How do I find the square root of a decimal number without using built in functions in C?

#include"stdio.h" #include"conio.h" void main() { float n,sqrt,temp; clrscr(); printf("Enter a number\n"); scanf("%f",&n); temp=0; sqrt=n/2; while(sqrt!=temp) { temp=sqrt; sqrt=(n/sqrt+sqrt)/2; } printf("square root=%f",sqrt); getch(); } Say entered number is 4 i.e, n=4. Now, by line 10, sqrt=4/2=2. By line 11, compiler enters while loop. First loop.. temp=2(line 13) sqrt=(4/2 +2)/2=2 temp=sqrt Loop terminates and sqrt=2.000000 Hope this helps! :) Another Way : You can use Newton-Raphson technique of successive approximation to get the root of a number as follows: double squareroot(float n) { double k=n/2; // initial guess while((k*k-n) > 0.0000000001 || (n-k*k) > 0.0000000001) { k=(k+n/k)/2; } return k; } You have to first come up with a initial guess. I chose n/2. Then you calculate how far off the error is by squaring it and subtracting n. If it’s too far off, you modify guess slightly and try again. Repeat until error is small. Babylonian method for square root Algorithm: This method can be derived from (but predates) Newton–Raphson method. 1 Start with an arbitrary positive start value x (the closer to the root, the better). 2 Initialize y=1. 3. Do following until desired approximation is achieved. a) Get the next approximation for root using average of x and y b) Set y=n/x Implementation: /*Returns the square root of n. Note that the function */ float squareRoot(float n) { /*We are using n itself as initial approximation This can definitely be improved */ float x=n; float y=1; float e=0.000001; /* e decides the accuracy level*/ while(x - y > e) { x=(x + y)/2; y=n/x; } return x; } /* Driver program to test above function*/ int main() { int n=50; printf ("Square root of %d is %f", n, squareRoot(n)); getchar(); } Example: n=4 /*n itself is used for initial approximation*/ Initialize x=4, y=1 Next Approximation x=(x + y)/2 (= 2.500000), y=n/x (=1.600000) Next Approximation x=2.050000, y=1.951220 Next Approximation x=2.000610, y=1.999390 Next Approximation x=2.000000, y=2.000000 Terminate as (x - y) > e now. If we are sure that n is a perfect square, then we can use following method. The method can go in infinite loop for non-perfect-square numbers. For example, for 3 the below while loop will never terminate. /*Returns the square root of n. Note that the function will not work for numbers which are not perfect squares*/ unsigned int squareRoot(int n) { int x=n; int y=1; while(x > y) { x = (x + y)/2; y = n/x; } return x; } /* Driver program to test above function*/ int main() { int n = 49; printf (" root of %d is %d", n, squareRoot(n)); getchar(); } read less