n + 1
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1. let n = the odd integer then n + 2 must be the next odd integer 2. n + (n + 2) = 56 3. 2n + 2 = 56 4. 2n = 54 5. n = 27 6. 27 + 2 = 29 is the next odd integer
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.
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.
2n + 2 = 2(n+1)
Why Zero is EvenDivision : By definition, an even number is divisible by 2, with no remainder. 0/2 = 0 with no remainder.Number Line : on the integer number line, even and odd numbers alternate, and any odd number increased or reduced by 1 is an even number.Addition : additive rules state that even + even = even, and odd + odd = even. Under these rules, zero can be even, but cannot be odd.With the obvious solution n=0, zero fits the criteria that for any integer n, the integer 2n is even. (multiplication or division by 2)