Ask Uncle Colin: Some number theory

Dear Uncle Colin,

I need to show that $\sqrt{7}$ is in $\mathbb{Q}[\sqrt{2}+\sqrt{3}+\sqrt{7}]$ and I don't really know where to start.

We Haven't Approached Tackling Such Questions

Hi, WHATSQ, and thanks for your message!

I am absolutely not a number theorist, although I must admit to getting a bit curious about it recently.

What does that notation mean?

In number theory, you can extend a field1 by throwing extra numbers or symbols into it. For example, the complex numbers can be written as $\mathbb{R} [i]$ - the real numbers, with the extra number $i$ added in such that $i^2 = -1$. It's all of the numbers $z = a + bi$, with $a$ and $b$ in the real numbers.

Another example: $\mathbb{Q}[\sqrt{2}]$ is the extension of the rational numbers to include the number $\sqrt{2}$ - that is, all of the numbers $z$ that can be written as $a + b\sqrt{2}$, with $a$ and $b$ both rational.

We have $\mathbb{Q}[\sqrt{2} + \sqrt{3}+\sqrt{7}]$, which contains all of the numbers that can be written as $z = a + b\br{\sqrt{2} + \sqrt{3}+\sqrt{7}}$ with $a$ and $b$ being rational.

This extended field is also a field -- which means (in particular) that any power of the extended number also belongs to the field, and that's how we're going to attack this problem.

Messing around and seeing what we get

We know that $s = \sqrt{2} + \sqrt{3}+\sqrt{7}$ is in the field $\mathbb{Q}[\sqrt{2} + \sqrt{3}+\sqrt{7}]$, by definition - and I'm going to call the field $F$ to save me typing it out every time.

Because $F$ is a field, $s^2 \in F$ as well; that is $s^2 = 12 + 2\br{\sqrt{6}+\sqrt{14}+\sqrt{21}}$, which doesn't look like it helps us much. However, we can use algebraic operations to say $\frac{s^2 - 12}{2} = \sqrt{6}+\sqrt{14}+\sqrt{21}$ is also in $F$. Let's call that number $t$.

Now, $t^2 \in F$ as well, and that works out to be $41 + 2\br{\sqrt{84} + \sqrt{126} +\sqrt{294}}$, which is $41 + 2\sqrt{42}\br{\sqrt{2}+\sqrt{3}+\sqrt{7}}$.

Now we're getting somewhere: this tells us that $\sqrt{42}\br{\sqrt{2}+\sqrt{3}+\sqrt{7}}$ is in the field. Dividing by $\sqrt{2}+\sqrt{3}+\sqrt{7}$ (which is a non-zero element of the field) tells us that $\sqrt{42} \in F$.

Keep on messing!

If we multiply $t$ by $\sqrt{42}$, we get $21 \sqrt{2} + 14\sqrt{3} + 6 \sqrt{7}$.

If we subtract $6s$ from this, we find that $15\sqrt{2} + 8 \sqrt{3}$ is in $F$.

Squaring this gives $642 + 240\sqrt{6}$ - which means that $\sqrt{6} \in F$.

And finally, since $\sqrt{42}$ and $\sqrt{6}$ are both in $F$, so is $\frac{\sqrt{42}}{\sqrt{6}} = \sqrt{7} \blacksquare$

I don't know that that's the simplest way to do it - I'd be delighted to hear of a less convoluted method! - but I hope it helps all the same.

- Uncle Colin

Colin

Colin is a Weymouth maths tutor, author of several Maths For Dummies books and A-level maths guides. He started Flying Colours Maths in 2008. He lives with an espresso pot and nothing to prove.

  1. A set of numbers where you have the four basic operators defined and following the usual rules of arithmetic []

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6 comments on “Ask Uncle Colin: Some number theory

  • Barney Maunder-Taylor

    Doesn’t look convoluted to me. Thanks Colin, I did not know about the extensions like C=R(i) but I do now!

  • Tom

    Thanks for this – I enjoyed that and I understood it, which always makes me feel a little bit cleverer than I often do.

    One question that did pop into my head though – I remember bits of number theory (and the notation) from my studies long ago, but I couldn’t remember how to “say” some of this in my head.

    For example, I totally get the notation for the rationals being extended by some values as a blackboard Q followed by those values in square brackets, but I’m not sure what a recognisable verbal/ mental shorthand for this is (as “a blackboard Q followed by some values in square brackets” is a bit cumbersome.)

    What should my brain be saying to me when I read it?

    Thanks!

    • Colin

      No idea, I’m afraid! I shall ask the Twitters.

    • Evelyn

      I think people usually use the term adjoin. It’s still a bit cumbersome, but “Q adjoin root 2 plus root 3 plus root 7” is probably what I’d say. I’ve more often seen Q adjoin just one root, so “Q adjoin the square root of 2” is a phrase that I have heard more frequently.

      • Colin

        Thanks, Evelyn!

      • Tom

        That’s great, thank you Evelyn!

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