To explore why [tex]\(c\)[/tex] could be an irrational number in the context of the Pythagorean theorem given rational numbers [tex]\(a\)[/tex] and [tex]\(b\)[/tex], let's delve into each statement provided and analyze their validity.
Consider the Pythagorean theorem formula:
[tex]\[ a^2 + b^2 = c^2 \][/tex]
### Analysis:
1. "The square of rational numbers is irrational, and the sum of two irrational numbers is irrational."
- This statement is incorrect because the square of rational numbers is always rational. For example, if [tex]\(a\)[/tex] and [tex]\(b\)[/tex] are rational, then [tex]\(a^2\)[/tex] and [tex]\(b^2\)[/tex] are also rational.
2. "The product of two rational numbers is rational, and the sum of two rational numbers is irrational."
- This statement contains partial truths: the product of two rational numbers is indeed rational. However, the sum of two rational numbers is not necessarily irrational; it can also be rational.
3. "The left side of the equation will result in a rational number, which is a perfect square."
- This statement is incorrect because the left side [tex]\(a^2 + b^2\)[/tex] does not necessarily result in a perfect square, even if both [tex]\(a\)[/tex] and [tex]\(b\)[/tex] are rational.
4. "The left side of the equation will result in a rational number, which could be a non-perfect square."
- This statement is correct. When both [tex]\(a\)[/tex] and [tex]\(b\)[/tex] are rational, [tex]\(a^2 + b^2\)[/tex] will always be a rational number. However, this rational sum [tex]\(a^2 + b^2\)[/tex] could be a non-perfect square (i.e., a number that is rational but not the square of any rational number).
### Conclusion and Elaboration:
When [tex]\(a\)[/tex] and [tex]\(b\)[/tex] are rational, their sum [tex]\(a^2 + b^2\)[/tex] is a rational number. However, this sum does not have to be a perfect square of a rational number. A classic example is when [tex]\(a = 1\)[/tex] and [tex]\(b = 1\)[/tex]:
[tex]\[ 1^2 + 1^2 = 1 + 1 = 2 \][/tex]
Here, [tex]\(a^2 + b^2 = 2\)[/tex].
The value of [tex]\(c\)[/tex] will be [tex]\(c = \sqrt{2}\)[/tex], which is irrational since [tex]\(\sqrt{2}\)[/tex] is not a rational number.
Thus, the reason why [tex]\(c\)[/tex] could be irrational is that the left side [tex]\(a^2 + b^2\)[/tex], while always rational if [tex]\(a\)[/tex] and [tex]\(b\)[/tex] are rational, can result in a non-perfect square, leading to an irrational [tex]\(c\)[/tex]. Therefore, the correct statement is:
"The left side of the equation will result in a rational number, which could be a non-perfect square."
