The Diagonal of a Square Has No Exact Decimal

Because it involves $\sqrt{2}$, and $\sqrt{2}$ has a decimal that never ends and never repeats. A square of side $1$ has a diagonal of exactly $\sqrt{2}$ — a length you can draw perfectly but never write down in full.

That is a genuinely strange fact, and this chapter earns it. This page covers everything in the CBSE Class 8 Mathematics chapter's first part: the theorem and missing sides, the diagonal of a square, the geometric justification, and why $\sqrt{2}$ is irrational.

Because a diagonal splits a square into two isosceles right-angled triangles, whose two legs are equal.

Take a square of side $a$. Its diagonal is the hypotenuse of a right triangle with both legs equal to $a$:

$$c^2=a^2+a^2=2a^2\Rightarrow c=a\sqrt{2}$$

So the diagonal of any square is its side multiplied by $\sqrt{2}$, whatever the size.

Worked examples, taking $\sqrt{2}\approx 1.414$:

- Side $1$ cm gives a diagonal of $\sqrt{2}\approx 1.414$ cm
- Side $5$ cm gives $5\sqrt{2}\approx 7.07$ cm
- Side $10$ cm gives $10\sqrt{2}\approx 14.14$ cm

Running it backwards, a square with a diagonal of $8$ cm has side

$$a=\frac{8}{\sqrt{2}}\approx \frac{8}{1.414}\approx 5.66\text{ cm}$$

The angles are fixed too. Since the two legs are equal, the two acute angles are equal, and being a right triangle they share the remaining $90°$:

$$\frac{180-90}{2}=45°\text{ each}$$

A folded square sheet of paper, creased corner to corner, produces exactly this triangle.

The result is worth recognising instantly, because it appears throughout later geometry. The ratio of a square's diagonal to its side is always $\sqrt{2}$ — it does not depend on the square's size, which is why $\sqrt{2}$ turns up so often.

By showing that the squares drawn on the two shorter sides can be cut up and reassembled into the square on the hypotenuse.

The doubling argument, for an isosceles right triangle. Take a square of side $1$, so its area is $1$. Draw the square on its diagonal.

The original square's two diagonals cut it into four equal triangles. The square built on one diagonal is made of eight such triangles — so its area is exactly double:

$$\text{area of square on diagonal}=2\times 1=2$$

Since that area is 2, its side must be $\sqrt{2}$ — which is the theorem for this triangle, since $1^2+1^2=2$.

The halving argument. Run the same picture backwards. A square of area $2$ cut along both its diagonals gives four triangles, and two of them reassemble into a square of area $1$ — so the square on the diagonal is twice the original, once again.

The combining argument, in general. For a right triangle with legs $a$ and $b$, draw squares of area $a^2$ and $b^2$ on the two legs. Those two squares can be cut and fitted together to fill exactly the square of area $c^2$ on the hypotenuse, with no gap and nothing left over. That is what $a^2+b^2=c^2$ says in pictures.

Checking with the $3,4,5$ triangle: squares of $9$ and $16$ unit tiles together make $25$ tiles, which is precisely the square on the side of length 5.

The point of the geometric version is what the algebra hides. The theorem is a statement about areas, not about lengths — the squares on the two legs literally hold as much surface as the square on the hypotenuse, which is why it can be demonstrated with scissors and paper rather than calculation.

Because squaring those two decimals brackets the number 2 from both sides.

$$1.414^2=1.999396,1.415^2=2.002225$$

Since $1.999396<2<2.002225$, it follows that

$$1.414<\sqrt{2}<1.415$$

Narrowing further: $1.4142^2=1.99996164$ and $1.4143^2=2.00024449$, so $\sqrt{2}$ lies between $1.4142$ and $1.4143$. The squeezing can be continued for ever, and it never closes onto an exact value.

Why it cannot be a fraction. Suppose $\sqrt{2}$ could be written as $\frac{p}{q}$ in lowest terms, with $p$ and $q$ positive integers having no common factor. Then

$$2=\frac{p^2}{q^2}\Rightarrow p^2=2q^2$$

So $p^2$ is even, which forces $p$ to be even, since the square of an odd number is odd. Write $p=2k$:

$$(2k{)}^2=2q^2\Rightarrow 4k^2=2q^2\Rightarrow q^2=2k^2$$

Now $q^2$ is even too, so $q$ is also even. But then $p$ and $q$ share a factor of 2 — contradicting "lowest terms".

No such fraction can exist, so $\sqrt{2}$ is irrational. And since every terminating or recurring decimal can be written as a fraction, $\sqrt{2}$ can be neither.

The distinction worth holding is between drawing and writing. The diagonal of a unit square is a perfectly definite, finite length you can construct with a ruler — it simply has no exact decimal or fractional name, which is a limitation of our number notation rather than of the length.