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Lemma: Let $ k $ be a real number. Then $ k^{2} \geq 0 $.

Proof: I will do this by considering cases and using the following axioms:

Axiom 1: Every $ x\in\mathbb{R} $ has a negative $ -x $ such that $ x + (-x) = 0 $.

Axiom 2: The real numbers are closed under addition (and hence subtraction).

Axiom 3: Every non-integer lies between two consecutive integers.

Axiom 4: Multiplication of two real numbers is commutative.

Axiom 5: Addition of two non-negative numbers yields a non-negative number.

Axiom 6: A line having positive gradient implies it is increasing from left to right.

We will now begin the proof.

Case 1: $ k = 0 $

Since $ 0^{2} = 0 $, we have that $ k^{2} = 0 \geq 0 $. Therefore true for $ k = 0 $.

Case 2: $ k \neq 0 $

Fix $ k \neq 0 $. By Axiom 1, $ \exists-k\in\mathbb{R} $ such that

$ k + (-k) = 0 $

Multiplying both sides by $ k $:

$ k^{2} + k(-k) = 0 $ $ (1) $

Multiplying both sides by $ -k $:

$ (-k)k + (-k)^{2} = 0 $ $ (2) $

Equating $ (1) $ and $ (2) $:

$ k^{2} + k(-k) = (-k)k + (-k)^{2} $

But $ k(-k) = (-k)k $ by Axiom 4, so by cancellation we are left with $ k^{2} = (-k)^{2} $. Therefore, it suffices to show that the square of any positive real number is non-negative.

Fix $ k > 0 $. If $ k\in\mathbb{N} $, then, by definition, $ k^{2} = k\times{k} = \underbrace{k+k+\dots+k}_{k \textit{ times}} $. But, using Axiom 5,

$ k > 0 $

$ \Longrightarrow k+k > k > 0 $

$ \Longrightarrow k+k+k > k+k > k > 0 $

.

.

.

$ \Longrightarrow $ $ \underbrace{k+k+\dots+k}_{k \text{ times}} $ $ > ... > k > 0 \geq 0 $.

$ \therefore $ $ k^{2} \geq 0 $.

If $ k\notin\mathbb{N} $, then by Axiom 3 there exists two integers $ [k] $ and $ [k] + 1 $ such that $ [k] < k < [k]+1 $. Note that $ [k] \geq 0 $.

Now consider the points $ ([k], [k]^{2}) $ and $ (k, k^{2}) $ in the plane. Let $ m $ denote the gradient of the line segment joining these points. Then

$ m = \frac{k^{2}-[k]^{2}}{k-[k]} = \frac{(k-[k])(k+[k])}{k-[k]} = k+[k] \geq k > 0 $.

Hence, the line joining these two points has positive gradient. So by Axiom 6 the $ y $-values of this line are increasing as $ x $ increases. $ \therefore $ $ k^{2} > [k]^{2} $ since $ k > [k] $. But $ [k]^{2} = \underbrace{[k]+[k]+\dots+[k]}_{[k] \text{ times}} \geq 0 $ by repeated application of Axiom 5.

$ \therefore k^{2} \geq 0 $ for $ k\notin\mathbb{N} $ as well as for $ k\in\mathbb{N} $, which completes the proof. $ \square $

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    $\begingroup$ You cannot define multiplication as repeated addition for $k$ that isn't a non-negative integer. How do you add something $2.3$ times, or $\pi$ times? (ETA: Sorry, I see how you dealt with negative numbers. But you still have to deal with positive non-integers.) $\endgroup$ – Brian Tung Aug 14 '18 at 5:58
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    $\begingroup$ If you know (or can prove) that $\,1 \gt 0\,$ and $\,(-1)\cdot(-1)=1\,$ then the rest follows. $\endgroup$ – dxiv Aug 14 '18 at 6:07
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    $\begingroup$ You also have to mention what axioms of the real numbers you are allowing answerers to use without proof. For example, is $1 > 0$ obvious, or is it to be proved using some definition of order on the real numbers that you have? $\endgroup$ – астон вілла олоф мэллбэрг Aug 14 '18 at 6:33
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    $\begingroup$ Prove it for $k<0$, then $k=0$, then $k>0$ and you're done. $\endgroup$ – David G. Stork Aug 14 '18 at 6:39

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