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MCS013 - Assignment 8(d)

A function is onto if and only if for every y$y$ in the codomain, there is an x$x$ in the domain such that f(x)=y$f(x)=y$.

So in the example you give, f:R→R,f(x)=5x+2$f:\mathbb{R}\to \mathbb{R},f(x)=5x+2$, the domain and codomain are the same set: R.$\mathbb{R}.$ Since, for every real number y∈R,$y\in \mathbb{R},$ there is an x∈R$x\in \mathbb{R}$ such that f(x)=y$f(x)=y$, the function is onto. The example you include shows an explicit way to determine which x$x$ maps to a particular y$y$, by solving for x$x$ in terms of y.$y.$ That way, we can pick any y$y$, solve for f′(y)=x$f^{\prime }(y)=x$, and know the value of x$x$ which the original function maps to that y$y$.

Side note:

Note that f′(y)=f−1(x)$f^{\prime }(y)=f^{-1}(x)$ when we swap variables. We are guaranteed that every function f$f$ that is onto and one-to-one has an inverse f−1$f^{-1}$, a function such that f(f−1(x))=f−1(f(x))=x$f(f^{-1}(x))=f^{-1}(f(x))=x$.

#include<stdio.h>

#include<conio.h>

#include<string.h>

void main()

{

char str1[21],str2[21];

int i,j,a[20],b[20];

int add[21]={0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0};

printf("\n Enter number 1:");

for(i=0; i<21; i++)

{

str1[i]=getchar();

}

printf("\n Enter number 2:");

for(i=0; i<21; i++)

{

str2[i]=getchar();

}

//converting char to integer

for(i=0; i<20; i++)

{

a[i]=(int)str1[i]-48;

b[i]=(int)str2[i]-48;

}

printf("\n Print Number 1:");

for(i=0; i<20; i++)

{

printf("%d",a[i]);

}

printf("\n Print Number 2:");

for(i=0; i<20; i++)

{

printf("%d",b[i]);

}

for(i=20; i>-1;i--)

{

j= a[i]+b[i];

if(j<10)

{

add[i] +=j;

}

else

{

add[i]+=(j-10);

add[i-1]+=1;

}

}

printf("\n Print Addition:");

for(i=0; i<20; i++)

{

printf("%d",add[i]);

}

 

getch();

}