The statement is not true. Disprove by counter-example: 3 is an integer and 5 is an integer, their product is 15 which is odd.
The product of an odd and even number will always have 2 as a factor. Therefore, it will always be even.
To prove this, we need to be able to prove two things:The product of two integers is always even if at least one integer is evenThe product of two integers is never even if neither integer is evenFirst, let's prove that the product of two integers is always even if at least one integer is even:Let's say that n is an even integer, and that k is some integer (even or odd). We want to prove that k*n has to be even. Since even is even, we can think of it as 2m, where m is some integer. In other words, n is equal to (2 + 2 + 2 + 2 ...) for however many twos it takes to get there.We can re-write k*n as k*2m. We can factor out a 2 to get:2*k*2m-1.Since we are multiplying an integer quantity by two, this proves that the quantity has to be divisible by 2, which, by definition, means that the number must be even.Now let's prove that the product of two integers is never even if neither integer is even. Let's call the two integers k and n. Since nither k and n are even, they must both be odd.k*n can be re-written as k*n - n + n, which can be factored into (k-1)n + n.Now we know that (k-1)n has to be even, since k-1 is even, and we have proved that a number times an even number is even above.So now we have an even number plus and odd number. We can re-write n as 2m + 1, where m is an integer. Since (k-1)n is even, that quantity can be written as 2p, where p is an integer.To recap so far, this means that we can say that if k and n are both odd:k*n = (k-1)n + n = 2p + 2m + 1.This can be re-written as 2p+m + 1. Since 2p+m is a multiple of two, it must be even, and an even number plus one must be odd, so the product of two odd numbers must be odd.
Yes the product of two odd integers is odd. The proof lies in recognizing that 2 times an integer is an even integer. Like, given two arbitrary integers a and b, 2a+1 and 2b+1 are odd. And the product of (2a+1)(2b+1) can be represented as 2c+1, where c might be even or odd - it doesn't matter. c = 2ab+a+b, in fact (check it out.) However, 2c+1 is clearly an odd integer.
Another even integer.
A positive integer.
Yes 3320 is an even number as the product of dividing it by 2 is an integer.
Yes. Any even number can be expressed as 2x (where x is an integer). (2x)2 = 4x2. We don't know exactly what x2 is, and we don't need to know, because we know that the square of any integer is also an integer, and any integer multiplied by 4 is an even number.
Another even integer.
The product of two even numbers is always an even number.Here is the proof:We define an even number as a number of the form 2n for some integer n.Now let 2n be one even number and 2m be another.The product is (2n)(2m)=2(2mn) and of course 2mn is an integer since the integers are closed under multiplication. Hence, 2(2mn) is an even number.
581116 is an even integer.581116 is an even integer.581116 is an even integer.581116 is an even integer.
even integer.
The product of an even integer with any other integer is an even integer: (2a)*b =2ab = 2(ab) Any integer evenly divisible by 2 is even.