something wrong with irrational numbers

Question

$$(5-\sqrt2)(5+\sqrt2)=5^2-\left(\sqrt2\right)^2=25-2=23$$

But as some general rule of mathematics, if a rational number is added or subtracted from or to an irrational number the result is an irrational number but in above example there's something wrong

As per rules $(5-\sqrt2)$ and $(5+\sqrt2)$, both would be irrational and if we multiply both these irrational terms we get an irrational number which do not match to our result $23$.

How's that possible??

Answer 1

Yes, if a rational number is added or subtracted from or to an irrational number the result is an irrational number.

But if you multiply 2 irrational numbers the result isn't always irrational!

For example: $\sqrt{3}*\sqrt{3}=\sqrt{9}=3 $ and $3\in \Bbb{R}$.

Answer 2

if a rational number is added or subtracted from or to an irrational number the result is an irrational number

Added yes, but that's not what you're doing.

Note that the rational numbers are closed under addition and multiplication since: $$\frac{a}{b}+\frac{c}{d} = \frac{ad+bc}{bc} \in \mathbb{Q} \quad \mbox{and} \quad \frac{a}{b}\frac{c}{d} = \frac{ac}{bd} \in \mathbb{Q}$$

This doesn't hold for irrational numbers.

You are multiplying two irrational numbers and the result need not be irrational, since for example: $\sqrt{2}\sqrt{2} = 2 \in \mathbb{Q}$. The same goes for addition: $\left(1-\sqrt{2}\right)+\sqrt{2} = 1\in \mathbb{Q}$.

Answer 3

The rules are

• rational + rational $\to$ rational

• rational + irrational $\to$ irrational

• irrational + irrational $\to$ any (think of $a+(-a)=0$)

• rational $\times$ rational $\to$ rational

• rational $\times$ irrational $\to$ irrational

• irrational $\times$ irrational $\to$ any (think of $a\times a^{-1}=1$)