Let me first confirm you that this question is not a duplicate of either this, this or this or any other similar looking problem.
Here in the current problem I'm asking to disprove me(most probably I'm wrong).
As you can see in this problem as answered by Nicolas that if a map is from $A \to B$ and is bijective then the cardinality of $A$ and $B$ is same.
Logarithmic map is from $\mathbb{R^+} \to \mathbb{R}$ and it is a bijective map and therefore it implies that the cardinality of $\mathbb{R^+}$ and $\mathbb{R}$ is same.
My logic
We can rewrite $\mathbb{R}=\mathbb{R^-} \cup \{0\} \cup \mathbb{R^+}$
Now we can see that $\mathbb{R}$ has all the elements of $\mathbb{R^+}$ and over that it has {0} and elements of $\mathbb{R^-}$. Now using pigeonhole principle, if we pair each element of $\mathbb{R^+}$ to itself from $\mathbb{R^+} \to \mathbb{R}$ (eg. 5.124 is paired to 5.124 and so on) now when the pairing gets over then you have elements of $\mathbb{R^-}$ which have not been paired.
Now one can say that since they are infinite sets therefore we cannot talk about pairing as I did above. When we are dealing with the pigeonhole principle then at that time it is not necessary to know the exact numbers involved.
Now whatever method you use for pairing you will always end with some elements of $\mathbb{R}$ which have not been paired (acc to pigeonhole principle).
Most probably I'm wrong but how?.
Kindly make me understand that I'm wrong and the above used logic by me is inappropriate.
Best Answer
The definition of equicardinal is that there exists a bijection between the sets.
You are trying to define "not equicardinal" as "there exists a bijection between one set and a strict subset of another". This definition is not a good one, as all Dedekind infinite sets (such as $\mathbb{Z}, \mathbb{R}$) have the property that they are bijective with strict subsets of themselves; hence all Dedekind-infinite sets are "not equicardinal" with themselves by your definition.
In answer to OP's comment, the specific problem with the pigeonhole principle argument in the OP is that this proves that some attempts at a bijection fail. But as discussed above, and in the other solution, and in the comments, is that if ANY bijection exists, then the two sets are equicardinal.