Question #184450

Consider the following two functions;

1. f: R-R defined by f(x) = 4x-15.

2. g: RR-defined by f(x) = 15x3.

Prove that both f and g are one-to-one correspondence.

1
Expert's answer
2021-05-07T09:04:01-0400
  1. Given f:RRf:R\rightarrow R defined by f(x)=4x5f(x)=4x-5

Require to prove that ff is a one-to-one correspondence.

To prove that ff is a one-to-one correspondence, we need to prove that ff is one-one and onto.


One - one:

Let a,bRa,b\in R such that f(a)=f(b)f(a)=f(b)

We prove that a=ba=b

Now f(a)=f(b)4a5=4b5f(a)=f(b)\Rightarrow 4a-5=4b-5

4a=4ba=b\Rightarrow 4a=4b\Rightarrow a=b

So, we proved that f(a)=f(b)a=bf(a)=f(b)\Rightarrow a=b for all a,bRa,b\in R

Therefore, ff is one - one.


Onto:

Let bRb\in R (codomain)

We need to find bRb\in R (domain) such that f(a)=bf(a)=b

Now f(a)=b4a5=b4a=b+5a=b+54Rf(a)=b\Rightarrow 4a-5=b\Rightarrow 4a=b+5\Rightarrow a=\frac{b+5}{4}\in R

So, we proved that for each bRb\in R (codomain) a=b+54R\exists a=\frac{b+5}{4}\in R such that f(a)=bf(a)=b

Therefore, ff is onto.

ff is one - one and onto implies ff is one - to - one correspondence.

Hence, ff is a one-to-one correspondence.

  1. Given: f:RRf:R\rightarrow R defined by f(x)=15x3f(x)=15x^3

Require to prove that ff is a one - to - one correspondence.


One - one:


Let a,bRa,b\in R such that f(a)=f(b)f(a)=f(b)

Now f(a)=f(b)15a3=15b3a3=b3a=bf(a)=f(b)\Rightarrow 15a^3=15b^3\Rightarrow a^3=b^3\Rightarrow a=b

We proved that f(a)=f(b)a=bf(a)=f(b)\Rightarrow a=b for all a,bRa,b\in R

Therefore, ff is one - one.


Onto:


Let bRb\in R (codomain)

We require to find aRa\in R (domain) such that f(a)=bf(a)=b

Now f(a)=b15a3=ba3=b15a=(b15)13Rf(a)=b\Rightarrow 15a^3=b\Rightarrow a^3=\frac{b}{15}\Rightarrow a=(\frac{b}{15})^{\frac{1}{3}}\in R

So, we proved that for each bRb\in R a=(b15)13R\exists a=(\frac{b}{15})^{\frac{1}{3}}\in R such that f(a)=bf(a)=b

Therefore, ff is onto.


Hence, ff is a one - to one correspondence.


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