Answer: Irrational
I'm going to write 'x' in place of "square root of 2" just to save time. Let's consider the possibility that 7-x is rational. I'm going to show there's a contradiction which ultimately concludes that the expression is irrational.
If 7-x was rational, then 7-x = p/q for some integers p,q with q being nonzero. Solve for x to get
7-x = p/q
-x = (pq) - 7
x = -(p/q) + 7
x = 7 - (p/q)
x = (7q/q) - (p/q)
x = (7q-p)/q
So if 7-x were rational, then this forces x to be rational because 7q-p is an integer over q (also an integer). The format is rational = integer/integer. But this is where the contradiction lies. Remember that x is standing in place for "square root of 2", which is an irrational number. There is no way to write sqrt(2) as a ratio of two integers. So x cannot be both rational and irrational at the same time.
Therefore, 7-x must be irrational making 7-sqrt(2) to be irrational as well.
note: sqrt is shorthand for "square root"
another note: adding any rational number to an irrational number will always result in an irrational number.
