Which of the following descriptions best fits the problem you have?

I want to prove that there is a real solution to an equation.

• In that case, there are two possibilities, with a slightly fuzzy boundary between them. If the equation is particularly simple, then you may be able to solve it algebraically. For example, if your equation is then you can rearrange it and deduce that Here, you are implicitly using algebraic facts about the real numbers, such as that every non-zero real number has a multiplicative inverse, that the sum or product of two real numbers is a real number, etc. A slightly more complicated example is which you could solve using the formula for solving a quadratic. But if you do that, then you are implicitly using not just the algebraic facts (basically, that is a field), but also the fact that you can take square roots of non-negative numbers. If you want to justify that, then you need to use more than just that is a field: the simplest approach is to use the intermediate value theorem. And these considerations apply to more complicated equations involving exponential and trigonometric (and other) functions as well: sometimes you can use the fact that it has already been established that these functions have inverses (as you would if, for example, you solved the equation by saying ), but if you have not established the existence of these inverses, then you can fall back on the intermediate value theorem (arguing, say, that if then there must exist and with and and also proving that is continuous). For more complicated equations, you may not be able to solve them analytically and will have to rely on the intermediate value theorem to prove that a solution exists: consider the equation for example.

I want to prove that a real number exists with a certain property , and property can be naturally formulated as the property of having an infinite sequence of simpler properties all at once.

• It is here that the real numbers really come into their own. A simple example is the existence of a non-algebraic number. One can formulate this as follows. Let be an enumeration of all the algebraic numbers. Then let be the property of not equalling A number is transcendental if and only if it satisfies for every How do we use the basic theory of real numbers to prove that a transcendental number exists? We use the nested-intervals property: that if we have a sequence of closed intervals with each one containing the next, then they have a non-empty intersection. It is easy to create a sequence of such intervals such that and so on, and any number that belongs to all these intervals must therefore not equal any of the This is not the only tool available: others are the convergence of bounded monotone sequences, the convergence of Cauchy sequences, and the Bolzano-Weierstrass theorem. See creating real numbers using limiting arguments for more information.

I want to prove that a real number exists with a certain property , and the property can be naturally formulated as having every property , where ranges over some uncountable set of real numbers.

• An example of such a property is the property of being the maximum of a set. If you want to prove that some uncountable set has a maximum, then you need to find such that for every (and also such that ). That is, if is the property of being at least as big as then you need such that holds for every This may seem like a lot to ask, but very often it turns out to be possible to choose a countable subcollection of the properties in such a way that if every property in the subcollection holds, then all the properties hold. Alternatively, one can use the property that every non-empty set that is bounded above has a supremum. (That would be the natural first step towards showing that had a maximum: one would prove that it was non-empty and bounded above, take the supremum, and prove that the supremum was an element of ) More details about this can be found in creating real numbers using limiting arguments.