Answer:
Consider f: N → N defined by f(0)=0 and f(n)=n-1 for all n>0.
Step-by-step explanation:
First we will prove that f is surjective. Let y∈N be any natural number. Define x as the number x=y+1. Then x∈N, and f(x)=x-1=(y+1)-1=y. We conclude that f is surjective.
However, f is not injective. Take x1=0 and x2=1. Then x1≠x2 but f(x1)=0 and f(x2)=x2-1=1-1=0. We have shown that there are two natural numbers x1,x2 such that x1≠x2 but f(x1)=f(x2), that is, f is not injective.
Note:
If 0∉N in your definition of natural numbers, the same reasoning works with the function f: N → N defined by f(1)=1 and f(n)=n-1 for all n>1. The only difference is that you consider x1=1, x2=2 for the injectivity.
Answer and Step-by-step explanation:
The two statements that are true are statement 1 and statement 4:
- The sum of pi and 0 is irrational.
- The product of 
and 
is irrational.
Pi is an irrational number, and the square root of 19 is irrational. That's because these numbers can't be put into a simplified fraction to encompass all the digits it contains.
#teamtrees #PAW (Plant And Water)
Clearly, alternative B
y = 0.01x + 7.5
y = (0.01)*5000 + 7,5
y = 50 + 7.5
y = 57.50
The apothem of this is 13
