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Is the odd one?
Any number of the form 2*n+1 where n is an integer.
Which is the odd one?
Any number of the form 2*n+1 where n is an integer.
Definition of odd number?
An odd number is an integer of the form 2*n+1 where n is an integer. Equivalently, an odd number is an integer which leaver a remainder of 1 when divided by 2.
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.
Is the sum of one odd number and one even number even or odd prove and explain in words?
The sum of an odd and an even number is odd. Any odd number can be expressed as 2n + 1 (for some integer "n"). Any even number can be expressed as 2m (for another integer, "m"). The sum of the two is 2(m+n) + 1. Since the expression in parentheses is an integer, multiplying it by 2 gives you an even number. Adding 1 makes the entire expression odd.
What is the uneven number?
An uneven number, more commonly referred to as an odd number, is any integer that cannot be evenly divided by 2. This means that when you divide an odd number by 2, there will always be a remainder of 1. Examples of odd numbers include -3, 1, 5, and 21. In general, any integer of the form (2n + 1) (where (n) is an integer) is considered an odd number.
What would be an algebraic representation of an odd number?
Assuming that n is an integer, 2n + 1 is an odd number.
Why is the sum of an even and odd number alsways odd?
A number a is even if there exists an integer n such that a = 2n A number b is odd if there exists an integer m such that b = 2m + 1. So: a+b = (2n) + (2m +1) = 2 (n+m) + 1 Since n and m are integers, n+m is also an integer. So a+b satisfies the definition of an odd number.
How is even times odd equal even?
In mathematics, an even number is defined as any integer that can be expressed as (2n), where (n) is an integer, while an odd number can be expressed as (2m + 1) for some integer (m). When you multiply an even number by any integer (even or odd), the result is still even, because the product will still have a factor of 2. Therefore, when you multiply an even number by an odd number, the product remains an even number. Thus, even times odd equals even.
What is a odd number in algebra?
If n is equal to any natural number, then 2n-1 = Odd. This is algebraic presentation of add numbers. Traditionaly (non-algebraic) an odd number is an integer that is not divisible by 2.
When you multiply an odd number and an even number is your answer always odd?
No, it's always even, and here's the proof: All even numbers can be expressed as 2n, where n is any integer. All odd numbers can be expressed as 2p + 1, where p is any integer. Multiply those two together: 2n(2p + 1) = 2(2np + n). Since both 2np and n are integers, that means 2np + n is an integer; and since that integer is being multiplied by 2, it must be even.
How would you prove that the sum of an even number and an odd number is always an odd number?
Suppose x is an even number and y is an odd number. Then x = 2*n for some integer n and y = 2*m + 1 for some integer m Therefore x + y = 2*n + 2*m + 1 = 2*(n + m) +1 Now, since n and m are integers, (n + m) is also an integer [by the closure of integers under addition]. Thus, x + y = 2*p + 1 where p = n + m is an integer. ie x + y is an odd integer.
