Answer to Question #184450 in Linear Algebra for Patrick Buthelezi

Question

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.

Answer 1

1. Given $f:R\rightarrow R$ 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 $a,b\in R$ such that $f(a)=f(b)$

We prove that $a=b$

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

$\Rightarrow 4a=4b\Rightarrow a=b$

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

Therefore, $f$ is one - one.

Onto:

Let $b\in R$ (codomain)

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

Now $f(a)=b\Rightarrow 4a-5=b\Rightarrow 4a=b+5\Rightarrow a=\frac{b+5}{4}\in R$

So, we proved that for each $b\in R$ (codomain) $\exists a=\frac{b+5}{4}\in R$ 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.

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

Require to prove that $f$ is a one - to - one correspondence.

One - one:

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

Now $f(a)=f(b)\Rightarrow 15a^3=15b^3\Rightarrow a^3=b^3\Rightarrow a=b$

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

Therefore, $f$ is one - one.

Onto:

Let $b\in R$ (codomain)

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

Now $f(a)=b\Rightarrow 15a^3=b\Rightarrow a^3=\frac{b}{15}\Rightarrow a=(\frac{b}{15})^{\frac{1}{3}}\in R$

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

Therefore, $f$ is onto.

Hence, $f$ is a one - to one correspondence.