Introduction
Consider the reaction:

In chemistry, oxidation refers to a process where electrons are lost or removed from a particular element.  Often, people will assume the word oxidation refers to any reaction with oxygen.  This is not wrong, as oxygen will react with species that will give them (or lose) electrons; however oxygen is not the only element that can do that.  The word oxidation has stuck because reactions with oxygen, particularly in the context of a combustion reaction, are common and well known.  It is important to note that oxygen is NOT oxidized, the compound it reacts with is oxidized.

What about the oxygen, or for that matter, any element that forces another to give it electrons?  What term describes what happens to it?  It is said that those compounds are reduced.  This may seem counterintuitive as it has gained electrons, but the word reduced refers to oxidation number.
Identifying Redox Reactions
Unlike other reaction types, there is no set pattern for redox reactions. There are some clues that a reaction you are working with is a redox reaction. First, look for free state elements (e.g. Fe or Cl2). Their oxidation number is zero. This means that once they appear elsewhere in the reaction combined with other elements there is a guarantee that their charge is no longer zero. Second, look for polyatomic ions that exist on one side of the equation, but not the other. This is indicative of an element within the polyatomic ion has been oxidized or reduced.
Example

First, let's examine why this is a redox reaction.  There are two good reasons.  The presence of free state elements (Bi and Al) means there are elements uncombined with a zero charge.  Once combined in compounds, as in Bi(NO3)2 and NaAlO2, the charges of these elements cannot be zero.  Another reason this reaction must be redox is because the polyatomic ion nitrate (NO3-) is present on one side of the reaction, but not the other.
Now, let's break down the charges for each element.  Consider the figure above to see the breakdown for the nitrogen and oxygen in nitrate.  Note that the sum of the oxidation numbers is not zero, but the charge of the ion.  This method should be used to determine the oxidation numbers of any element within a polyatomic ion.  Since the charge of nitrate is -1 and there are three of them giving a total of -3, bismuth must be a +3 to cancel out the nitrate ions and give Bi(NO3)2 an overall charge of zero.  NaOH is unique since the positively charged hydrogen is the last element in the formula.  NH3 is also unique since the negative oxidation state nitrogen (-3) is written first, and the positive oxidation state hydrogen (+1) is second.  Within compounds that have three elements like NaAlO2, the first element is generally positive, the last is negative, and the middle element must be determined from the other two.  Hence, Na is a +1 (as an alkali metal should) and oxygen is a -2 (as a nonmetallic chalcogen should), leaving aluminum to be a +3.  Remember, there are two oxygen atoms, so the total charge for oxygen is -4:

Na + Al + 2(O) = 0

+1 + Al + 2(-2) = 0

Al - 3 = 0; Al = +3
Overall, Bi goes from a +3 to a 0, and is therefore reduced (gained electrons).  Aluminum goes from a 0 to a +3 and is oxidized (lost electrons).  Nitrogen goes from a +5 to a -3 and is also reduced (gained electrons).  There can be more than one oxidation or reduction in a reaction, but there must always be at least one of each.