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wat is the answer to p(3
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Are all perfect square number are even numbers?
No, all perfect square numbers are not even numbers. Eg. the square of 3 is 9. (32=9) To generalize the proof: If p is odd then p=2n+1 and p2=(2n+1)2=4n2+4n+1=2(2n2+2n)+1 So odd numbers have odd square
What is an odd number multiplied by an odd number?
A definition of an odd number is 2n-1 where n is any integer. The product of two odd numbers is thus (2n-1)(2p-1) for any numbers n and p. Expanding the expression yields 2(2pn -(p+n)) + 1. The term on the left is even by definition. Adding 1 to any even number is, by definition, an odd number. Therefore, the conclusion is elementary (Watson); the product of two two odd numbers is an odd number
Is p2 plus p an odd or even number?
Even
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.
Why do two even numbers always equal an even number?
On a number line, adding an even number to another number (or zero) results in an even displacement, which must end in the same type of number as the original. If the beginning number is odd, adding an even number produces an odd sum. If the beginning number is even, adding an even number produces an even number. A corollary is that: Adding two like numbers produces an even number. Adding two unlike numbers produces an odd number. ---------------------------------- Algebraically: let x be an even number, Then y = 2x for some value of x Two even numbers would be 2m and 2n Adding them gives: 2m + 2n = 2(m + n) = 2p where p = m + n; 2p is of the form y = 2x, so 2p is an even number. Thus adding two even numbers results in an even number. Similarly for odd numbers: If y = 2x is an even number then z = y + 1 = 2x +1 is an odd number. Two odd numbers would be 2m+1 and 2n+1 Adding them gives: 2m+1 + 2n + 1 = 2m + 2n + 2 = 2(m + n + 1) = 2p where p = m + n + 1 Thus adding two odd numbers results in an even number. Similarly for one even and one odd number. An even number would be 2m and an odd number would be 2n+1 Adding them gives: 2m + 2n + 1 = 2(m + n) + 1 = 2p + 1 where p = m + n Thus adding an even number and an odd number results in an odd number.
What is the average of two odd numbers?
Let us say two odd numbers are p and q. And we know that sum of two odd numbers is always even. But the average of two odd numbers is not always even. Average of p and q = (p+q)/2 = even number/2 = odd or even. Two examples will make it clear. e.g. Average of 5 and 1 = (5+1)/2 = 6/2 = 3(odd number) Average of 3 and 5 = (3+5)/2 = 8/2 = 4(even number) So, average of two odd numbers can be even or odd.
Are all perfect square number are even numbers?
No, all perfect square numbers are not even numbers. Eg. the square of 3 is 9. (32=9) To generalize the proof: If p is odd then p=2n+1 and p2=(2n+1)2=4n2+4n+1=2(2n2+2n)+1 So odd numbers have odd square
Which pair of numbers has an odd product and an even sum 6 9 12 15?
Must be the 2 odd numbers, 9 & 15 (P = 135, S = 24)
Is odd times odd always odd?
Yes, it's always odd, and here's the proof: All odd numbers can be expressed as 2p + 1, where p is any integer. Multiply two of those together: (2n + 1)(2p + 1) = 4np + 2n + 2p + 1 = 2(np + n + p) + 1. Since both np, n, and p are integers, that means np + n + p is an integer; and since that integer is being multiplied by 2, it must be even. Thus, by adding 1 to that even number, the result will be odd.
What is an odd number multiplied by an odd number?
A definition of an odd number is 2n-1 where n is any integer. The product of two odd numbers is thus (2n-1)(2p-1) for any numbers n and p. Expanding the expression yields 2(2pn -(p+n)) + 1. The term on the left is even by definition. Adding 1 to any even number is, by definition, an odd number. Therefore, the conclusion is elementary (Watson); the product of two two odd numbers is an odd number
How do you prove that if the sum of two prime numbers is prime then one of the numbers is 2?
We know that a prime number is a positive integer greater than 1, whose divisors are 1 and itself. We know that the only even prime number is 2. That means that all other prime numbers are odd numbers.We know that when we add two odd numbers the result is an even number, which are not prime numbers (expect 2, and 2 = 1 + 1 where 1 is odd but is neither prime nor composite). Thus adding two odd prime numbers cannot give us another prime number. We show that the conclusion follows from the premise:Assume that r = p + q where all r, p and q are prime numbers, then we have that ris either even or not:* If r is even then r is at least 4 (the smallest number which is the sum of two primes) and thus not a prime number. This contradicts the assumption that r is a prime number, and therefore we conclude that r is not even. * If ris odd then either p or q must be odd and the other one must be even, since both p and q are prime numbers one of them must be 2 (the only even prime number). 2 + 3 = 5, 2 + 17 = 19, are examples of such numbers. See http://en.wikipedia.org/wiki/Twin_prime for more.
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 p2 plus p an odd or even number?
Even
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.
Is it true that the sum of two sequential prime number number is not prime?
We know that a prime number is a positive integer greater than 1, whose divisors are 1 and itself. We know that the only even prime number is 2. That means that all other prime numbers are odd numbers.We know that when we add two odd numbers the result is an even number, which are not prime numbers (expect 2, and 2 = 1 + 1 where 1 is odd but is neither prime nor composite). Thus adding two odd prime numbers cannot give us another prime number.We show that the conclusion follows from the premise:Assume that r = p + q where all r, p and q are prime numbers, then we have that r is either even or not:If r is even then r is at least 4 (the smallest number which is the sum of two primes) and thus not a prime number. This contradicts the assumption that r is a prime number, and therefore we conclude that r is not even.If r is odd then either p or q must be odd and the other one must be even, since both p and q are prime numbers one of them must be 2 (the only even prime number).2 + 3 = 5, 2 + 17 = 19, are examples of such numbers.
Why do two even numbers always equal an even number?
On a number line, adding an even number to another number (or zero) results in an even displacement, which must end in the same type of number as the original. If the beginning number is odd, adding an even number produces an odd sum. If the beginning number is even, adding an even number produces an even number. A corollary is that: Adding two like numbers produces an even number. Adding two unlike numbers produces an odd number. ---------------------------------- Algebraically: let x be an even number, Then y = 2x for some value of x Two even numbers would be 2m and 2n Adding them gives: 2m + 2n = 2(m + n) = 2p where p = m + n; 2p is of the form y = 2x, so 2p is an even number. Thus adding two even numbers results in an even number. Similarly for odd numbers: If y = 2x is an even number then z = y + 1 = 2x +1 is an odd number. Two odd numbers would be 2m+1 and 2n+1 Adding them gives: 2m+1 + 2n + 1 = 2m + 2n + 2 = 2(m + n + 1) = 2p where p = m + n + 1 Thus adding two odd numbers results in an even number. Similarly for one even and one odd number. An even number would be 2m and an odd number would be 2n+1 Adding them gives: 2m + 2n + 1 = 2(m + n) + 1 = 2p + 1 where p = m + n Thus adding an even number and an odd number results in an odd number.
When you add an odd and even number together is the sum odd or even?
It will always be odd. A proof: Call the even number m, call the odd number p. 'p' is equal to an even number 'n' + 1. Adding m and p is equal to m + n + 1; since m and n are both even, their sum is also even (see the related questions). m + n + 1 is one more than an even number, so it is odd, and therefore an odd number plus an even number is also always odd. QED
