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Is the sum of an even whole number and an odd whole number always odd?
Yes --------------------------------------------- Let n be an integer. Then 2n is an even number Let m be an integer. Then 2m is an even number and 2m + 1 is an odd number. Then: even + odd = (2n) + (2m + 1) = (2n + 2m) + 1 = 2(n + m) + 1 = 2k + 1 (where k = m + n) which is an odd number.
The product of an odd and an even number always odd?
odd x even = even (2n + 1)(2n) = 2[n(2n + 1)] let n(2n + 1) = t = 2t (even)
What is the Chromatic number of a cycle graph?
3 if n is odd 2 if n is even where n is the number of vertices.
Program for check a number is even or odd using PYTHON language?
This program checks whether a number is odd or even. NOTE: This site removes formatting from answers. Replace (tab) with a tab or four spaces. #!/usr/bin/python print("Type a number.") n = input(": ") l = len(n)-1 even = ("02468") if n[l] in even: (tab)print(n, "is even.") if n[l] not in even: (tab)print(n, "is odd.")
What number is both odd and even?
There is no number that is both even (divisible by 2) and odd (not divisible by 2). Even numbers are of the form 2m for all integer m; Odd numbers are of the form 2n + 1 for all integer n; Assume there is an even number is also an odd number, then for some integer m and n: 2m = 2n + 1 � 2m - 2n = 1 � 2(m - n) = 1 � m - n = 1/2 But as m and n are both integers, their difference cannot be a fraction. Thus there are no integer m and n that satisfy 2m = 2n + 1, which means that the original assumption that there is an even number that is also an odd number is false. Thus there is no number that is both even and odd. This question is actually a riddle. The answer is 6 or 9, since flipping either number will give you the other.
Can you prove n2 - n is even for any natural number n?
There are two cases here: one where n is even and one where n is odd. Let's consider the case where n is even: If n is even then n2 has to be even (since multiple of an even number must always be even.) In this case, we are subtracting an even number from and even number, the result must be even. This proves than n2 - n is even when n is even. Now let's consider the case where n is odd: If n is odd, then n2 must be odd. This is because an odd number times an odd number is always odd. (You could think if this as an odd number times and even number and then adding an odd number. For example, say that n is odd. n-1 is then even, and n2 = n(n-1) + n. n(n-1) must be even, since it is a multiple of an even number. And even number plus and odd number then has to be odd.) Now we know we have and odd number minus and odd number, which has to be even. So this proves that n2 - n is even when n is odd. Since we have proved that n2 - n is even for both when n is even and when n is odd, and there are no other cases, n2 - n must be even for any natural number n. or Let n be a natural number. Then n can be even or odd. We want to show that n2 - n = 2m where m is any positive nteger (by the def. of even). Case 1: Let n be even. Then n = 2k (def. of even), where k is any positive integer. Then, n2 - n = (2k)2 - (2k) = 2(2k2 - k); let 2k2 - k = m = 2m Therefore, n2 - n is even. Case 2: Let n be odd. Then n = 2k +1 (def. of odd), where k is any positive integer. Then, n2 - n = (2k - 1)2 - (2k - 1) = 4k2 - 4k + 1 - 2k + 1 = 4k2 - 6k + 2 = 2(2k2 - 3k + 1); let 2k2 -3k + 1 = m = 2m Therefore, n2 - n is even. Therefore, for any natural number n, n2 - n is even.
When you add an odd and even number together is the sum odd or even?
It will always be odd. A proof: Call the even number m, call the odd number p. 'p' is equal to an even number 'n' + 1. Adding m and p is equal to m + n + 1; since m and n are both even, their sum is also even (see the related questions). m + n + 1 is one more than an even number, so it is odd, and therefore an odd number plus an even number is also always odd. QED
Can identify the odd and the even in the java script using HTML?
