Y = 3.
Odd divided by even will always be a decimal. The answer will not be an integer.
An even number can be divided by 2 evenly. An odd number will have a remainder of 1 when divided by 2. Even plus odd is odd.
No. In fact, no odd integer is irrational. They are all rational.
Yes, the product of an odd integer and an even integer is always even. This is because an even integer can be expressed as 2 times some integer, and when multiplied by any other integer (odd or even), the result will still be a multiple of 2, thus making it even. For example, multiplying 3 (odd) by 4 (even) gives 12, which is even.
The difference between two odd numbers is always even because odd numbers can be expressed in the form of (2n + 1), where (n) is an integer. When you subtract one odd number from another, the equation looks like this: ((2m + 1) - (2n + 1) = 2m - 2n = 2(m - n)). Since (m - n) is an integer, the result is a multiple of 2, which defines an even number. Thus, the difference is always even.
It cannot be done. The basic rules of math. odd integer plus odd integer = even integer. odd integer plus even integer = odd integer. Always. odd integer plus odd integer plus odd integer = odd integer. Always.
Odd divided by even will always be a decimal. The answer will not be an integer.
The product of an odd and even number will always have 2 as a factor. Therefore, it will always be even.
An even number can be divided by 2 evenly. An odd number will have a remainder of 1 when divided by 2. Even plus odd is odd.
No. In fact, no odd integer is irrational. They are all rational.
Yes, the product of an odd integer and an even integer is always even. This is because an even integer can be expressed as 2 times some integer, and when multiplied by any other integer (odd or even), the result will still be a multiple of 2, thus making it even. For example, multiplying 3 (odd) by 4 (even) gives 12, which is even.
The difference between two odd numbers is always even because odd numbers can be expressed in the form of (2n + 1), where (n) is an integer. When you subtract one odd number from another, the equation looks like this: ((2m + 1) - (2n + 1) = 2m - 2n = 2(m - n)). Since (m - n) is an integer, the result is a multiple of 2, which defines an even number. Thus, the difference is always even.
An even integer is a multiple of 2 so that if x is the even integer then there is some other integer a such that x = 2a. An odd integer is one which, when you divide it by 2, leaves a remainder of 1. That is, if y is the odd integer, then y = 2b + 1 for some other integer b. Now, the sum of the even and odd integer is x + y = 2a + 2b + 1 = 2(a+b) + 1 By the closure of integers under addition, a and b are integers implies that a+b is an integer. So 2(a+b) is even and so the sum is odd.
all even numbers can be written in the form 2k where, k is an integer. all odd numbers can be written in the form 2j+1, where j is an integer. 2k + 2j + 1 = 2(k+j) + 1. You can see that this is in the form of an odd number. hence, the sum of an odd number and an even number is always odd. Test the theory. Pick any even and any odd number. Add them. The total will always be odd.
Yes - while multiplying odd numbers by even numbers will always produce an even result.
When we add an even number and an odd number, the result is always an odd number. This is because an even number can be expressed as 2n (where n is an integer), and an odd number can be expressed as 2m + 1 (where m is also an integer). Adding these two forms results in 2n + (2m + 1) = 2(n + m) + 1, which is the definition of an odd number. Therefore, the sum is always odd.
Adding four odd integers will produce an even integer, and 13 is odd; so it can't be done.