Answer:
The product of two rational numbers is rational, and the sum of two rational numbers is irrational.
Step-by-step explanation:
Let be [tex]a[/tex] and [tex]b[/tex] rational numbers whose forms are, respectively:
[tex]a = \frac{m}{n}[/tex] and [tex]b = \frac{q}{r}[/tex], [tex]m[/tex], [tex]n[/tex], [tex]q[/tex], [tex]r[/tex] [tex]\in \mathbb{Z}[/tex]
We proceed to demonstrate if the Pythagorean sum of two rational numbers leads to a rational number.
1) [tex]a = \frac{m}{n}[/tex], [tex]b = \frac{q}{r}[/tex], [tex]m[/tex], [tex]n[/tex], [tex]q[/tex], [tex]r[/tex] [tex]\in \mathbb{Z}[/tex] Given.
2) [tex]c^{2} = a^{2}+b^{2}[/tex] Given.
3) [tex]c^{2} = \left(\frac{m}{n} \right)^{2}+\left(\frac{q}{r} \right)^{2}[/tex] 1) in 2)
4) [tex]c^{2} = (m\cdot n^{-1})^{2}+(q\cdot r^{-1})^{2}[/tex] Definition of division.
5) [tex]c^{2} = [m^{2}\cdot (n^{-1})^{2}]+[q^{2}\cdot (r^{-1})^{2}][/tex] [tex](a\cdot b)^{c} = a^{c}\cdot b^{c}[/tex]
6) [tex]c^{2} = m^{2}\cdot n^{-2}+q^{2}\cdot r^{-2}[/tex] [tex](a^{b})^{c} = a^{b\cdot c}[/tex]
7) [tex]c^{2} = [m^{2}\cdot (n^{2})^{-1}]+[q^{2}\cdot (r^{2})^{-1}][/tex] [tex](a^{b})^{c} = a^{b\cdot c}[/tex]
8) [tex]c^{2} = \frac{m^{2}}{n^{2}}+\frac{q^{2}}{r^{2}}[/tex] Definition of division.
9) [tex]c^{2} = \frac{m^{2}\cdot r^{2}+q^{2}\cdot n^{2}}{n^{2}\cdot r^{2}}[/tex] [tex]\frac{x}{y} +\frac{w}{z} = \frac{x\cdot z+w\cdot y}{y\cdot z}[/tex]
10) [tex]c^{2} = \frac{(m\cdot r)^{2}+(q\cdot n)^{2}}{(n\cdot r)^{n}}[/tex] [tex](a\cdot b)^{c} = a^{c}\cdot b^{c}[/tex]/Result.
If [tex]c[/tex] is a rational number, then [tex](m\cdot r)^{2}+(q\cdot n)^{2} = a^{2}[/tex], so that [tex]a \in \mathbb{Z}[/tex]. Which means that [tex]c[/tex] will be rational number if and only if the square root of [tex](m\cdot r)^{2}+(q\cdot n)^{2}[/tex] is an entire number.
Which means that correct answer is B.
