contestada

You are given that the following set of statements are true. (a) If an integer is even, then it is cosmological. (b) If a real number is cosmological, then it is not mystical. (c) If two real numbers are not mystical, then their product is mystical. Provide a proof by contradiction for the following statement: It is true that at least one of V2 or 18 is mystical.

Respuesta :

DeanR

18 is even, so by rule (a) cosmological, let's call it C. 18 is real and C and so by (b) not mystical, not M.  

So V2 which I guess is [tex]\sqrt 2[/tex] is our only hope.

Assume [tex]\sqrt 2[/tex] is not M.

Then [tex]\sqrt 2 \times \sqrt 2 = 2[/tex] is M, as by (c) it's the product of two not M numbers.

2 is even, so 2 is C (rule (a)).  So 2 is not M (rule (b)).  That's a contradiction.

Therefore our assumption is false.  We conclude

[tex]\sqrt 2[/tex] is mystical.