Adding or Multiplying Transcendentals Is it possible to add or multiply (no subtraction) only positive transcendental numbers and yield a solution that is algebraic?
Exponential manipulation is excluded from this question, as $e^{\ln2} = 2$
EDIT: Also, excluding reciprocals of other transcendental numbers and subtraction?
 A: Let's choose a transcendental number $T$. Let's try to define the additive group $S$ generated by $T$. It must be closed, so all natural multiples of $T$ must be in $S$. It must have the identity, so $0 \in S$. It must have the inverses, so $-T \in S$, just as any natural multiple of $-T$. So now $S$ is isomorph to the additive group of the integers, $(\mathbb{Z}, +)$, which means S is countable. Moreover, $S$ is finitely generated and it's generator is $<T>$. All of this is trivial, now comes the interesting part of which I'm not 100% sure, neither able to give a formal proof, just an intuition.
Any rational multiple of $T$ that is not a integer multiple of $T$ is not in $S$. So we can extend our group to have all rational multiples of $T$, just observing that all properties still hold. Call this new group $S_{\mathbb{Q}}$. It's easy to see that it is isomorph to $(\mathbb{Q}, +)$, and in this sense it can no longer be finitely generated. So the generating group of $S_{\mathbb{Q}}$ is infinite, but it cannot be bigger the $\mathbb{Q}$ itself, so it must be still a countable infinity.
The same can be done to extend $T$ to all algebraic multiples of $T$ not in $S$. All properties still hold, all of them (except the identity 0) are still transcendental and this new group called $S_{\mathbb{A}}$ is isomorph to $(\mathbb{A}, +)$, no longer can be finitely generated, but it's generating group $S_{\mathbb{A}}$ also cannot be bigger then $\mathbb{A}$ itself, so it's also a countable infinity.
But the set of all transcendental numbers is an uncountable infinity, so there exists transcendental numbers (say $T'$) which are not in $S_{\mathbb{A}}$, so this relation is not valid: $T' = aT$ for some algebraic $a$. Which means (I'm guessing here) that two transcendental numbers satisfy $\dfrac{T'}{T} = a$ if and only if $T' \in S_{\mathbb{A}}$. So that's an answer for the division case (which is not on the question but it's the first result I've found). This doesn't mean $\pi / e$ isn't an algebraic number. It could happen that some unlucky undiscovered algebraic number transforms $e$ into $pi$. What is notable here is that taken $e$, a transcendental number, there exists at least one (or, by my guess from all this, a infinite number, $S_{\mathbb{A}}$, or more, a uncountable infinite number) of transcendental numbers $T$ such that $e / T$ is not an algebraic number.
Maybe the same reasoning can be applied to multiplicative groups? Then we would form transcendental fields, and all transcendental numbers outside a transcendental field $(\mathbb{F}_\mathbb{a}, +, *)$ would be transcendental numbers that could not be generated by the common operations +, *, - and / from our initial number $T$ - aka the "trivial" transcendental numbers exemplified in the answers. If the cardinality of this field is still countable, then we know that there is no example other than the trivial ones.
This thinking took me the last hour, here at my city it's already 06:03AM and I have to go to college. I'll give more thought to it during the day.
Also a disclaimer, I never studied abstract algebra formally, so my sincere apologies if I'm speaking great nonsense. I would really appreciate a veteran opinion on this.
A: Yes, take Joseph Liouville number
$$\sum_{k=1}^\infty 10^{-k!}$$
we can construct a slightly different one
$$-\sum_{k=2}^\infty 10^{-k!}$$
both being trancendental but their sum is $0.1=\frac{1}{10}$
With your change in question the answer changes to
we construct another positive real trancendental number, namely
$$1-\sum_{k=1}^\infty 10^{-k!}$$
and adding this to our original number gives 1
We know $r\pm t$, $r$ is rational and $t$ is transcendental, is transcedental because if not
$$r_1\pm t = r_2$$
implies that
$$\pm t = r_2-r_1$$
and rational minus rational is rational so our $t$ must be rational as well
A: Yes, you can even get integer results. Even if you exclude the obvious examples you're out of luck.
Just take any algebraic number $A$ and consider. $$A = A-t + t, t$$ where $t\in\mathbb R$. Now the set of $t$'s that make $A-t$ or $t$ algebraic is countably infinite, but $\mathbb R$ is uncountable so there will definitely exist uncountable number of $t$ such that $A-t$ and $t$ are transcendental yet their sum being algebraic. 
Similar reasoning applies to products.
The reasoning can even be extended to the case where $A-t$ and $t$ are both positive as long as $A$ is. And the case where $A/t>1$ and $t>1$ as long as $A$ is for the product case.
