Can a posterior probability be >1? In Bayes' formula:
$$P(x|a) = \frac{P(a|x) P(x)}{P(a)}$$
can the posterior probability $P(x|a)$ exceed 1? 
I think it is possible if for example, assuming that $0 < P(a) < 1$, and $P(a) < P(x) < 1$, and $P(a)/P(x) < P(a|x) < 1$. But I'm not sure about this, because what would it mean for a probability to be greater than one? 
 A: The assumed conditions do not hold- it can never be true that $P(a)/P(x) < P(a|x)$ by the definition of conditional probability:
$P(a|x) = P(a\cap x) / P(x) \leq P(a) / P(x)$
A: No, it is not possible for the posterior probability to exceed one.  That would be a breach of the norming axiom of probability theory.  In your question you specify that $\mathbb{P}(a)/\mathbb{P}(x) < \mathbb{P}(a | x)$ as part of your example.  However, using the rules of conditional probability, you must have:
$$\mathbb{P}(a | x) = \frac{\mathbb{P}(a,x)}{\mathbb{P}(x)} \leqslant \frac{\mathbb{P}(a)}{\mathbb{P}(x)}.$$
This means that you cannot have the inequality conditions you have specified.  (Incidentally, this is a good question: it is good that you are probing the probability laws looking for problems.  It shows that you are exploring these matters with a greater degree of rigour than most students.)
An additional point: It is worth making one additional point about this situation, which is about the logical priority of different characteristics of probability.  Remember that probability theory starts with a set of axioms that characterise what a probability measure actually is.  From these axioms we can derive "rules of probability" which are theorems derived from the axioms.  These rules of probability must be consistent with the axioms to be valid.  If you ever found that a rule of probability leads to a contradiction with one of the axioms (e.g., the probability of the sample space is greater than one), this would not falsify the axiom - it would falsify the probability rule.  Hence, even if it were the case the Bayes' rule could lead to a posterior probability greater than one (it doesn't), this wouldn't mean that you can have a posterior probability greater than one; it would simply mean that Bayes' rule is not a valid rule of probability.
A: The Bayes formula $\displaystyle P(B \mid A) = \frac{P(A\mid B)P(B)}{P(A)}$ cannot give values for $P(B\mid A)$ exceeding $1$. An intuitive way to see this is to express $P(A)$ via the law of total probability as
$$P(A) = P(A\mid B)P(B) + P(A\mid B^c)P(B^c)$$ giving that
$$P(B \mid A) = \frac{P(A\mid B)P(B)}{P(A)} = \frac{P(A\mid B)P(B)}{P(A\mid B)P(B) + P(A\mid B^c)P(B^c)}$$ which shows that the numerator is just one of the terms in the sum in the denominator, and so the fraction cannot exceed $1$ in value.
