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Month: March 2016
MCS023 Assignment 3
MCS023 Assignment 3
MCA2 assignment
We will upload all assignment of MCA2 before 28th march
MCA2 assignment
We will upload all assignment of MCA2 before 28th march
MCA2 assignment
We will upload all assignment of MCA2 before 28th march
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…
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…
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…
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]);…