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:
-5 when ...
Step-by-step explanation:
The rules of exponents can help you simplify the given product.
<h3>Rules</h3>
(a/b)^c = (a^c)/(b^c)
(ab)^c = (a^c)(b^c)
(a^b)(a^c) = a^(b+c)
(a^b)/(a^c) = a^(b-c)
<h3>Application</h3>
This expression does not match any of those offered.
When x=-1 and y=5, this becomes ...
Answer:
3√7
Step-by-step explanation:
√63
√3x21
√3x3x7
3√7
