Squares and Square Roots | IT

Question 13

Will we ever get a number with a terminating decimal representation whose square is 2?

If there is such a terminating decimal starting with 1.414... whose square is 2, then it must have a non-zero last digit. If this is the case, then the decimal representation of its square will also have a non-zero last digit after the decimal point. For example, if $\sqrt{2}$ is of the form 1.414...4, then its square will be of the form—

Solution

We will show that no terminating decimal can have a square of 2.

Step 1 — Understanding terminating decimals

A terminating decimal has a finite number of digits after the decimal point. We can write it as a fraction $\frac{A}{10^k}$ . Here $A$ is an integer. $k$ is the number of decimal places. If $x$ is an integer, $k=0$ . If $x$ is not an integer, $k \ge 1$ . We can always choose $A$ so its last digit is not zero. For example, $1.50$ is the same as $1.5$ . So, for a non-integer terminating decimal $x$ : $x = \frac{A}{10^k}$ . Here $k \ge 1$ . $A$ is an integer not ending in zero.

Step 2 — Squaring a terminating decimal

Let us find the square of $x$ . $x^2 = \left(\frac{A}{10^k}\right)^2$ $= \frac{A^2}{10^{2k}}$ The last digit of $A$ is not zero. So, the last digit of $A^2$ is also not zero. Let us check the last digits of squares: $1^2$ ends in 1. $2^2$ ends in 4. $3^2$ ends in 9. $4^2$ ends in 6. $5^2$ ends in 5. $6^2$ ends in 6. $7^2$ ends in 9. $8^2$ ends in 4. $9^2$ ends in 1. So, $A^2$ ends in $1, 4, 5, 6,$ or $9$ . Since $k \ge 1$ , $2k \ge 2$ . $x^2 = \frac{A^2}{10^{2k}}$ will have $2k$ decimal places. The last digit of $x^2$ after the decimal point will be non-zero. This is because $A^2$ does not end in zero. For example, let $x = 1.2$ . $x^2 = 1.44$ The last digit after the decimal point is 4. Let $x = 1.4$ . $x^2 = 1.96$ The last digit after the decimal point is 6.

Step 3 — Completing the sentence

The question asks to complete a sentence based on a hypothetical scenario. "If $\sqrt{2}$ is of the form $1.414...4$ , then its square will be of the form—" If a terminating decimal ends in 4, its square will end in 6. We saw this with $1.4^2 = 1.96$ . So, the square would end in 6. The completed sentence is: "If $\sqrt{2}$ is of the form $1.414...4$ , then its square will be of the form . _ ... _ 6."

Diagram 1

Step 4 — Checking for contradiction

We are looking for a number $x$ such that $x^2 = 2$ . If $x$ were a terminating decimal, then $x^2$ must be 2. The number 2 is an integer. Its decimal representation is $2.000...$ . All digits after the decimal point are zero. So, the last digit of $x^2$ after the decimal point would be 0. However, we know that $1^2 = 1$ and $2^2 = 4$ . So, $x$ cannot be an integer. From Step 2, if $x$ is a non-integer terminating decimal, its square $x^2$ must have a non-zero last digit after the decimal point. This is a contradiction. The last digit of $x^2$ after the decimal point cannot be both zero and non-zero. Therefore, our initial assumption is false.

Answer

No, we will never get a number with a terminating decimal representation whose square is 2. The completed sentence is: "If $\sqrt{2}$ is of the form $1.414...4$ , then its square will be of the form . _ ... _ 6."

More questions in IT

How can one construct a square having double the area of a given square?

Why does the new dotted square have double the area of the original square?

In many of the constructions in the Śulba-Sūtra, it is desirable to construct, where needed, what Baudhāyana calls ‘east-west’ and ‘north-south’ lines, i.e., horizontal and vertical lines that are perpendicular to each other. Can you draw some horizontal and vertical lines to see why the new square has double the area of the original square? You could draw some horizontal and vertical lines as shown on the right.

Why should the extension of the vertical and horizontal sides of the original square pass through the vertices of the dotted square?

[Hint: From the diagonal property of a square, the line that bisects an angle passes through the opposite vertex. Argue why the vertical and horizontal sides of the original square bisect the two angles of the dotted square.**]

Context: So, the new square has double the area of the original square, because the original square is made up of two small triangles, while the new square is made up of four small triangles.

Q. Moreover, all these small triangles are congruent to each other. Can you explain why?

Now suppose we are given a square, and we want to construct a square whose area is half that of the original square. How would you do it?

Why is the smaller inside square half the area of the larger square?

Again, adding some east-west and north-south lines can explain it:

Why is PQRS a square? Why is its area half that of the original paper?

Explain by connecting QS and PR, finding the different angles formed, and then using tringle congruence.

Find the hypotenuse of this isosceles right triangle.

What is the value of $\sqrt{2}$ ?

Q10

Is $\sqrt{2}$ less than or greater than 1?

Q11

Is $\sqrt{2}$ less than or greater than 2?

Q12

Can we find closer bounds for $\sqrt{2}$ ?

Q13

Will we ever get a number with a terminating decimal representation whose square is 2?

Q14

Use this formula to check your answers in the Figure it Out on page 39.

Q15

What if we wish to combine two squares of 'different' sizes to make a large square whose area is the sum of the two smaller squares?

Q16

Why does Baudhāyana’s method work?

Q17

Can you see why the method works in the case where the two squares are the same size? Does it agree with the method we used earlier to combine two same sized squares into a bigger square?

Q19

Explain why all the angles of this new 4-sided figure are right angles and so it is a square.

Q20

List down all the Baudhāyana triples with numbers less than or equal to 20.

Q21

Is there an unending sequence of Baudhāyana triples?

Q22

Is (30, 40, 50) a Baudhāyana triple?

Is (300, 400, 500) a Baudhāyana triple?

Q23

Context: The list of Baudhāyana triples having numbers less than or equal to 20 contains the following triples — (3, 4, 5), (6, 8, 10), (9, 12, 15), (12, 16, 20).

Q. Do you see any pattern among them?

Q24

Context: All these triples can be obtained by multiplying each term of (3, 4, 5) by a certain positive integer.

Q. Can we form a conjecture on Baudhāyana triples based on this observation?

Q25

Context: Conjecture: (3k, 4k, 5k) is a Baudhāyana triple, where k is any positive integer.

Q. Is this true?

Q26

Is $(5, 12, 13)$ a primitive Baudhayana triple? What are the other primitive Baudhayana triples with numbers less than or equal to 20?

Q27

Generate 5 scaled versions of each of these primitive triples. Are these scaled versions primitive?

Q28

If $(a, b, c)$ is non-primitive, and the integers have $f$ — greater than 1 — as a common factor, then is $\left(\frac{a}{f}, \frac{b}{f}, \frac{c}{f}\right)$ a Baudhayana triple? Check this statement for $(9, 12, 15)$ . Justify this statement.