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all perfect square numbers
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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.
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.
The sum of four consecutive odd integers is -72 Write an equation to model this situation and find the values of the four integers?
let n = first odd number; n + 2 = 2nd odd number ,etc. n + n+2 + n+4 + n +6 = 4n + 12 = -72 4n = -84 n = -21 -21,-19,-17 and -15
Why odd number minus a odd number even?
Every odd number leaves a remainder of 1 when divided by 2. Therefore, every odd number is of the form 2k +1 where k is some integer.Suppose 2m + 1 and 2n + 1 are two odd numbers.Then (2m + 1) - (2n + 1) = 2m + 1 - 2n - 1 = 2m - 2n = 2*(m - n)By the closure property of integers under addition (and subtraction), m and n being integers implies than (m - n) is an integer. Therefore 2*(m - n) is an even integer.
The n in x'' indicatiing the number of factors of x?
exponent
What is an odd number of factors?
Perfect squares have an odd number of factors. If f is a factor of N, then so is N/f. (For example, 3 is a factor of 30, and so is 10). Thus, it seems factors come in pairs, and there should always be an even number of them. But if f = N/f, then these two factors are the same, resulting in an odd number of factors. For example, 3 is a factor of 9, and it's "mate" is also 3. So for the numbers with an odd number of factors, there is some f where f=N/f. Multiplying both sides by f, we have f^2 = N. So this happens when N is a perfect square.
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"); }
Choose the true biconditional statement that can be formed from the conditional statement If a natural number n is odd then n2 is odd and its converse.?
An integer n is odd if and only if n^2 is odd.
What is the six greatest odd number?
There is no such number. Given any odd number, n, the number (n + 2) is a greater odd number. You can go on, for ever, finding larger odd numbers.
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.
Why when you add odd and even numbers the sum is odd?
Factors can be listed in pairs. Square numbers will have one pair that is the same number. When put into a list, that number will only be listed once and will result in an odd number of factors.
Why is 23 a odd number?
Because it isn't divisible by 2. Or..Because it's not even. An even number has 2 as one of it's factors (same as divisible by 2), or you could say that an even number can be determined by 2*n, where n is a positive integer, while an odd number can be determined by 2*n - 1, where in is a positive integer. Even (2*n) ---> 2,4,6,8,10,12,.....
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.
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.
How many positive factors square numbers have?
Any positive odd number.For example for p = 17 positive factors, consider n^(p-1) for any integer n.Since p is odd, p-1 is even and so n^(p-1) is a perfect square number.
Is 1 even or odd its either?
It is odd. The definition of an odd number is "a number a is odd if there exists an integer n such that a = 2n + 1." Let a = 1 and let n = 0, so 1 = (2x0) + 1.
