Original Theorem : Given two real numbers $a$ and $b$ with $a < b$, where $a \geq 0$ there exists a rational number $r$ satisfying $a < r < b$.
To be proven from above theorem :
Without doing too much work, show how to prove Theorem above in the case where a < 0 by converting into the case already proven (which is when $a \geq 0$ in the above theorem)
Case when $a < 0$ (To be Proven )
Now Case1
Consider $b > 0$. Since $a < 0$ and $b > 0$, so we have $a < b$ and we have reduced to original theorem
Case 2
When $b < 0$. How do i do this ?
Thanks
In Case $1$, you did not "reduce to the original theorem". The original theorem only tells you something if $a<b$ and $a\geq 0$. You only proved $a<b$, while $a\geq 0$ is still not true, and therefore, the original theorem tells you nothing in this case. Your argument is incorrect.
However, in case $1$, it should be very easy to find a rational number satisfying $a<r<b$. Think about it. $a$ is smaller than $0$. $b$ is larger than $0$. That is, $a$ is on the left of $0$. $b$ is to the right of $0$. Can you think of any number that is in between $a$ and $b$? Is that number rational?
For case $2$, you can actually reduce to the original theorem. Think about what happens if you define $a'=-b$ and $b'=-a$. Do $a', b'$ satisfy the conditions of the original theorem?