In fact, the statement is true. Consequently, there is not a proper counterexample. The fallacy is in asserting that a terminating decimal is not a repeating decimal.
First, there is the trivial argument that any terminating decimal can be written with a repeating string of trailing zeros. But, Cantor or Dedekind (I can't remember which) proved that any terminating decimal can also be expressed as a repeating decimal.
For example, 2.35 can be written as 2.3499...
Or 150,000 as 149,999.99...
Thus, a terminating decimal becomes a recurring decimal. As a consequence, all real numbers can be expressed as infinite decimals. And that proves closure under addition.
Division by 0, which can also be written as 0.000... (repeating) is not defined.
There cannot be a counterexample since the assertion is true. This requires you to accept the true fact that the terminating decimal 1.25, for example, is equivalent to the repeating decimal 1.25000... (or even 1.24999.... ).
There is no counterexample because the set of whole numbers is closed under addition (and subtraction).
-2 - (-5) = -2 + +5 = +3. (+3 is not in the set of negative numbers.)
You can give hundreds of examples, but a single counterexample shows that natural numbers are NOT closed under subtraction or division. For example, 1 - 2 is NOT a natural number, and 1 / 2 is NOT a natural number.
No; here's a counterexample to show that the set of irrational numbers is NOT closed under subtraction: pi - pi = 0. pi is an irrational number. If you subtract it from itself, you get zero, which is a rational number. Closure would require that the difference(answer) be an irrational number as well, which it isn't. Therefore the set of irrational numbers is NOT closed under subtraction.
Yes they are closed under multiplication, addition, and subtraction.
Yes. They are closed under addition, subtraction, multiplication. The rational numbers WITHOUT ZERO are closed under division.