Is $\omega-1$ finite? I saw some videos and read some stuff about ordinals, and it came to me that $\omega-1$ should be finite.
My logic is that $\omega$ is the smallest transfinite number, so $\omega-1$ should be finite... right?  And that would also make $\omega-1$ the largest finite number, right?
 A: Especially when dealing with infinities, it's important to ask exactly what the thing you're trying to define should mean. 
In the case of ordinal subtraction, I'd argue that the only reasonable definition is "$\alpha-\beta$ is the least ordinal $\gamma$ (if one exists) such that $\gamma+\beta=\alpha$." There's two comments to make about this definition: 


*

*First, note that "$\gamma+\beta$" and "$\beta+\gamma$" are different in general, so we have to make a choice here - but presumably $(\omega+1)-1=\omega$, which justifies the choice of ordering.)

*Second, note the use of the word "least" in the definition. Ordinal addition is not injective even once we fix all but one variable - see BrianO's answer - so we have to pick a "natural" candidate from the set of possible ones. For various reasons, I'd argue that "least" is the best choice here.
Under this definition, what is $\omega-1$? Well, we ask which ordinals $\alpha$ satisfy $\alpha+1=\omega$; unfortunately, there are no such $\alpha$ (as noted by the other answers).
There is a slightly less natural definition which has the advantage of being total: set $\alpha\ominus\beta$ to be the least $\gamma$ such that $\gamma+\beta\ge\alpha$. This allows us to "overshoot" the original ordinal; for example, under this definition, $\omega\ominus1=\omega$. I've actually used $\ominus$ in various arguments, so it is mathematically useful to a small degree; but it's not very useful, and I don't think it's natural, either. At the end of the day, I think we're stuck with the following situation: the "natural" subtraction of ordinals is not always defined. And that's okay - neither is division on the real numbers!
A: No, if $\omega-1$ was finite, so would obviously $\omega=(\omega-1)+1$ be.
A: No, because $\omega-1$ (the predecessor of $\omega$) does not exist: $\omega$ is a limit ordinal.
A: Which integer is the largest? There is no such thing. Assuming "subtraction" means an inverse of ordinal addition "$+\,1$", $\omega - 1$ is not defined, as $\omega$ is a limit ordinal not a successor ordinal: it has no immediate predecessor. Similarly, division is not defined for infinite ordinals. 
For any $\alpha$, the ordinal functions $\beta\mapsto \alpha + \beta$ and (for $\alpha \ne 0$) $\beta\mapsto \alpha * \beta$ are 1-1 but not onto the ordinals (except for $\alpha = 0, 1$ respectively), so for any $\alpha, \gamma$, in general there isn't any $\beta$ such that $\gamma = \alpha + \beta$ or $\gamma = \alpha * \beta$. 
Furthermore, there can be distinct $\alpha_1, \alpha_2$ such that $\alpha_1 + \beta = \alpha_2 + \beta$: for instance, $0 + \omega = 17 + \omega$, and $0 + \omega^2 = \omega + \omega^2$. If $\gamma = \alpha_1 + \beta = \alpha_2 + \beta$, then "$\gamma - \beta$" is ambiguous and thus, for a different reason, ill-defined. Similarly for ordinal multiplication and "division".
As @GEdgar mentions, you might want to check out the surreal numbers, where something called $\omega-1$ does exist (and still isn't an integer).
