Question

8. H={all numbers of the form 3n+1, for n a natural number} under multiplication?

Answer 1

Answer and Step-by-step explanation:

H= {all numbers of the form 3n+1, for n natural number} under multiplication:

Sol: all number of the form 3n+1, for n natural number so, put n=2, 3(2) +1 = 7, which is a natural number. A natural number under multiplication does not form a group because its inverse does not exist.

J= {all number of the form 3n+2, for n natural number} under multiplication:

Sol: all the number of the form 3n+2, n is the natural number does not form a group under multiplication because the identity element does not form in this set, and inverse does not exist.

K= {perfect squares of integers} under addition:

Sol: perfect square of integers under addition does not form a group because the inverse does not exist. For example, for any element x, inverse (-x) does not exist because of a square integer.

L= {perfect squares of integers} under multiplication:

Sol: a set of the perfect square of integers under multiplication is not a set, because it does not have the inverse property

M= {all numbers of the form 2, for n natural number} under addition:

Sol: natural number in the form of 2n, n is a natural number under addition is not a group as it does not have identity property. Identity property, under addition, is zero.

N= {all number of the form 2n, n is natural number} under multiplication:

All number of the form 2n, n is the natural number under multiplication is not a group. Because its inverse does not exist. For example, an element x belongs to a natural number; its inverse (1/x) does not exist in this set.