√2 as an irrational number

Draw a unit square and measure its diagonal — you get a real, physical length, $\sqrt2$ . Try to write that length down exactly, though, and something strange happens: no whole number works, no terminating decimal works, and no fraction $\dfrac{p}{q}$ ever lands on it exactly. Every decimal you compute — $1.4$ , $1.41$ , $1.414$ , $1.4142\ldots$ — only gets closer to $\sqrt2$ without ever reaching it, and the digits never fall into a repeating pattern. Numbers with this property are called irrational. This article shows, using nothing more than squaring and comparing, how we pin $\sqrt2$ down between ever-narrower decimal bounds, why that process never terminates, and how to use a rounded value safely in calculations without mistaking it for the exact one.

The core insight

$\sqrt2$ is irrational because squaring shows it always lies strictly between two nearby decimals, however finely you narrow the gap, so its exact decimal expansion never terminates and never repeats.

The core idea

Step 1 — locate √2 between whole numbers. From the isosceles right triangle with both legs equal to $1$ , the Baudhāyana–Pythagoras relation gives $c^2 = 1^2+1^2 = 2$ , so $c=\sqrt2$ . Since $1^2=1$ and $2^2=4$ , and $1<2<4$ , the value $\sqrt2$ must sit strictly between $1$ and $2$ — it cannot be a whole number.

Step 2 — squeeze with decimals. The same squaring test works at any level of precision:

1.4^2=1.96,\qquad 1.41^2=1.9881,\qquad 1.414^2=1.999396,\qquad 1.4142^2=1.99996164

where: each number on the left is a trial decimal for $\sqrt2$ , and each square on the right is compared against the target value $2$ . As more digits are added, the square creeps closer to $2$ from below but never actually equals $2$ — there is always a tiny gap left over.

Step 3 — name what this means. However many digits you add, this squeeze never finishes: the decimal expansion of $\sqrt2$ keeps going forever, and the digits never settle into a repeating block. A number whose decimal form is non-terminating and non-repeating in this way is an irrational number — it cannot be written exactly as $\dfrac{p}{q}$ for whole numbers $p,q$ (with $q\neq0$ ). Compare this with rational numbers such as $\dfrac13=0.333\ldots$ (repeating) or $\dfrac12=0.5$ (terminating): both settle into a fixed, predictable pattern, while $\sqrt2$ never does.

Step 4 — use it, don't pretend it's exact. For everyday work we quote a rounded value, most often $\sqrt2\approx1.414$ , always remembering it is an approximation. Ancient computers of this value, working within the Baudhāyana tradition, already reached a remarkably close fraction, $1+\tfrac13+\tfrac1{3\cdot4}-\tfrac1{3\cdot4\cdot34}\approx1.4142$ , thousands of years before "irrational" was a defined term — proof that the puzzle of $\sqrt2$ 's exactness is not a modern one. This trapping habit generalises beyond $\sqrt2$ : the same squaring-and-comparing test can bracket $\sqrt3$ , $\sqrt5$ , or any square root of a non-perfect square, always confirming the same pattern — a shrinking gap that never closes to a single terminating decimal.

Figure

A number line showing root two trapped in a shrinking gap between decimal bounds that never closes

Derivation ATrapping √2 between narrowing decimal bounds

The goal is to show, using nothing beyond squaring and comparing decimals, that we can locate $\sqrt2$ inside a gap between two decimal numbers as small as we like, while also seeing why this narrowing process can never actually finish and hand us an exact terminating value.

Step 1. Start from the whole-number bracket. Because $1^2=1<2$ and $2^2=4>2$ , and squaring is increasing for positive numbers, $\sqrt2$ lies strictly between $1$ and $2$ : a gap of width $1$ .

Step 2. Narrow to one decimal place by testing values between $1$ and $2$ . Squaring shows $1.4^2=1.96<2$ and $1.5^2=2.25>2$ , so $\sqrt2$ lies between $1.4$ and $1.5$ : a gap of width $0.1$ .

Step 3. Narrow again to two decimal places inside that smaller gap. Testing further, $1.41^2=1.9881<2$ and $1.42^2=2.0164>2$ , so $\sqrt2$ lies between $1.41$ and $1.42$ : a gap of width $0.01$ — ten times narrower than before.

Step 4. Repeat once more. $1.414^2=1.999396<2$ and $1.415^2=2.002225>2$ , so $\sqrt2$ lies between $1.414$ and $1.415$ : a gap of width $0.001$ . Each repetition of this squaring-and-comparing step shrinks the gap by a factor of ten, and nothing in the process ever produces a decimal whose square lands on exactly $2$ — so the narrowing continues without end, which is exactly why $\sqrt2$ 's decimal expansion never terminates.

