I need to find a real number with a certain property

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I need to find a real number with a certain property

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[QUICK DESCRIPTION]
Clicking on answers to questions below will take you to further text and/or further questions.

[PREREQUISITES]
Basic concepts of real analysis such as limits, least upper bounds, Cauchy sequences, etc.

===What is the problem?===

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

[add] 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 $3x-2=\pi,$ then you can rearrange it and deduce that $x=(\pi+2)/3.$ 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 $x^2=x+1,$ which you could solve using the formula for [[How to solve quadratic equations|solving a quadratic]]. But if you do that, then you are implicitly using not just the algebraic facts (basically, that $\R$ is a [[w:Field_(mathematics)|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 $\R$ is a field: the simplest approach is to use the [[w: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 $e^x=t$ by saying $x=\log t$), 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 $t>0$ then there must exist $a$ and $b$ with $e^a<t$ and $e^b>t,$ and also proving that $e^x$ 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 $e^x+x^2=20,$ for example.}}[/add]

[add] I want to prove that a real number exists with a certain property $P$, and property $P$ can be naturally formulated as the property of having an infinite sequence of simpler properties $P_1,P_2,P_3,\dots$ 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 $a_1,a_2,\dots$ be an enumeration of all the algebraic numbers. Then let $P_n$ be the property of not equalling $a_n.$ A number is transcendental if and only if it satisfies $P_n$ for every $P.$ 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 $[u_1,v_1],[u_2,v_2],\dots$ 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 $a_1\notin[u_1,v_1],$ $a_2\notin[u_2,v_2],$ and so on, and any number that belongs to all these intervals must therefore not equal any of the $a_i.$ This is not the only tool available: others are the convergence of bounded monotone sequences, the convergence of [[w:Cauchy sequence|Cauchy sequences]], and the [[w:Bolzano-Weierstrass theorem]]. See [[creating real numbers using limiting arguments]] for more information.}}[/add]

[add] I want to prove that a real number exists with a certain property $P$, and the property can be naturally formulated as having every property $P_t$, where $t$ 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 $A$ has a maximum, then you need to find $x$ such that $x\geq y$ for every $y\in A$ (and also such that $x\in A$). That is, if $P_y$ is the property of being at least as big as $y,$ then you need $x$ such that $P_y$ holds for every $y\in A.$ 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 $A$ 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 $A.$) More details about this can be found in [[creating real numbers using limiting arguments]].}}[/add]

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