Question

if u can solve this↑↑↑↑↑↑↑↑↑↑
❌❌no spaming ❌❌​

Answer 1

sEmpty relation

Universal relation

Reflexive relation

Symmetric relation

Transitive relation

Equivalence relation

A function f : X → Y is one-one (or injective) if f (x1) = f(x2) ⇒ x1 = x2 ∀ x1, x2 ∈ X.

A function f : X → Y is onto (or surjective) if given any y ∈ Y, ∃ x ∈ X such that f(x) = y.

A function f : X → Y is one-one and onto (or bijective), if f is both one-one and onto.

A function f : X → Y is invertible if and only if f is one-one and onto.

oh fine! its ok

Answer 2

Answer:

Showing that limn→∞(∑k=1n1k−log(n)) exists

Note that

1n+1≤∫n+1n1xdx≤1n(1)

Summing the right hand inequality shows that

∑k=1n1k−log(n+1)(2)

is increasing. Summing the left hand inequality shows that

∑k=1n1k−log(n)(3)

is decreasing.

Since limn→∞log(n+1n)=0, (2) and (3) show that γ=limn→∞(∑k=1n1k−log(n)) exists and is between 0 and 1.

I'm fine wbu?