1
$\begingroup$

$\text{Let} $ $a^b$ $(a \in Q_+$ and $ b \in Q$ \ {${0}$}$) $ $=$ $P_{(a,b)}$ and If $c=\dfrac{P_{(a_1,b_1)}}{P_{(a_2,b_2)}}$ is a rational number then we say :$P_{(a_1,b_1)}$ and $P_{(a_2,b_2)}$ is the same kind

I think about this two propositions

1) If $P_{(a_i,b_i)}$ and $P_{(a_j,b_j)}$ isn't the same kind

then $k_1P_{(a_1,b_1)}+k_2P_{(a_2,b_2)}+...+k_nP_{(a_n,b_n)}=0$ $\Rightarrow$ $k_1=k_2=...=k_n=0$

2) If $P_{(a_i,b_i)}$ and $P_{(a_j,b_j)}$ isn't the same kind

and not all $a_i=0$

then $\sum_1^n a_iP_{(a_j,b_j)}$ is a irrational number

If the proposition has proof before? Thanks for any help.

$\endgroup$
  • $\begingroup$ $P_{(2,3)}/P_{(2,4)}=1/2$ is rational, but $P_{(2,3)}\ne P_{(2,4)}$. Also, if you mean for the $k_i$ and the $a_i$ to be rational, please edit that into your question. $\endgroup$ – Gerry Myerson Apr 17 '13 at 13:08
  • $\begingroup$ What does "same kind" mean? $\endgroup$ – Thomas Andrews Apr 17 '13 at 13:20
  • $\begingroup$ I dont mean the normal equal...I mean a equivalence class..To prevent the occurrence of certain circumstances like $\sqrt2+(-\sqrt2)=0$ $\endgroup$ – Xiaolang Apr 17 '13 at 13:20
  • $\begingroup$ I dont know how to express it exactly ,if anyone can know what I mean ,help edit it ,thanks. $\endgroup$ – Xiaolang Apr 17 '13 at 13:21

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.