The introduction to proof book used by the university i'm going to is very hard for me to understand.

For instance(at the beginning of the chapter for the transition from rational to real numbers):

The author defines the set of rational numbers including $\sqrt2$ as

$\mathbb{Q}(\sqrt2)=\{a+b\phi:a,b \in \mathbb{Q}\}$

He then defines order in this set as:

$s+t\phi \lessdot u+v \phi $ when $|s-u|(s-u)<2|v-t|(v-t)$

He defines addition in the set as:

$(s+t \phi) \oplus(u+v\phi)=(s+u)+(t+v)\phi$

and multiplication as:

$(s+t \phi) \odot(u+v\phi)=(su+2tv)+(sv+tu) \phi$

At the end of the chapter exercise he asks

Q.Consider the field $\mathbb{Q}(\sqrt2)=\{a+b\phi:a,b \in \mathbb{Q}\}$

and the order on it given by:

$s+t\phi \lessdot u+v \phi $ iff $|s-u|(s-u)<2|v-t|(v-t)$

Then he asks questions like

Prove that if $a,b,c,d \in \mathbb{Q} $ then exactly one of the following holds:

$|a-c|(a-c)<2|d-b|(d-b);\\ |a-c|(a-c)>2|d-b|(d-b);\\ |a-c|(a-c)=2|d-b|(d-b).$

How am I supposed to answer these questions?

  • $\begingroup$ Comments are not for extended discussion; this conversation has been moved to chat. $\endgroup$ – Aloizio Macedo Jul 9 '19 at 18:04

The key point is that those expressions are the definitions of those symbols - you'll prove the things you need to by manipulating those definitions directly. (That said, see below the fold ...)

"$\bigoplus$" for example is the operation which takes in two things $s+t\phi$ and $u+v\phi$ and outputs the the thing $(s+u)+(t+v)\phi$, where the "$+$" is understood in the usual sense.

For example, here's a proof of the associativity of $\oplus$: we have $$[(a+b\phi)\oplus(m+n\phi)]\oplus(x+y\phi)=((a+m)+(b+n)\phi)\oplus(x+y\phi)$$


$$=(a+b\phi)\oplus[(m+n\phi)\oplus(x+y\phi)].$$ The key point to understand (and see also this old answer of mine) is that properties of the new operations/relations are being deduced using properties of the old operations/relations*: the black equalities come from applying the definition of $\oplus$, but the crucial step is the red equality, and that comes from the associativity of the already-understood operation $+$.

$^*$Now there is actually a vagueness here, but it's not around the definitions of the new symbols; rather, it's around what the elements of our new field literally are. The way the author has defined them makes it seem like we already know what $\mathbb{R}$ is, but then there's no need to define the new operations/relations since they're just the ones coming from $\mathbb{R}$. A fully formal approach would be something like:

  • An element of $\mathbb{Q}(\sqrt{2})$ is an ordered pair $\langle p,q\rangle$ with $p,q\in\mathbb{Q}$ (which we'll think of as "$p+q\sqrt{2}$").

  • Addition $\oplus$ of elements of $\mathbb{Q}(\sqrt{2})$ is defined by $$\langle p,q\rangle\oplus\langle a,b\rangle:=\langle p+a, q+b\rangle$$ where "$+$" is the usual notion of addition on $\mathbb{Q}$.

    • I've used "$:=$" for clarity to emphasize that the left hand side is being defined to be the right hand side.
  • Multiplication $\odot$ of elements of $\mathbb{Q}(\sqrt{2})$ is defined by setting $$\langle p,q\rangle\odot\langle a,b\rangle:=\langle p\cdot a+2\cdot (q\cdot b), p\cdot b+q\cdot a\rangle$$ where "$\cdot$" and "$+$" are the usual notions of multiplication and addition on $\mathbb{Q}$ respectively.

  • And so on.

  • 1
    $\begingroup$ It's particularly confusing to write $\langle p,q\rangle$ as "$p+q\phi$" where $\phi$ is apparently supposed to represent $\sqrt 2$ rather than $\frac12(1+\sqrt5)$. $\endgroup$ – hmakholm left over Monica Jul 8 '19 at 22:57
  • 1
    $\begingroup$ @HenningMakholm Oh yeah, I hadn't even thought of that. That is annoying indeed. $\endgroup$ – Noah Schweber Jul 8 '19 at 22:59
  • $\begingroup$ @NoahSchweber does the definition of the order in this case, $\endgroup$ – user686544 Jul 9 '19 at 0:07
  • 1
    $\begingroup$ @Eastcoastdancer Indeed. The goal is to give an equivalent condition to $s + t\sqrt{2} < u+v\sqrt{2}$ without using $\sqrt{2}$ as a number (since it hasn't been properly defined yet). $\endgroup$ – Sambo Jul 9 '19 at 3:26
  • 3
    $\begingroup$ @Eastcoastdancer Note that $|x|x = x^2 \operatorname{sgn}(x)$. The absolute values are here to allow the sign to matter in this expression. We do have that $s + t\sqrt{2} < u+v\sqrt{2}$ implies $(s-u)^2<(\sqrt2)^2(v-t)^2$, but the reverse implication is not true (take $s=v=0, t=1, u=-1$). To fix this we need to introduce the absolute value. $\endgroup$ – Sambo Jul 9 '19 at 3:38

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy