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I'm having trouble understanding the following set:

$$\mathbb{Z}[\sqrt{2}] := \{ a + \sqrt{2}b : a,b \in \mathbb{Z}\}$$ What elements does it actually contain? I thought $\mathbb{Z}$ only contains integers, so what does the root of $2$ behind $\mathbb{Z}$ mean?

Thank you.

Chubby Chef
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HORZ
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  • It is the set of all integers or integer multiples of the square root of two or sums of integers and integer multiples of square root of two. – JMoravitz Dec 08 '20 at 13:33
  • Compare to the polynomial ring $\Bbb Z[x]$ which contains terms of the form $a_0+a_1x+a_2x^2+a_3x^3+a_4x^4+\dots+a_kx^k$, the set of polynomials with integer coefficients. $\Bbb Z[\sqrt{2}]$ is the same, just replacing $x$ in the above with $\sqrt{2}$, but since $(\sqrt{2})^2=2$ the even terms can all combine and all the odd terms can combine, hence why every element can be written as the sum of up to two terms. – JMoravitz Dec 08 '20 at 13:42
  • Compare also to $\Bbb Z[\sqrt[3]{2}]$ which is the set of terms of the form $a_0+a_1\sqrt[3]{2}+a_2\sqrt[3]{4}$ – JMoravitz Dec 08 '20 at 13:43
  • the comparison to the polynomial ring made me understand it even better, thank you for your help! – HORZ Dec 08 '20 at 13:53

1 Answers1

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You are right: $\mathbb{Z}$ only contains integers. The set displayed here, though, is a subset of $\mathbb{R}$ and can thus contain $\sqrt{2}$.

The notation says it contains all elements of the form $a+\sqrt{2}b$, where $a$ and $b$ are integers. Let me give you a few examples:

$a=b=1$: Then $a+\sqrt{2}b = 1+\sqrt{2}$.

$a=b=0$: Then $a+\sqrt{2}b = 0$.

$a=3, b=-2$: Then $a+\sqrt{2}b = 3-2\sqrt{2}$.

$b = 0$: Then $a+\sqrt{2}b = a$ and because of this any integer is in your set. But as you have seen also nonintegers like $1+\sqrt{2}$ can be part of that set.

I hope this helps clearing things up a bit.

J. W. Tanner
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Guenterino
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