No, sqrt(5)*sqrt(16*5)=20. More trivially, there's always a number y such that z = x*y for a given irrational x. You can give similar examples for all the other basic operations.

Take any irrational a where 1/a is also irrational. Then a * 1/a = 1.

Even moving from addition and multiplication to exponentials won’t save you: there are irrational numbers to irrational powers that are raational.

> Take any irrational a where 1/a is also irrational.

In other words: any irrational at all

The nonconstructive proof of that is simple and fun: either sqrt(2)^sqrt(2) or (sqrt(2)^sqrt(2))^sqrt(2) is just such an example.

Definitely not; consider the formula for calculating the log of any base given only the natural logarithm. That can result e.g. in two irrational numbers, the ratio of which are integers.

pi and -pi are both irrational and their sum is zero.

There is an important distinction to be made here. Examples in this thread show cases of irrational numbers multiplied by or added to other irrational numbers producing real numbers, but in the special case of a rational number added to or multiplied by an irrational number, the result is always irrational.

Otherwise, supposing for instance that (n/m)x is rational for integers n, m, both non-zero, and irrational x, we can express (n/m)x as a ratio of two integers p, q, q non-zero: (n/m)x = p/q if and only if x = (mp)/(qn). Since integers are closed under multiplication, x is rational, against supposition; thus by contradiction (n/m)x is irrational for any rational r = (n/m), with integers n, m both non-zero. Similarly for the case of addition.

irrational number + rational number = irrational number [1]

irrational number + irrational number could be rational or irrational.

5 - sqrt(2) is irrational

sqrt(2) is irrational

Add them up you get 5, which is rational

[1] If it were rational, you will be able to construct a rational representation of the irrational number using this equation.

To put the first equation more formally, we know that ℚ is closed under addition¹, so given k∊ℝ\ℚ, l∊ℚ then if k+l=m∊ℚ, then m-l=m+(-l)∊ℚ, but m-l=k which is not in ℚ so k+l∉ℚ.

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1. For p,q∊ℚ, let p=a/b, q=c/d, a,b,c,d∊ℤ, then p+q=(ad+bc)/bd, but the products and sums of integers are integers, so p+q∊ℚ

x - (x - floor(x)) == x truncated to an integer

when x is an irrational number > 1:

"x - floor(x)" is just the fractional part of x, so it's an irrational number which is not equal to x.

Subtracting the fractional part from the original leaves only the integer part, which is obviously rational.