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I have decided to set up a bounty for one outstanding solution to this problem. The winner with the most clear, confident, solution manual-esque response will be awarded 50 of my participation points. The answer must be clear and unambiguous. You should attempt to solve the problem as if this were a midterm/final exam environment. It must show me clearly that either my attempt was completely off or on the right track. Good luck!

I am completely lost on how to answer this question. I have a feeling this is somewhat of an awkward proof. If I am missing something all hints will help!

Prove that for all $n \in \mathbb{N}$, if $a_1 \in F,a_2 \in F,...,a_n \in F$,where $F$ is an ordered field

then $$\sum_{k=1}^{n} a_k^2 \geq 0$$

I know I have to attempt this to receive a competent answer, so here it goes:

$$\sum_{k=1}^{1} a_1^2 \geq0$$

since $|a_1|^2 \geq 0$

This looks like an induction problem but I want to argue that anything in an ordered field squared is positive or $0$.

Assume $$\sum_{k=1}^{n_0} a_k^2 \geq 0$$

Therefore

$$\sum_{k=1}^{n_0} a_{k}^2 +a_{n_0+1}^2 \geq 0$$

Since $|a_1|^2+\ldots+|a_{n_0}|^2+|a_{n_0+1}|^2 \geq 0$

So by induction

$$\sum_{k=1}^{n} a_k^2 \geq 0$$ for all $n \in \mathbb{N}$

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    $\begingroup$ I think you have a good idea to focus on whether one square term is nonnegative. Setting aside the trivial case $a^2=0$ one can assume either $a$ or $-a$ is positive, and either way we get $a^2$ is positive. So the base case is easy if you choose to think of it that way. $\endgroup$ – hardmath Jul 12 at 1:51
  • $\begingroup$ @hardmath does this mean I should have three cases for both the inductive and base steps. One for all terms equaling zero, one for all positive terms and one for all negative terms? $\endgroup$ – user686544 Jul 12 at 1:57
  • $\begingroup$ Once the single square term is shown to be nonnegative, you don't need trichotomy for the proof that a finite sum of nonnegative of terms is nonnegative. I'll try and find a previous Question that does this argument, but the base case is really a sum of two nonnegative terms. $\endgroup$ – hardmath Jul 12 at 2:19
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    $\begingroup$ @hardmath thanks. Also are the absolute values unnecessary? I think that the answer to this problem should lie in the theorem that if $a \in F$ and $a \neq 0$ then $a^2>0$ I'm just not sure how to implement it. $\endgroup$ – user686544 Jul 12 at 2:24
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The absolute value is a way of abbreviating the argument from trichotomy. Likely the purpose of the exercise is to practice making an argument from the definition (axioms) of an ordered field.

You should review how the definition was presented in your study materials. The historical development was originally by imposing a binary relation $\le$ as a total order on field elements so that the sum of two positive elements is positive and the product of two positive elements is positive. (Here "positive" means being greater than and not equal to zero.)

An alternative approach is to focus on a subset of the field elements that are greater than or equal to zero, ascribing to that subset properties that amount to creating a total order (trichotomy) and closure under taking sums and products.

The crux of the proof you've identified is:

Lemma For any element $a\in F$ an ordered field, $0\le a^2$.

The way I sketched in my original comment is to use trichotomy to argue one of three cases must hold. If $a=0$ then $a^2=0$ and we are done. If $a\gt 0$, then $a^2$ is positive (by closure under taking products of two positive elements). Finally the only other possibility is that $a\lt 0$, then $-a \gt 0$, and we arrive at the same conclusion because $a^2 = (-a)^2 \gt 0$.

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Definition of ordered filed

A field $(F, +, \cdot)$ together with a (strict) total order $<$ on $F$ is an ordered field if the order satisfies the following properties for all $a, b$ and $c$ in $F$:

  • if $a < b$ then $a + c < b + c$, and

  • if $0 < a$ and $0 < b$ then $0 < a\cdot b$.


Lemma: $\forall a \in F$, $a^2 \ge 0$.

Proof:

  • If $a=0$, then $a^2 = 0 \cdot (1-1)=0\cdot 1 - 0 \cdot 1 = 0-0=0$ where I have used definition of additive inverse and distributive law.
  • If $a>0$, then $a^2=a\cdot a>0$ by definition of ordered field.
  • If $a<0$, adding $-a$ to both sides, we have $0<-a$. Also $1+(-1)=0$ by squaring, $$(1+(-1))\cdot (1+(-1)) = 0$$ which can be expanded to $$1+(-1)+(-1)+(-1)^2=0$$ $$-1+(-1)^2=0$$

That is $$(-1)^2=1.$$

Hence $$a^2=1\cdot a^2 = (-1)^2 \cdot a^2 = (-a)^2>0$$

That is $$\forall a \in F, a^2 \ge 0. \square$$


Claim: $\forall n \in \mathbb{N}$, $\sum_{i=1}^n a_i^2 \ge 0.$

Base case: For $n=1$, this has been shown by the lemma.

Suppose we have $k \in \mathbb{N}$, $\sum_{i=1}^k a_i^2 \ge 0$

we want to show that $\sum_{i=1}^{k+1} a_i^2 \ge 0$.

This is true because

\begin{align}\sum_{i=1}^{k+1} a_i^2 &= \left(\sum_{i=1}^{k} a_i^2\right) + a_{k+1}^2 \ge 0\end{align}

because $\left(\sum_{i=1}^{k} a_i^2\right) \ge 0$ and $a_{k+1}^2 \ge 0$.

Hence, by mathematical induction, $$\forall n \in \mathbb{N}, \sum_{i=1}^n a_i^2 \ge 0.$$

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