State whether 7-the sq root of 2 is rational or irrational. explain your 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.

-3

step-by-step explanation:

a drawing really does here:

so i draw the horizontal line y=2 and the vertical line x=-2 at this intersection i put b (this is where my right angle is going to be).   so point b is at (-2,2).   point a is 5 units long and is also on y=2 so this point has to be (3,2) or (-7,2) -doesn't matter i will go with (3,2).

now we know c is at (-2,y) we don't know the y yet bet we know that c is on x=-2.

we have the area of the triangle is .5bh=12.5

we know the base length is 5 (ab)

we have the height is (2-y) if you look at your

plug in

.5(5)(2-y)=12.5

solve for y to get the y-coordinate for c

2.5(2-y)=12.5

divide both sides by 2.5 giving:

2-y=5

now subtract 2 on both sides

-y=3

y=-3