Question

1)Complete each of the following statements using the words “greater than”, “less than” or “equal to”

a) The cardinality of the even numbers is _________________ the cardinality of the natural numbers.

b) The cardinality of the natural numbers is _________________ the cardinality of the positive rational numbers.

c) The cardinality of the natural numbers is _________________ the cardinality of the rational numbers.

d) The cardinality of the real numbers is _________________ the cardinality of the natural numbers.

e) The cardinality of the real numbers on the interval (0,1) is _________________ the cardinality of the natural numbers.

f) The cardinality of the integers is _________________ the cardinality of the real numbers.

g) The cardinality of the real numbers on the interval (0,1) is _________________ the cardinality of the real numbers.

2) Given each of the following functions, determine if they are surjective, injective or bijective. Write yes or no for each of the following (you don’t have to show any work to prove your conclusions).

a) ?(?):ℝ→ℝ given by ?(?)=3?+5 , where ℝ is the set of real numbers.

Injective _______

Surjective ______

Bijective _______

b) ?(?):ℝ+→ℝ+ given by ?(?)=?2 , where ℝ+ is the set [0,∞).

Injective _______

Surjective ______

Bijective _______

c) ?(?):ℝ→ℝ+ given by ?(?)=?2 , where ℝ is the set of real numbers and ℝ+ is the set [0,∞).

Injective _______

Surjective ______

Bijective _______

d) ?(?):ℝ+→ℝ given by ?(?)=?2 , where ℝ is the set of real numbers and ℝ+ is the set [0,∞).

Injective _______

Surjective ______

Bijective _______

Answer 1

I a) L Equal to os filNZIN F(31) = 28 - f is bjective So Cardinality of IN & 21N is equal. b) e) tess than Equal (as at is countable) Equal (o is countable) SIR & / IN FR is Greater than uncountable ? e) Greater than h Less than (ZER & IR is uncountable g) Equal 9 f(1) = 3x+5 f is injective & sudjective also 7 so f is bijective

of be F()=ve? % f is not injective as f(1) = f(-) also f is not suojecture (no pore mali so f is not Bijective. by få lRt R* +-(+1)= x² f is injective f is suajectie So f is bjective FAIR+ IR f is injective f is not suspective a f is not Bijective, estre d)