Question

move the options and the explanation to show whether the sum of a rational number and an irrational number is always irrational?
if ( x = sqrt{7} + \frac{3}{5} )
assume ( x ) is a rational number.
by subtracting ( \frac{3}{5} ) from both sides of the equation, we find ( sqrt{7} = x - \frac{3}{5} ).
because ( x ) and ( \frac{3}{5} ) are both rational numbers, ( x - \frac{3}{5} ) is ______.
however, ( sqrt{7} = x - \frac{3}{5} ) and ( sqrt{7} ) is ______ , so ( x - \frac{3}{5} ) must be an irrational number.
since we know the assumption ( x ) is a rational number is false.
thus, ( sqrt{7} + \frac{3}{5} ) is ______.
options:
a rational number
an irrational number

Explanation:

Step1: Analyze \( x - \frac{3}{5} \)

Given \( x \) is rational and \( \frac{3}{5} \) is rational. The difference of two rational numbers is rational (by the property of rational numbers: if \( a,b \in \mathbb{Q} \), then \( a - b \in \mathbb{Q} \)). So \( x - \frac{3}{5} \) is a rational number.

Step2: Analyze \( \sqrt{7} \)

\( \sqrt{7} \) is known to be an irrational number (since 7 is not a perfect square, and the square root of a non - perfect - square positive integer is irrational). But we have \( \sqrt{7}=x - \frac{3}{5} \), which would imply an irrational number equals a rational number, a contradiction. So the first blank (for \( \sqrt{7} \)): an irrational number; the second blank (for \( x-\frac{3}{5} \)): a rational number; the third blank (for \( \sqrt{7}+\frac{3}{5} \)): an irrational number (because the sum of an irrational number \( \sqrt{7} \) and a rational number \( \frac{3}{5} \) is irrational, as the sum of a rational and an irrational number is irrational).

Answer:

First blank: an irrational number; Second blank: a rational number; Third blank: an irrational number