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 1.4141.414\dots 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 2\sqrt{2} is of the form 1.41441.414\dots4, then its square will be of the form—

Question diagram 1
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Solution
Understand the Question
  • A non-integer terminating decimal with kk decimal places can be expressed as A10k\dfrac{A}{10^k}, where AA is an integer whose last digit is non-zero.
  • When squared, (A10k)2=A2102k\left(\dfrac{A}{10^k}\right)^2 = \dfrac{A^2}{10^{2k}}, which has 2k2k decimal places and its last digit must be non-zero because the square of any non-zero digit (11 to 99) never ends in 00.
  • Since 2=2.0002 = 2.000\dots is a whole number with no non-zero digits after the decimal point, no terminating decimal can ever square to equal 22.

Step 1 · Properties of Terminating Decimals

A terminating decimal has a finite number of decimal places and can be written as:

x=A10kx = \dfrac{A}{10^k}

where kk is the number of decimal places (k1k \ge 1 for non-integers) and AA is an integer not ending in zero.

Step 2 · Square of a Terminating Decimal

Squaring xx:

x2=(A10k)2=A2102k\begin{aligned} x^2 &= \left(\dfrac{A}{10^k}\right)^2 \\[0.6em] &= \dfrac{A^2}{10^{2k}} \end{aligned}

Since AA does not end in zero, its last digit must be 1,2,3,4,5,6,7,8,1, 2, 3, 4, 5, 6, 7, 8, or 99. Checking their squares:

  • 121^2 ends in 11
  • 222^2 ends in 44
  • 323^2 ends in 99
  • 424^2 ends in 66
  • 525^2 ends in 55
  • 626^2 ends in 66
  • 727^2 ends in 99
  • 828^2 ends in 44
  • 929^2 ends in 11

Thus, A2A^2 always ends in 1,4,5,6,1, 4, 5, 6, or 99 (non-zero). Since k1k \ge 1, x2x^2 has 2k22k \ge 2 decimal places with a non-zero last digit:

  • For x=1.2    x2=1.44x = 1.2 \implies x^2 = 1.44 (last digit is 44)
  • For x=1.4    x2=1.96x = 1.4 \implies x^2 = 1.96 (last digit is 66)

Step 3 · Complete the Given Form

Diagram 1

If a terminating decimal ends in 44, its square must end in 66 (since 42=164^2 = 16).

Therefore, if 2\sqrt{2} is of the form 1.41441.414\dots4, then its square will be of the form . _ ... _ 6.

Step 4 · Conclusion by Contradiction

Assume a terminating decimal xx exists such that x2=2x^2 = 2.

  • xx cannot be an integer since 12=1<2<4=221^2 = 1 < 2 < 4 = 2^2.
  • If xx is a non-integer terminating decimal, x2x^2 must have a non-zero digit at its last decimal place.
  • However, the decimal representation of 22 is 2.0002.000\dots, where all decimal digits are zero.

This is a contradiction. Hence, no terminating decimal representation can have a square equal to 22.

Answer

No, we will never get a number with a terminating decimal representation whose square is 22.

The square will be of the form . _ ... _ 6.

Common Mistakes
  • Approximation Confusion: Assuming that because 1.4142=1.99939621.414^2 = 1.999396 \approx 2, an exact terminating decimal exists. A terminating decimal squared always yields more non-zero decimal places.
  • Last Digit Behavior: Overlooking that the square of any non-zero digit (11 through 99) never ends in 00, which proves that the square of any terminating decimal can never become a whole integer.

More questions in IT

Q1

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

Q2

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.

Q3

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.**]

Q4

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?

Q5

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?

Q6

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:

Q7

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

Explain by connecting QSQS and PRPR, finding the different angles formed, and then using tringle congruence.

Q8

Find the hypotenuse of this isosceles right triangle.

Q9

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

Q10

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

Q11

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

Q12

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

Q13

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

If there is such a terminating decimal starting with 1.4141.414\dots 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 2\sqrt{2} is of the form 1.41441.414\dots4, then its square will be of the form—

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)(30, 40, 50) a Baudhāyana triple?

Is (300,400,500)(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)(3, 4, 5), (6,8,10)(6, 8, 10), (9,12,15)(9, 12, 15), (12,16,20)(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)(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)(3k, 4k, 5k) is a Baudhāyana triple, where kk is any positive integer.

Q. Is this true?

Q26

Is (5,12,13)(5, 12, 13) a primitive Baudhayana triple? What are the other primitive Baudhayana triples with numbers less than or equal to 2020?

Q27

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

Q28

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

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