Respuesta :
Sum/difference:
Let
[tex] x = 5 + (-3\sqrt{8}) = 5-3\sqrt{8} [/tex]
This means that
[tex] 3\sqrt{8} = 5-x \iff \sqrt{8} = \dfrac{5-x}{3} [/tex]
Now, assume that [tex] x [/tex] is rational. The sum/difference of two rational numbers is still rational (so 5-x is rational), and the division by 3 doesn't change this. So, you have that the square root of 8 equals a rational number, which is false. The mistake must have been supposing that [tex] x [/tex] was rational, which proves that the sum/difference of the two given terms was irrational
Multiplication/division:
The logic is actually the same: if we multiply the two terms we get
[tex] x = -15\sqrt{8} [/tex]
if again we assume x to be rational, we have
[tex] \sqrt{8} = -\dfrac{x}{15} [/tex]
But if x is rational, so is -x/15, and again we come to a contradiction: we have the square root of 8 on one side, which is irrational, and -x/15 on the other, which is rational. So, again, x must have been irrational. You can prove the same claim for the division in a totally similar fashion.
