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:
r=7
Step-by-step explanation:
5r-7=2r+14
add sevento both sides
5r=2r+21
Subtract 2r from both sides.
3r=21
divide both sides by 3
r=7
Answer:
Step-by-step explanation:
Given: 
.
we need to find the correct form for 
if the equation is solve using undetermined coefficients.
A first order differential equation 
is said to be homogeneous if 
for all t.
Consider homogeneous equation 
Let 
be the solution .
We get 
Since 
, 
.
So, we get solution as 
As constant term and 
are already in the R.H.S of equation
, we can take as
The answer in the simplest form : 3/4
