Answer to Question #95978 in Mechanics | Relativity for Bsaad Abdul Aziz

Question

Show that numbers of the form a squared plus b squared, where a and b are integers are closed Under multiplication

Answer 1

Suppose that the given numbers a and b are fractional, then will be fractional a squard and b squard.

 Therefore, their sum is also a fractional number.

The operation of the product of fractional numbers will also return the result to the set of fractional numbers.

Therefore, the initial condition is true only if the entered variables are integer.

Provided that the initial numbers are integers, we use the following identity:

$(A+B)^2=A^2+2AB+B^2, AB=((A+B)^2-A^2-B^2)/2.$ ,

.

,

If the given numbers are even, then the learned equality will be equal to the integer.

Let a=2n, b=2n+1, then

 $AB= ((2N+2N+1)^2-4N^2-(2N+1)^2)/2=((4N+1)^2-4N^2-4N^2-4N-1)/2=(16N^2+8N+1-8N^2-4N-1)/2=(8N^2+4N)/2=4N^2+2N.$

Arguing in a similar way, we can prove the statement for the case of two odd numbers.