Leelavati or Lilavati - Indian mathematician

 Who was Leelavati? or Lilavati?

Lilavati is one of the earliest eminent women mathematicians of India. The legend behind the name Lilavati is quite interesting. Bhaskar Acharya II had a beautiful and intelligent daughter named Lilavati. She was the apple of his eye, and he had great hopes for her future. Being an astrologer himself, he eagerly studied his daughter’s horoscope to know what the future held for her. What he saw shocked and upset him. He could clearly see that Lilavati would remain single and childless. A girl remaining unmarried was unheard of those days. It was considered a fate worse than death. Bhaskar Acharya decided to take a risk and get his daughter married during a wonderfully auspicious muhurta. He looked for a suitable groom and made all the arrangements. Those were the days when there were no clocks. So, to make sure that he knew the exact time when the auspicious muhurta occurred, he made a hole in an empty cup and let it float in a vessel filled with water. As the water slowly entered the cup through the hole, the cup would become full and sink. The minute the cup sank, the wedding would be conducted.

After making all these arrangements, Bhaskar Acharya warned the girls not to go near the place where the cup and water were kept. The usually obedient Lilavati could not contain her curiosity that day. She wanted to see exactly when the cup sank and went near the water and investigated it. As she bent over the vessel, without her knowledge, a tiny pearl from her bridal dress (some say, it was from her nose ring) fell into the cup and blocked the hole. And so, the cup never became full and never sank. The auspicious time passed by, but none knew about it. Bhaskar Acharya was greatly upset. The wedding was conducted anyway, but very soon, Lilavati’s husband died.

Leelavathi or Lilavati was a Hindu Indian woman. She was an Indian mathematician.  Bhāskara II's treatise on mathematics was written in 1150 AD. It is the first volume of his main work, the Siddhānta Shiromani Lilavati, the daughter of Bhaskar Acharya, was destined to die a spinster. To avoid this fate, Bhaskar Acharya predicted an auspicious moment for her wedding. To alert her at the right time, he nailed a small hole in a cup and allowed it to float in a vessel filled with water. Bhaskar Acharya was devastated. Lilavati became inconsolable. She was also guilty because the muhurta could not be noticed because of her.

After a few days, Bhaskar Acharya said, “My child! It is immensely sad that you have lost your husband. But how long will you be grieving over your loss? You must get over this and try to lead a normal life.”

Lilavati said, “Yes, father! What you are saying is true. But I don’t know what to do. The thought of my loss keeps coming back to my mind and I am not able to focus on anything.”

Bhaskar Acharya decided to find a way to prevent his daughter from getting out of her grief.

One day, Bhaskar Acharya had an idea. He knew that his daughter had a sharp intellect. He posed a mathematical puzzle to her and said, “Lilavati! Why don’t you solve this problem? You have always been interested in solving problems. Take it as a challenge and do it.”

Lilavati agreed immediately. As she got involved in solving the problem, she was totally drawn into it. She felt jubilant when she solved it. The dejection and depression disappeared magically. It was as if she was born again. Life became interesting and filled with possibilities.

Bhaskar Acharya collected all the problems he had created for his daughter and arranged them into thirteen chapters. He named this work Lilavati, after his beloved daughter.  Thus, a treatise on mathematics got the name of a girl.

A puzzle from Lilavati:

Of the group of elephants, half and one-third of the half went into a cave, one-sixth and one-seventh of one-sixth were drinking water from a river. One-eighth and one-ninth of one-eighth were sporting in a pond full of lotuses. The lover king of the elephants was leading three female elephants; how many elephants were there in the flock?

Solution by ChatGPT.

Let's break down the information provided step by step:

Half and one-third of the half went into a cave:

Half of the elephants went into a cave.

One-third of the half (1/3 * 1/2 = 1/6) also went into the cave.

One-sixth and one-seventh of one-sixth were drinking water from a river:

One-sixth of the elephants were drinking water from a river.

One-seventh of one-sixth (1/7 * 1/6 = 1/42) also went to the river.

One-eighth and one-ninth of one-eighth were sporting in a pond full of lotuses:

One-eighth of the elephants were sporting in a pond full of lotuses.

One-ninth of one-eighth (1/9 * 1/8 = 1/72) also joined them.

Now, let's calculate the total number of elephants by adding up the fractions of elephants mentioned for each activity:

Elephants in the cave: 1/2 + 1/6 = 4/6 = 2/3

Elephants at the river: 1/6 + 1/42 = 7/42 = 1/6

Elephants in the pond: 1/8 + 1/72 = 9/72 = 1/8

Now, we have fractions representing the number of elephants in each activity, and they should add up to the total number of elephants (1 whole):

Total = 2/3 + 1/6 + 1/8

To add these fractions, we need a common denominator, which is 24. So, let's convert all fractions to have a denominator of 24:

Total = (16/24) + (4/24) + (3/24)

Now, add the fractions:

Total = (16 + 4 + 3) / 24

Total = 23 / 24

So, the total number of elephants is 23, and the lover king leading three female elephants makes it 24.

Therefore, there are 24 elephants in the flock.

Pythagoras Theorem - Leelawati Proof- An Indian mathematician

Given. ABC is a right-angled triangle.

To prove. AC² = AB² + BC² 

               B² = c² + a²          

Const. Produce BC to D, such that CD= AB=c

Let AB = c, BC = a, AC = b.

On BD and AC draw Squares BF & AE and complete figure as shown.

Proof. In the figure, there are 4 right-angled triangles and two squares.

Area of ▲ ABC    = [Base x Altitude] /2

= a.c/2

There are 4 such ∆s.

Therefore, the area of 4 ∆s = a.c÷2 x4 =2ac…(i)

Area of big square BDF H = (a + c) ² = a² + c² +2ac

Area of Small Square ACEG = b²

The whole figure BDFH contains 4 triangles & one small square.

or (a+c) ² = 2ac + b²

a² + c² + 2ac = 2ac + b²

or BC² + AB² = AC² [cancelling 2ac]