No. The easiest counter-example to show that the product of two irrational numbers can be a rational number is that the product of √2 and √2 is 2. Likewise, the cube root of 2 is also an irrational number, but the product of 3√2, 3√2 and 3√2 is 2.
An even number
Reciprocals, like 2 and 1/2, have a product of 1.
4*(n + 2)
Twice that number!
The first prime number is 2. The product of 2, since there is no other number by which to multiply it, is 2.
There is no such number. If someone claimed that x was such a number then x+2 would be (a) a greater number and (b) a product of 2. So then x could not be such a number.
No. The easiest counter-example to show that the product of two irrational numbers can be a rational number is that the product of √2 and √2 is 2. Likewise, the cube root of 2 is also an irrational number, but the product of 3√2, 3√2 and 3√2 is 2.
A product is the answer to a multiplication problem. In the problem 2 x 2 = 4, 4 is the product.
product of prime number of 96 = 2 x 2 x 2 x 2 x 2 x 3
The product is a^2 + b^2.
An even number
The answer to this convoluted question is as follows: The product of prime numbers, for the composite number 64, is 2*2*2*2*2*2 or 26
Reciprocals, like 2 and 1/2, have a product of 1.
The number 76 as a product of its prime factors is: 2 x 2 x 19
The number 4 is a product of 2 x 2, for example.
4*(n + 2)