Question

Suppose x and y are different irrational numbers. Mark each statement as ALWAYS,

Answer 1

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 $x=\sqrt[3]{2}$

Here, $x$ is an irrational number.

$x^2=(\sqrt[3]{2})^2=2^{\frac{2}{3} }$ is also an irrational number

Now take $x=\sqrt{2}$

Here, $x$ is an irrational number.

$x^2=(\sqrt{2})^2=2$ is a rational number

So,

$x^2$ is a rational number: sometimes true

Take $x=\sqrt{2} \,,y=\sqrt{3}$

Here, $x,y$ are irrational numbers.

$xy=\sqrt{2}\sqrt{3}=\sqrt{6}$ is also an irrational number.

Now take $x=\sqrt{2} \,,y=\sqrt{8}$

Here, $x,y$ are irrational numbers.

$xy=\sqrt{2} \sqrt{8}=\sqrt{16}=4$ is a rational number.

So,

$xy$ is a rational number: sometimes true

As sum of a rational number and an irrational number is always irrational,

$x+3$ is an irrational number: always true

Take $x=\sqrt{2} \,,y=-\sqrt{2}$

Here, $x,y$ are irrational numbers.

$x+y=\sqrt{2} +(-\sqrt{2})=0$ is a rational number

Now take $x=\sqrt{2}\,,\,y=\sqrt{3}$

Here, $x,y$ are irrational numbers.

$x+y=\sqrt{2}+\sqrt{3}$ is an irrational number.

So,

$x+y$ is a rational number: sometimes true

As difference of two irrational numbers is always irrational,

$x-y$ is an irrational number: always true