Worked example

Problem: A student claims $\sqrt2=\dfrac{99}{70}$ exactly. Use squaring to check how close this fraction really is to $2$ , and use the result to explain, in the spirit of the trapping method above, why $\sqrt2$ still counts as irrational.

Step 1. Square the claimed fraction: $\left(\dfrac{99}{70}\right)^2=\dfrac{99^2}{70^2}=\dfrac{9801}{4900}$ .

Step 2. Convert to a decimal for comparison: $\dfrac{9801}{4900}\approx2.0002$ .

Step 3. Compare with $2$ : the square is $2.0002$ , extremely close to $2$ but not exactly $2$ — so $\dfrac{99}{70}$ actually gives an upper bound, with $\sqrt2$ sitting just below $\dfrac{99}{70}\approx1.41429$ .

Step 4. Conclude, as in the derivation: $\dfrac{99}{70}$ is an excellent approximation to $\sqrt2$ , accurate to about four decimal places, but the trapping process above shows no fraction or terminating decimal, however cleverly chosen, can ever be exact. $\sqrt2\ne\dfrac{99}{70}$ ; the fraction is only a very close approximation, and $\sqrt2$ remains irrational.

Competitive stretch

Push the trapping method one step further to squeeze $\sqrt2$ between $1.4142$ and $1.4143$ : $1.4142^2=1.99996164$ and $1.4143^2=2.00024449$ , so since $1.99996164<2<2.00024449$ , $\sqrt2$ lies between $1.4142$ and $1.4143$ — a gap of just $0.0001$ , and still shrinking with every extra digit. Notice the pattern across every stage of the squeeze: whichever decimal we test, its square lands strictly above or strictly below $2$ , never exactly on it, and there is no way to predict in advance where the next digit will fall — which is exactly the informal signature of an irrational number at this level.

Common traps

Treating a very accurate decimal or fraction as "being" √2: values like $1.414$ or $\dfrac{99}{70}$ are approximations, however good — the true value never terminates or repeats.
Confusing "cannot be written exactly" with "does not really exist": $\sqrt2$ is a real, drawable length (the unit square's diagonal); irrational refers only to how its value is expressed numerically.
Assuming every square root is irrational: $\sqrt4=2$ and $\sqrt9=3$ are whole numbers (rational); it is square roots of non-perfect squares like $2,3,5,\ldots$ that turn out irrational.
Stopping the squeeze too early and calling it exact: always keep the exact form $\sqrt2$ in working, and substitute a rounded decimal only for the final numeric answer.

Ask the tutor: Want to practise trapping $\sqrt3$ or $\sqrt5$ between closer and closer decimals using this same squeeze method?

If we can trap √2 as tightly as we like, why can't we ever reach the exact value?

Because each trapping step only narrows the gap by roughly a factor of ten — it never closes it to zero. However many digits you compute, there is always a next digit still undetermined, and no digit pattern ever settles into a repeating block. That endless, non-repeating narrowing is precisely what "irrational" means: you can get arbitrarily close to $\sqrt2$ using decimals or fractions, but you can never write down a finite or repeating decimal that equals it exactly.

Is √2 = 1.414 close enough to use in a real calculation?

Yes, for most practical and exam purposes $\sqrt2\approx1.414$ (or $1.41$ ) is perfectly usable — the error is smaller than one part in a thousand. The key habit is to keep the exact form $\sqrt2$ (or $a\sqrt2$ ) through your working and only substitute the rounded decimal at the very last step, when a numeric answer is actually required. That way you never accidentally treat the approximation as if it were exact earlier in a proof or a longer calculation.

How is √2 different from a repeating decimal like 1/3?

$\dfrac13=0.333\ldots$ does go on forever, but its digits form a completely predictable, repeating block — the "3" simply repeats, so it can be written exactly as the fraction $\dfrac13$ . $\sqrt2$ 's decimal digits, by contrast, never fall into any repeating block at all; they simply continue unpredictably. That distinction — repeating versus non-repeating — is exactly the line between rational numbers (always expressible as $\dfrac{p}{q}$ ) and irrational ones like $\sqrt2$ . Every terminating decimal, like $0.5$ , is also rational, since it can be written as a fraction such as $\dfrac12$ — so "rational" really means any number expressible exactly as a ratio of whole numbers, however that decimal behaves.

√2 as an irrational number

The core idea

Worked example

Competitive stretch

Common traps

Before this

Where this leads