1 Our present way of representing numbers is due to the 2 Fi

1. Our present way of representing numbers is due to the?

2. Find all right triangles with consecutive integers sides. That is, find all right triangles with sides a, a+1, a+2, where a is a positive integer.

3. The Indian mathematician Brahmagupta considered the equation x² - 3y² = 1. Find three distinct solutions for this equation in integers x,y with xy does not equal 0.

Solution

1)

The way we write our numbers is based on a system of tens - the decimal system. Each column is worth ten times the one on its right so that the columns indicate powers of ten.

Example : 3405 is written as : 3000 + 400 + 00 + 5

3*10^3 + 4*10^2 + 0*10^1 + 5

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2)

Since these are right triangles, it must satisfy the pythagorean theorem.....

Longest side = a + 2 obviously

(a + 2)^2 = (a)^2 + (a + 1)^2

Expanding :

a^2 + 4a + 4 = a^2 + a^2 + 2a + 1

a^2 + 4a + 4 = 2a^2 + 2a + 1

a^2 - 2a - 3 = 0

(a - 3)(a + 1) = 0

a = 3 , -1

a = -1 ---> not considered because this would mean a side with a negative value....

a = 3
So, a + 1 --> 3 + 1 --> 4
a+2 --> 3 + 2 --> 5

So, the only right triangle with consecutive sides has side lengths ----> 3 , 4 , 5

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x^2 - 3y^2 = 1

x^2 = 1 + 3y^2

On the left, we have x^2, which is a perfect square

So, the right, 1 + 3y^2 must be a perfect square, lets find a few perfect squares of the form 1 + 3y^2....

When y = 0 , 1 , 2 and 3, it does not work.....

When y = 4, 1 + 3(4)^2 ---> 1 + 3(16) ---> 1 + 48 ---> 49, which is a perfect square

x^2 = 49

x = 7 or -7

So, we have two solutions namely ----> (7 , 4) and (-7 , 4)

Now, we have two more solutions, which are ----> (7 , -4) and (-7 , -4)

So, here are 4 possible solutions :

x = 7 , y = 4
x = 7 , y = -4
x = -7 , y = 4
x = -7 , y = -4