Respuesta :
Answer:
always True
sometimes true
sometimes true
always true
sometimes true
always true
Step-by-step explanation:
As product of non-zero rational number and irrational number is irrational,
3x is an irrational number : always True
Take [tex]x=\sqrt[3]{2}[/tex]
Here, [tex]x[/tex] is an irrational number.
[tex]x^2=(\sqrt[3]{2})^2=2^{\frac{2}{3} }[/tex] is also an irrational number
Now take [tex]x=\sqrt{2}[/tex]
Here, [tex]x[/tex] is an irrational number.
[tex]x^2=(\sqrt{2})^2=2[/tex] is a rational number
So,
[tex]x^2[/tex] is a rational number: sometimes true
Take [tex]x=\sqrt{2} \,,y=\sqrt{3}[/tex]
Here, [tex]x,y[/tex] are irrational numbers.
[tex]xy=\sqrt{2}\sqrt{3}=\sqrt{6}[/tex] is also an irrational number.
Now take [tex]x=\sqrt{2} \,,y=\sqrt{8}[/tex]
Here, [tex]x,y[/tex] are irrational numbers.
[tex]xy=\sqrt{2} \sqrt{8}=\sqrt{16}=4[/tex] is a rational number.
So,
[tex]xy[/tex] is a rational number: sometimes true
As sum of a rational number and an irrational number is always irrational,
[tex]x+3[/tex] is an irrational number: always true
Take [tex]x=\sqrt{2} \,,y=-\sqrt{2}[/tex]
Here, [tex]x,y[/tex] are irrational numbers.
[tex]x+y=\sqrt{2} +(-\sqrt{2})=0[/tex] is a rational number
Now take [tex]x=\sqrt{2}\,,\,y=\sqrt{3}[/tex]
Here, [tex]x,y[/tex] are irrational numbers.
[tex]x+y=\sqrt{2}+\sqrt{3}[/tex] is an irrational number.
So,
[tex]x+y[/tex] is a rational number: sometimes true
As difference of two irrational numbers is always irrational,
[tex]x-y[/tex] is an irrational number: always true
