91Ó°ÊÓ

Let's assume that the sum \(\frac{6}{7}+\sqrt{2}\) is a rational number. We can denote this assumption as follows: \(\frac{6}{7}+\sqrt{2} = \frac{a}{b}\), where a and b are integers, and b is non-zero.

Now we want to isolate the square root term by moving \(\frac{6}{7}\) to the other side of the equation: \(\sqrt{2} = \frac{a}{b} - \frac{6}{7}\)

Find a common denominator for the fractions and simplify the equation. The common denominator is 7b: \(\sqrt{2} = \frac{7a - 6b}{7b}\)

Square both sides of the equation to eliminate the square root term: \((\sqrt{2})^2 = \left(\frac{7a - 6b}{7b}\right)^2\) This simplifies to: \(2 = \frac{(7a - 6b)^2}{(7b)^2}\)

Rearrange the equation by multiplying both sides by \((7b)^2\): \(2(7b)^2 = (7a - 6b)^2\)

Now we'll try to find a contradiction. Notice that the left side of the equation, \(2(7b)^2\), is an even number because it is divisible by 2. However, the right side of the equation, \((7a - 6b)^2\), is also an even number since it is the square of an even number (7a - 6b is even). Thus, we can say that the right side is divisible by 4: \((7a - 6b)^2 = 4k\), where k is some integer. Divide both sides by 2: \(\frac{(7a - 6b)^2}{2} = 2k\) Now compare this to our original equation: \(2(7b)^2 = (7a - 6b)^2\) Since the left side of the equation is even, the right side should also be even, and hence, divisible by 2: \(2(7b)^2 = 2k\), where k is some integer. However, this means that \((7b)^2\) must also be divisible by 2. This is a contradiction because it implies that b is an even number, and then a must be even as well. But then, the fraction \(\frac{a}{b}\) is reducible which contradicts the definition of a rational number. Thus, our assumption that \(\frac{6}{7}+\sqrt{2}\) is rational leads to a contradiction, and so \(\frac{6}{7}+\sqrt{2}\) must be an irrational number.