If the two numbers are x and y, thenthe square of the two numbers is one => (xy)^2 = 1 which implies that xy = +/-1.
Then xy is a maximum => xy = 1.
The product of four consecutive integers is always one less than a perfect square. The product of four consecutive integers starting with n will be one less than the square of n2 + 3n + 1
To find the numbers that maximize the product p, we can use the formula for a quadratic equation: x = -b / 2a. Let's call one number x and the other number (120-x). Therefore, the equation becomes x(120 - x^2), which simplifies to -x^3 + 120x. We can find x by setting the derivative equal to zero, resulting in x = 10. Therefore, the two numbers that maximize the product are 10 and 110, with a product of 12100.
Numbers whose product is one is called multiplicative inverses.
Prove the opposite.Assume that a square number is prime.A square number is one that is a product of a number multiplied by itselfA prime number is one that has no factors other than itself and 1.For a prime number to be square, the only choice is for it to be 1*1=1Since 1 is not a prime number, there is a contradiction, and the original premise is false.Therefore, all square numbers must be composite.â–
If the two numbers are a and b, then: √a √b = √(ab) = √36 → ab = 36 So any two [positive] numbers whose product is 36 will be a solution, except the pair 1 and 36 as that has been specifically excluded; thus possible solutions are: √4 x √9, √2 x √18, √(1/2) x √72, √(1/4) x √144, etc. (The first one being the only solution where the square roots are whole numbers.)
The product of four consecutive integers is always one less than a perfect square. The product of four consecutive integers starting with n will be one less than the square of n2 + 3n + 1
For the product to be zero, one of the numbers must be 0. So the question is to find the maximum sum for fifteen consecutive whole numbers, INCLUDING 0. This is clearly achived by the numbers 0 to 14 (inclusive), whose sum is 105.
To find the numbers that maximize the product p, we can use the formula for a quadratic equation: x = -b / 2a. Let's call one number x and the other number (120-x). Therefore, the equation becomes x(120 - x^2), which simplifies to -x^3 + 120x. We can find x by setting the derivative equal to zero, resulting in x = 10. Therefore, the two numbers that maximize the product are 10 and 110, with a product of 12100.
If the product of 2 numbers is one, than those 2 numbers are recipricals
Numbers whose product is one is called multiplicative inverses.
Prove the opposite.Assume that a square number is prime.A square number is one that is a product of a number multiplied by itselfA prime number is one that has no factors other than itself and 1.For a prime number to be square, the only choice is for it to be 1*1=1Since 1 is not a prime number, there is a contradiction, and the original premise is false.Therefore, all square numbers must be composite.â–
If the two numbers are a and b, then: √a √b = √(ab) = √36 → ab = 36 So any two [positive] numbers whose product is 36 will be a solution, except the pair 1 and 36 as that has been specifically excluded; thus possible solutions are: √4 x √9, √2 x √18, √(1/2) x √72, √(1/4) x √144, etc. (The first one being the only solution where the square roots are whole numbers.)
Not necessarily, but at least ''one'' of them must be even, unless the two numbers are both the (irrational) square root of the same even number.
The numbers are reciprocals of one another.
You should be able to find this out quickly with trial and error. Or One more than the product will be the square of the even number between the two odds.
1
The attribute that they have one square root which belongs to the set of natural numbers.