Question

3 a solution to an equation is $-2 + \sqrt{3}$. explain how adding 2 to the expression can help you know that the expression is irrational.

Explanation:

Step1: Add 2 to the expression

Take the expression \(-2 + \sqrt{3}\) and add 2: \((-2 + \sqrt{3}) + 2\). Simplifying this, the \(-2\) and \(+2\) cancel out, leaving \(\sqrt{3}\).

Step2: Recall properties of irrational numbers

We know that \(\sqrt{3}\) is an irrational number (a number that cannot be expressed as a fraction \(\frac{p}{q}\) where \(p\) and \(q\) are integers and \(q
eq0\)). Also, the set of rational numbers is closed under addition (adding a rational number to another rational number gives a rational number). The number 2 is rational (since \(2=\frac{2}{1}\)).

Step3: Use closure property to deduce

If \(-2 + \sqrt{3}\) were rational, then adding 2 (a rational number) to it would give a rational number (because of closure of rationals under addition). But we just saw that adding 2 to \(-2 + \sqrt{3}\) gives \(\sqrt{3}\), which is irrational. Therefore, \(-2 + \sqrt{3}\) must be irrational.

Answer:

By adding 2 to \(-2 + \sqrt{3}\), we get \(\sqrt{3}\) (an irrational number). Since 2 is rational and the sum of a rational and a rational is rational (closure property), if \(-2 + \sqrt{3}\) were rational, \(\sqrt{3}\) would be rational (which it's not). Thus, \(-2 + \sqrt{3}\) is irrational.