Is my proof that the square of any real number is non-negative correct?

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. $\square$

• 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.) – Brian Tung Aug 14 '18 at 5:58
• If you know (or can prove) that $\,1 \gt 0\,$ and $\,(-1)\cdot(-1)=1\,$ then the rest follows. – dxiv Aug 14 '18 at 6:07
• 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? – астон вілла олоф мэллбэрг Aug 14 '18 at 6:33
• Prove it for $k<0$, then $k=0$, then $k>0$ and you're done. – David G. Stork Aug 14 '18 at 6:39