- Given f:R→R defined by f(x)=4x−5
Require to prove that f is a one-to-one correspondence.
To prove that f is a one-to-one correspondence, we need to prove that f is one-one and onto.
One - one:
Let a,b∈R such that f(a)=f(b)
We prove that a=b
Now f(a)=f(b)⇒4a−5=4b−5
⇒4a=4b⇒a=b
So, we proved that f(a)=f(b)⇒a=b for all a,b∈R
Therefore, f is one - one.
Onto:
Let b∈R (codomain)
We need to find b∈R (domain) such that f(a)=b
Now f(a)=b⇒4a−5=b⇒4a=b+5⇒a=4b+5∈R
So, we proved that for each b∈R (codomain) ∃a=4b+5∈R such that f(a)=b
Therefore, f is onto.
f is one - one and onto implies f is one - to - one correspondence.
Hence, f is a one-to-one correspondence.
- Given: f:R→R defined by f(x)=15x3
Require to prove that f is a one - to - one correspondence.
One - one:
Let a,b∈R such that f(a)=f(b)
Now f(a)=f(b)⇒15a3=15b3⇒a3=b3⇒a=b
We proved that f(a)=f(b)⇒a=b for all a,b∈R
Therefore, f is one - one.
Onto:
Let b∈R (codomain)
We require to find a∈R (domain) such that f(a)=b
Now f(a)=b⇒15a3=b⇒a3=15b⇒a=(15b)31∈R
So, we proved that for each b∈R ∃a=(15b)31∈R such that f(a)=b
Therefore, f is onto.
Hence, f is a one - to one correspondence.
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