Let $R$ an ordered ring, with $a,b\ge 0$ and $a^n=b^n$ with $n\in\Bbb N_{>0}$. Prove that $b=a$

Question

Let $R$ an ordered ring, with $a,b\ge 0$ and $a^n=b^n$ with $n\in\Bbb N_{>0}$. Prove that $b=a$.

Ring, in this context, dont mean that exist a multiplicative identity. Some people call these structures as rng instead of ring to differentiate them.

The axioms of order for a ring are the same that for a field, i.e.

$$a>b\implies a+c>b+c\tag{OR1}$$

$$a,b>0\implies ab>0\tag{OR2}$$

where $a<b\iff a\le b\land a\neq b$ and $\le$ is a total order. And $a^ka=aa^k$ due to the fact that the multiply operation is associative.

My try: I tried by contradiction. Let $a^n=b^n$ and $a\neq b$, without lose of generality suppose that $a>b$. Then:

$$\begin{align}&a>b\implies a-b>0 &\text{(inverse of addition, OR}1)\\&a\ge0\implies a^k\ge0 &\text{(associativity, OR}2)\end{align}$$

But Im unable to get something useful from here. Can you help me please?

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EDIT: Eureka! I solved it. Check the proof please and let some alternative proof if you want/know.

Some important rules of ordered rings:

1. We have that $a0=0$ still holds in rings without multiplicative identity, due to the fact that

$$a0=a(0+0)=a0+a0\implies a0=0\tag{TH1}$$

2. Due to OR1 we have that if $c<0$ then $-c>0$. Then for some $a>0$ we have by OR1 that $a(-c)>0$ what implies by OR1 and distributivity that $ac<0$. Then we conclude that

$$a>0\text{ and } b<0\implies ab<0\tag{TH2}$$

3. If $a<0$ and $b<0$ by TH2 we can see that $ab>0$. Then we have that if $c>0$ then

$$c>c-c\implies c^2>c^2+c(-c)\implies c(-c)=(-c)c=-c^2\tag{TH3}$$

Proof by induction over n:

(Base case) Due to the notational rule $c^j c^k=c^{j+k}$ for $j,k\in\Bbb N_{>0}$ we have that $c^2=c\cdot c$, meaning that $c^1=c$ for all $c\in R$. Then

$$a^1=b^1\iff a=b$$

(Induction hypothesis) Assume that is true that if $a^n=b^n$ then $a=b$.

(Induction step) We want to prove that $a^{n+1}=b^{n+1}$ implies $a=b$ based in the hypothesis above and the properties of ordered rngs.

If $a^{n+1}=b^{n+1}$ then by associativity $a^na=b^nb$ and using the induction hypothesis, distributivity and TH3 we have

$$a^na=b^nb\iff b^na=b^nb\implies b^na-b^nb=0\implies b^n(a-b)=0\tag{*}$$

Then I will prove that (*) implies that $a=b$.

1. Note that the case where $b^n=0$ is trivial: $b^2$ cannot be positive or negative due to OR2 and TH2, so $b=0$. Then cause $b^n=a^n$ we get that $a=b=0$.

2. The case $b^n<0$ is impossible because $a,b\ge 0$ by definition and OR2.

3. If $b^n>0$ and $a-b>0$ we have that $b^n(a-b)>0$ due to OR2 that contradicts (*).

4. If $b^n>0$ and $a-b<0$ we have that $b^n(a-b)<0$ due to TH2 that contradicts (*).

Then in all cases the unique solution is that $a=b$, and the induction is complete. Q.E.D.

Answer 1

First I show that $b > a \ge 0$ implies $b^n > a^n$ for all $n \in \mathbb{N}_{>0}$.

It's clearly true for $n=1$. Suppose $b^n > a^n$ for some $n$. Then $$b^{n+1} - a^{n+1} = b^n b - b^n a + b^n a - a^n a = b^n (b-a) + (b^n - a^n) a > 0$$ since $b^n > 0$, $b-a > 0$, $b^n - a^n > 0$, and $a \ge 0$. Then the proof is complete by induction.

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Now if $a \ne b$, then either $a > b$ or $b > a$; in either case, the above shows that $a^n \ne b^n$.

Answer 2

Another way: $$0=a^n-b^n=(a-b)\left(\sum_{i+j=n-1}a^i b^j\right )$$ If $a\ne b$ you have WLOG $a-b\gt 0$ and the sum of positives cannnot be zero in a (complet) ordered ring (any finite ring is (complet) ordered) so you could have a product of positives equal to zero which is not true.