Answer:
See Explanation for proofs
Step-by-step explanation:
Given
See attachment for complete question
Solving (a):
[tex]\frac{\sqrt{2}}{2}[/tex] is irrational
[tex]\sqrt{2} =1.41421356237.....[/tex]
So, we have:
[tex]\frac{\sqrt{2}}{2} = \frac{1.41421356237.....}{2}[/tex]
[tex]\frac{\sqrt{2}}{2} = 0.70710678118....[/tex]
[tex]0.70710678118....[/tex] can not be represented as a fraction of whole numbers.
Hence:
[tex]\frac{\sqrt{2}}{2}[/tex] is irrational
Solving (b):
[tex]\sqrt{2} -2[/tex] is irrational
[tex]\sqrt{2} =1.41421356237.....[/tex]
So, we have:
[tex]\sqrt{2} -2 = 1.41421356237..... - 2[/tex]
[tex]\sqrt{2} -2 = -0.58578643763.....[/tex]
[tex]-0.58578643763.....[/tex] can not be represented as a fraction of whole numbers.
Hence:
[tex]\sqrt{2} -2[/tex] is irrational
Solving (c):
The sum of two positive irrational number is always irrational
Proof Below
If a is irrational and b is also irrational,
then
[tex]a + b = irrational[/tex]
Take for instance:
[tex]a = \sqrt{2[/tex]
[tex]b = \sqrt{8[/tex]
[tex]a + b = \sqrt{2} + \sqrt{8[/tex]
[tex]a + b = \sqrt{2} + \sqrt{4*2[/tex]
[tex]a + b = \sqrt{2} + \sqrt{4}*\sqrt{2[/tex]
[tex]a + b = \sqrt{2} + 2\sqrt{2[/tex]
[tex]a + b = 3\sqrt{2[/tex]
[tex]3\sqrt{2[/tex] is irrational and this is true for the sum of every positive irrational numbers
