true or False? Prove your answer!  a) Suppose h:R→R is not one-to-one, and suppose a, b∈R with a≠b. Then h(a) =h(b). b) Suppose g:R→R is a function such that for every x∈R, there exists y∈R such that g(x)≠y . Then g is not onto.

Advanced Engineering Mathematics
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ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
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true or False? Prove your answer! 

a) Suppose h:R→R is not one-to-one, and suppose a, b∈R with a≠b. Then

h(a) =h(b).

b) Suppose g:R→R is a function such that for every x∈R, there exists y∈R such that g(x)≠y . Then g is not onto.

Expert Solution
Step 1

(a) The given statement, "Suppose h:RR is not one-to-one, and suppose a, bR with ab then h(a) =h(b)." is false.

Counter example:- Let h:RR defend by hx=x2

Since function hx=x2 is not one-one

Take a=1 and b=2

At a=1, ha=1

At b=2, hb=4

Here, ab  and  hahb

Hence, statement is false.

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