SOLUTION: 1) If n is a real number, then 2n would never be part of which number set? A) a real number (B) an irrational number(C) a rational number (D) an imaginary number (E) an integer

Question 251989: 1) If n is a real number, then 2n would never be part of which number set?
A) a real number (B) an irrational number(C) a rational number (D) an imaginary number (E) an integer
2) If 2n^2 is an even integer, what is true about n + 1?
(A) It is an even integer.(B) It is an odd integer.(C) It can be an odd or even integer.(D) It is a prime number. (E) It is a perfect square.
can you give me explanations please

Answer by Theo(13342) (Show Source): You can put this solution on YOUR website!
if n is real, that means that it is not imaginary.

multiply it by 2 and it is still real.

all the other options are part of the real number set implying that it could be any of them.

if 2n^2 is an even integer, what can you say about (n+1)?

2 * n^2 will always be even because it will be always be divisible by 2 since it was multiplied by 2.

n+1 should therefore be either even or odd, because n could be either even or odd.

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