Answer:
See explanation below
Step-by-step explanation:
1) Prove or disprove that if [tex] x^y[/tex] is an irrational number, then x or y is also an irrational number.
Let's take the following instances:
i) When x= 2 and y=[tex] \sqrt{2} [/tex] we have: [tex] 2^\sqrt^{^2^} [/tex]
ii) When [tex] x=2\sqrt{2} [/tex] and y=3, we have: [tex] (x=2\sqrt{2})^3 [/tex]
iii) When [tex] x=2\sqrt{2} [/tex] and [tex] y = \sqrt{2}[/tex], we have: [tex] (2\sqrt{2})^\sqrt^{^2^}[/tex]
It is proven because, in scenario
i) x is rational and y is irrational
ii) x is irrational and y is rational
iii) x and y are irrational
2) Prove tha x² is irrational, then x is irrational.
Use contradiction here.
Thus, x² is irrational and x is rational.
[tex] x =\frac{b}{a} [/tex] when x is rational, a & b are integers.
Therefore, [tex] x^2 =\frac{b^2}{a^2} [/tex]. This x² is rational.
This contradicts the statement that x² is irrational.
Therefore, if x² is irrational, x is also irrational.
