Proving that $aRb \iff a = b \lor a = b^2$ is antisymmetric.

Question

Over $\mathbb{N}$, $aRb \iff a = b \lor a = b^2$.

I'm having problems determining if this relation is antisymmetric.

I think it is. I did the following:

---

Direct proof attempt (got stucked)

We have to prove that $aRb \land bRa \implies a = b$.

Our premise is the fact that $aRb \land bRa$.

Based on our premise, we know that the following occurs:

$$(a = b \lor a = b^2) \land (b = a \lor b = a^2)$$

Distribute:

$$[(a = b \lor a = b^2) \land b = a] \lor [(a = b \lor a = b^2) \land b = a^2]$$

Absortion:

$$(a = b) \lor [(a = b \lor a = b^2) \land b = a^2]$$

Clearly I need to get rid of the rightmost part. But I'm not sure how. How can I proceed here?

---

Proof by contradiction attempt (almost worked)

We have to prove that $aRb \land bRa \implies a = b$.

Our premise is the fact that $aRb \land bRa$.

Suppose that $a \not = b$.

Based on our premise, we know that the following occurs: $$(a = b \lor a = b^2) \land (b = a \lor b = a^2)$$

Since we're supposing that $a \not = b$, it is simplified to

$$a = b^2 \land b = a^2$$

Alright, so this is where I'm not sure: since $a \not = b$, can I affirm that the square of either number will always be different from the other number, in $\mathbb{R}?$ Because if that's the case, clearly $a = b^2 \land b = a^2$ won't hold, contradicting the premise, proving that it is indeed antisymmetric.

Answer 1

HINT: Note that in $\Bbb N$ if $a=b^2$ then either $a=0$ or $a=1$ or $b<a$.

Answer 2

The hypothesis “$a\mathrel{R}b$ and $b\mathrel{R}a$” splits into four cases:

• $a=b$ and $b=a$

• $a=b$ and $b=a^2$

• $a=b^2$ and $b=a$

• $a=b^2$ and $b=a^2$

The only case that needs work is the fourth one, where $a=a^4$, that is $a=0$ or $a=1$. But …