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 $
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$ \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 $