Posted on

Prove that the inverse of one-one onto mapping is unique.

MCS013 – Assignment 8(d)A function is onto if and only if for every yy in the codomain, there is an xx in the domain such that f(x)=yf(x)=y.So in the example you give, f:R→R,f(x)=5x+2f:R→R,f(x)=5x+2, the domain and codomain are the same set: R.R. Since, for every real number y∈R,y∈R, there is an x∈Rx∈R such that f(x)=yf(x)=y, the function is onto. The example you include shows an explicit way to determine which xx maps…

Read more

Posted on

Prove that the inverse of one-one onto mapping is unique.

MCS013 – Assignment 8(d) A function is onto if and only if for every yy in the codomain, there is an xx in the domain such that f(x)=yf(x)=y. So in the example you give, f:R→R,f(x)=5x+2f:R→R,f(x)=5x+2, the domain and codomain are the same set: R.R. Since, for every real number y∈R,y∈R, there is an x∈Rx∈R such that f(x)=yf(x)=y, the function is onto. The example you include shows an explicit way to…

Read more

Posted on

Prove that the inverse of one-one onto mapping is unique.

MCS013 – Assignment 8(d)A function is onto if and only if for every yy in the codomain, there is an xx in the domain such that f(x)=yf(x)=y.So in the example you give, f:R→R,f(x)=5x+2f:R→R,f(x)=5x+2, the domain and codomain are the same set: R.R. Since, for every real number y∈R,y∈R, there is an x∈Rx∈R such that f(x)=yf(x)=y, the function is onto. The example you include shows an explicit way to determine which xx maps…

Read more

Posted on

Write a program in C language for addition of two numbers which have at least 20 digits each

#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]);…

Read more