Given a decimal number $d$ we define the function $D$ to be the number where the integer part is the same and the decimal string is reversed.
For example, $D(5.879)=5.978$, $D(-800.5924)=-800.4295$ and $D(7.2)=7.2$.
If $a\neq b$ and $D(a)=b$ then can $a+b$ be an integer?
Are there two unequal decimal numbers where the integer part is the same and the decimal expansion is reversed such that there sum is an integer
algebra-precalculusdecimal-expansion
Best Answer
Suppose that some $a$ exists satisfying the requirements. First notice that $a$ must have a finite decimal expansion as otherwise $D$ is not defined. If the decimal part is zero clearly $a=D(a)$ which is a contradiction.
If the decimal part is non-zero then it is clear that the final non-zero digit added to the first digit must equal exactly $10$. If the decimal expansion is exactly $1$ digit long it must be $5$ and so $a=D(a)$ which is a contradiction.
Now consider that if the decimal expansion is greater than $1$ digit long the first digit added to the last digit equals $10$ and so if there is any carry into the first decimal place when doing $a+b$ we would immediately have a contradiciton as the total for the first decimal place would not be zero.
However if the second decimal place recieved any carry it would have to carry forward again to be able to end up at zero, therefore the second decimal place cannot recieve any carry.
Repeating this procedure we eventually reach the second to last decimal place, but we know that this place recieved carry from the last decimal place which is a contradiction.
The only way around this is if the first decimal place is the last decimal place, but we already showed that was a contradiction.
Therefore by contradiction no number $a$ meeting the requirements can exist.