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

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??

lioness99a
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Sahil Singh
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  • How does multiplying the two numbers give you a irrational number? – Soby Jun 14 '17 at 07:36
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    So according to your reasoning $\sqrt{2}\sqrt{2}=2$ must be irrational!!! – Anurag A Jun 14 '17 at 07:39
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    The numbers $\pi$ and $1/\pi$ are both irrational, but their product is $1$. – user49640 Jun 14 '17 at 07:44
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    Would love to see the source of this so-called rule – lioness99a Jun 14 '17 at 07:46
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    The only rule in this context, really, is that if $a+b = c$ or $ab = c$ (with $c\neq 0$), then we cannot have that exactly one of $a, b, c$ are irrational. We can have that none of them are, and we can have that two or three of them are, but not one. – Arthur Jun 14 '17 at 07:50
  • "as some general rule of mathematics": what rule? – Klangen Jun 14 '17 at 07:54
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    "...if a rational number is added or subtracted from or to an irrational number..." You didn't do this. What you did was to multiply an irrational number by another irrational number. – nickgard Jun 14 '17 at 07:57

3 Answers3

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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: $\left(1-\sqrt{2}\right)+\sqrt{2} = 1\in \mathbb{Q}$.

StackTD
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2

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 $a\times a^{-1}=1$)

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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 $3\in \Bbb{R}$.

Super-kenguru
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