Here is a JavaScript option for determining if a number is odd or even. It even lets you know if the number is zero (if you want zero to be neither odd nor even). var n = prompt("Enter a number to identify as odd or even", "Type your number here"); n = parseInt(n); if (isNaN(n)) { alert("Please Enter a Number"); } else if (n == 0) { alert("The number is zero"); } else if (n%2) { alert("The number is odd"); } else { alert("The number is even"); }
If n plus 4 represents an odd integer the next larger number odd integer is represented by?
Every integer is either even (divisible by 2) or odd (not divisible by 2). Since an even number plus 1 is odd and an odd number plus one is even, because 1 does not divide 2. We know (n + 4) is odd. The next integer is (n + 4 + 1) = (n + 5), because an odd number plus 1 is even, (n + 5) is even. The integer after (n + 5) is (n + 6), since (n + 5) we know is even, (n + 6) must be odd. Since (n + 6) is the smallest integer that is greater than (n + 4) and is odd, so (n + 6) is the next odd integer.
What is an odd number minus odd number?
All odd numbers are of the form 2n + 1, where n is an integer.So an odd number minus an odd number is (2n+1) - (2m+1) = 2n -2m = 2(n-m). Both n and m are integers, so while we don't know whether n-m is odd or even, we definitely know that it's an integer and that multiplying it by two cannot possibly give an odd number. So an odd number minus an odd number is an even number. For similar reasons, an odd number plus an odd number is also an even number.
Is the sum of an even whole number and an odd whole number always odd?
Yes --------------------------------------------- Let n be an integer. Then 2n is an even number Let m be an integer. Then 2m is an even number and 2m + 1 is an odd number. Then: even + odd = (2n) + (2m + 1) = (2n + 2m) + 1 = 2(n + m) + 1 = 2k + 1 (where k = m + n) which is an odd number.
Why is the sum of 2 consecutive integers always an odd number?
The sum of two even numbers is always an even number. The sum of two odd numbers is always an even number.The sum of an even number and an odd number is always an odd number.Because integers alternate between even and odd, adding two consecutive integers will always result the sum of an odd number and an even number, which, as stated above, will always be an odd number.Another way to look at it:An even number is divisible by 2, therefore, 2*n (where n is an integer) is always an even number.2*n+1 would be the next number after 2*n (so 2*n & 2*n+1 are consecutive integers).If we add these numbers together, we get 4*n+1 = 2(2*n)+1. Since 2(2*n) is an even number, adding 1 makes it an odd number._________________________________________________________________________I'm not going to but in or anything, seriously that was a good answer
C program to accept a sequence of number terminated by zero and sum all odd nos and even nos?
#include <stdio.h> main() { int n, odd=0, even=0; while (scanf("%d",&n)&&(n!=0)) (n%2)?++odd:++even; printf("odd: %d even: %d\n",odd,even); }
Odd minus even?
Let's do a small experiment with a few numbers to see we can find out.3 - 2 = 1, Odd when the first number is odd and larger than the second number which is even and smaller.5 - 12 = -7, Odd when the first number is odd and smaller than the second number which is even and larger.Based on the Inverse commutative property, switching the first number for an even would yield the same number except it's inverse.(a - b) = c(b - a) = -c2 - 3 = -112 - 5 = 7So it looks that an odd number minus an even number results in an odd number.
Turbo c program that will determine if the input is odd or even number?
#include <stdio.h> int main() { printf("Program to find ODD or Even Number\n"); while(1) { int n = 0; printf("\nEnter a number(-1 for Exit): "); scanf("%d",&n); if( n 0) { printf("%d is a EVEN number.\n", n); } else { printf("%d is a ODD number.\n", n); } } return 0; }
Write a single line code in C to find a given number is even or odd?
printf ("%d is %s\n", n, n%2 ? "odd": "even");
The product of an odd and an even number always odd?
odd x even = even (2n + 1)(2n) = 2[n(2n + 1)] let n(2n + 1) = t = 2t (even)
