Question

The Pythagorean theorem states that the sum of the squares of the legs of a right triangle is equal to the square of the

hypotenuse by the formula a2 + b2 = c2.

If a is a rational number and b is a rational number, why could c be an irrational number?

O The square of rational numbers is irrational, and sum of two irrational numbers is irrational.

O The product of two rational numbers is rational, and the sum of two rational numbers is irrational.

O The left side of the equation will result in a rational number, which is a perfect square.

O The left side of the equation will result in a rational number, which could be a non-perfect square.

Answer 1

The product of two rational numbers is rational, and the sum of two rational numbers is irrational.

Explanation:

Let be
`a` and
`b` rational numbers whose forms are, respectively:

`a = (m)/(n)` and
`b = (q)/(r)`,
`m`,
`n`,
`q`,
`r`
`\in \mathbb{Z}`

We proceed to demonstrate if the Pythagorean sum of two rational numbers leads to a rational number.

1)
`a = (m)/(n)`,
`b = (q)/(r)`,
`m`,
`n`,
`q`,
`r`
`\in \mathbb{Z}` Given.

2)
`c^(2) = a^(2)+b^(2)` Given.

3)
`c^(2) = \left((m)/(n) \right)^(2)+\left((q)/(r) \right)^(2)` 1) in 2)

4)
`c^(2) = (m\cdot n^(-1))^(2)+(q\cdot r^(-1))^(2)` Definition of division.

5)
`c^(2) = [m^(2)\cdot (n^(-1))^(2)]+[q^(2)\cdot (r^(-1))^(2)]`
`(a\cdot b)^(c) = a^(c)\cdot b^(c)`

6)
`c^(2) = m^(2)\cdot n^(-2)+q^(2)\cdot r^(-2)`
`(a^(b))^(c) = a^(b\cdot c)`

7)
`c^(2) = [m^(2)\cdot (n^(2))^(-1)]+[q^(2)\cdot (r^(2))^(-1)]`
`(a^(b))^(c) = a^(b\cdot c)`

8)
`c^(2) = (m^(2))/(n^(2))+(q^(2))/(r^(2))` Definition of division.

9)
`c^(2) = (m^(2)\cdot r^(2)+q^(2)\cdot n^(2))/(n^(2)\cdot r^(2))`
`(x)/(y) +(w)/(z) = (x\cdot z+w\cdot y)/(y\cdot z)`

10)
`c^(2) = ((m\cdot r)^(2)+(q\cdot n)^(2))/((n\cdot r)^(n))`
`(a\cdot b)^(c) = a^(c)\cdot b^(c)`/Result.

If
`c` is a rational number, then
`(m\cdot r)^(2)+(q\cdot n)^(2) = a^(2)`, so that
`a \in \mathbb{Z}`. Which means that
`c` will be rational number if and only if the square root of
`(m\cdot r)^(2)+(q\cdot n)^(2)` is an entire number.

Which means that correct answer is B.