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The easiest way to do this is to use a whole number and the square root of a number. Having the square root of a number present makes the total number irrational. So let's define X and Y and see what we get.

X = 4 + √3

Y = 4 - √3

X + Y = (4 + √3) + (4 - √3) = 4 + 4 + √3 - √3 = 8 + 0 = 8

X * Y = (4 + √3) * (4 - √3) = 4 * 4 + 4√3 - 4√3 + √3 * -√3

= 16 + 0 -3 = 13

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lol, we posted at the same time!

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One pair is 1 + sqrt(2) and 1 - sqrt(2). Their sum is 2 and their product is -1. You can come up with infinitely many pairs of irrational numbers that have a rational sum and product with the same form as above. Just take any integer c and any positive non-square integer d. The pair

c + sqrt(d)

c - sqrt(d)

add up to 2c and multiply out to c^2 - d, both of which are rational numbers (integers in fact). Now that I think of it, c and d don't have to be whole numbers; they could also be fractions so long as sqrt(d) does not equal another fraction.

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Thank you for the clear explanation.

The general solution would be any pair of numbers (A, B) where A is a complex number composed of a real number plus the square root of a real number, and b is the same complex number minus the square root of the same number, so long as the square root is an irrational number. Note that A does not need to be an integer, it can be a fraction.

The simplest would be where the real number is zero: sqrt(2) + -squrt(2) = 0, and sqrt(2) * -sqrt(2) = 2.

You can't use numbers like sqrt(4) because they are not irrational numbers.

You can use odd things like (19.5 + squrt(2)) and (19.5 - sqrt(2)), as you get a ratiional number (a fraction) from the sum and from the product.

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Thank you for mentioning possible solutions in the complex numbers.