a whole number. A value such as $2.5$ lies between integers and is not an integer.

Fractions and the two kinds of real numbers

The rational numbers, $\mathbb{Q}$, can be written as fractions of integers with a nonzero denominator. This includes integers, since $-3=-3\mathrm{/}1$, and decimals that terminate or repeat.

The irrational numbers are real numbers outside $\mathbb{Q}$. Examples include $\sqrt{11}$ and $-\pi$. A common notation is $ℝ\setminus \mathbb{Q}$, read as “the real numbers without the rational numbers.” There is no single universal letter for this set; some books use ${\mathbb{Q}}^{\mathrm{\prime }}$.

The real numbers, $ℝ$, combine all rational and irrational numbers. Each real number belongs to exactly one of those two groups.

Follow the nesting

Imagine a large box for the real numbers. Inside it is a box for rational numbers; inside that is a box for integers. Whole numbers fit inside the integers, and natural numbers fit inside the whole numbers. Irrational numbers occupy the part of the real-number box outside the rational-number box.

The main symbols show

Here $\subset$ means that the set on the left fits inside the set on the right. Whole numbers also lie between natural numbers and integers, even though this chain does not give them a letter.

Name every set that applies

Simplify a number before deciding where it fits:

Calling $3$ an integer is correct but does not name every set it belongs to. Calling $-2.5$ an integer is incorrect: being negative does not decide whether a number is an integer, and its fractional part matters.

Read membership symbols

The symbol $\in$ means “is an element of,” or “belongs to.” Its crossed-out form $\mathrm{\notin }$ means “does not belong to.” Thus $-3\in \mathbb{Z}$, $1\mathrm{/}2\mathrm{\notin }\mathbb{Z}$, $\sqrt{2}\mathrm{\notin }\mathbb{Q}$, and $\sqrt{2}\in ℝ$.

A negative number under a square-root sign needs special care. Since no real square is negative, $\sqrt{-4}$ is not real. It belongs to none of these real-number sets. This differs from $-\sqrt{4}=-2$, which is an integer, rational, and real.

Every real number has exactly one point on the number line. The spaces between marked ticks contain numbers too, including both rational and irrational values.

Locate a root approximately

Use nearby squares or a decimal estimate. Since $1^2<2<2^2$, $\sqrt{2}$ lies between $1$ and $2$. Its approximation $1.41$ puts it a little left of $1.5$. Likewise, $\sqrt{8}\approx 2.83$ lies near $3$, while $-\sqrt{3}\approx -1.73$ lies closer to $-2$ than to $-1$.

The mark belongs at the value of the root. Placing $\sqrt{8}$ at $8$ would locate the input, not the result. Halving $8$ would give $4$, whose square is $16$, so that is not the right location either.

Construct an exact irrational point

Draw a square of side $1$ above the segment from $0$ to $1$. Its diagonal from $0$ has length $\sqrt{2}$ by the Pythagorean theorem. Open a compass to that diagonal’s length, put its point at $0$, and draw an arc to the positive side of the number line. The intersection marks $\sqrt{2}$ exactly.

A pencil drawing has practical limits, but the geometric construction specifies the exact point. You do not need to finish an irrational decimal to define its position.

There is no next rational number

Between two different rational numbers, their mean is another rational number. Between $1\mathrm{/}3$ and $1\mathrm{/}2$, use

You could take the mean again between $1\mathrm{/}3$ and $5\mathrm{/}12$. Repeating this keeps finding new fractions. This property is called density: rational numbers occur between any two different real numbers, and infinitely many occur in every such interval.

Even so, not every point is rational. No fraction lands exactly at $\sqrt{2}$. The irrational numbers supply the real points that rational numbers miss. There are infinitely many irrational numbers between any two different real numbers as well.

To find examples, check candidate values. Between $3$ and $4$, $\sqrt{10}$ is irrational because $9<10<16$, and $\pi$ also lies there. Between $\sqrt{2}$ and $\sqrt{3}$, $1.5$ is rational: $2<{1.5}^2=2.25<3$ proves it lies between the positive roots.

Shade a set of real values

The inequality $-1\le x<3$ includes every real value from $-1$ up to $3$, including $-1$ but excluding $3$. Draw a filled circle at $-1$, an open circle at $3$, and shade the stretch between them. Interval notation writes this as $[-1,3)$.

The shaded stretch includes $\sqrt{2}$, $\sqrt{5}$, fractions, and all other real points inside it. It is not just the integers shown by ticks.

Check which set the variable comes from

If $x$ must be an integer, the same inequality has only the solutions $-1,0,1,2$. Show separate dots instead of a continuous shaded stretch. The inequality sets the bounds; the allowed number set tells you which points inside the bounds count.

For $x>\sqrt{2}$ with real $x$, draw an open circle at about $1.41$ and shade to the right with an arrow. The endpoint is excluded because the sign is strictly greater than. Finally, $1.\bar{9}$ is exactly $2$, so it shares the point labeled $2$ instead of sitting just to its left.