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I wanted to re-prove the statement given in the title above. To do this I used the following methods. Let's write an expression that is equivalent to $ a^r<b^r $. $$ \frac{1}{a^n} < \frac{1}{b^n} , \space \space (n = -r)$$ Then I multiplied $b^n$ to both sides. $$ \frac{b^n}{a^n} < 1 $$ From here, I assumed that $\frac{b^n}{a^n}$ is always small than 1 since we clearly know that $a > b > 0$. For more clarification, I just proved this claim as well.
Let's assume $a > b > 0$, and $\frac{b}{a} < 1$ is true. Here, a and b are integers. But $a \neq0 $
If multiply both sides by a, we will get the exact condition between a and b. $$b < a$$ I want to ask does my way of proving these statements make sense at all? And is it ok doing proofs in such a manner? I mean is this acceptable to the mathematical community?

Ilhom Sadriddinov
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  • Well, the title statement is wrong. Also, what is $r$? – peek-a-boo Jun 21 '22 at 05:29
  • Overall, proofs are done using some kind of background system in mind. If you have very clearly stated those assumptions and used a logical system appropriately then it should work. So in essence I'd ask what are axioms or assumptions you are using. We'd probably want axioms showing using how, order, multiplication and division work out and maybe if need addition and subtraction. From their we can probably make a case for your proof and or rework it till it works. The field axioms over the reals or rational with some additional axioms for convience seems like a great choice here. – Pymamba Jun 21 '22 at 05:31
  • Sorry, I have made a mistake. I have changed the title. – Ilhom Sadriddinov Jun 21 '22 at 05:37
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    Although it's a little thing, a problem I have with this proof is the way you justify that $a > b > 0$ implies $\frac{b}{a} < 1$: you can't just assume both to be true and then show that it's consistent, you would need to assume $a > b > 0$ and then show that $\frac{b}{a} < 1$ whenever this is true. Also, in this context I don't think the jump from $\frac{b}{a} < 1$ to $\frac{b^n}{a^n} < 1$ can be ignored, it's the essence of the entire proof. – Stephen Donovan Jun 21 '22 at 05:52

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