2.1 Real numbers, axiom of completeness

completeness axiom
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one of the things we alluded to in one of our previous videos is that the restaurant numbers do not actually cover the entire set of things that wed like to think of as real numbers we said that there are holes in it that there are certain numbers that you cant actually represent with a rational number numbers that you might better measure on a ruler for example so we didnt really provide much justification for this yet so by means of that were going to show that there is no for example rational number that is equal to the square root of 2 or in other words there is no rational number that when you square it will give you 2 so we can state this is a theorem there is no rational number whose square is 2 now to prove this directly we would have to examine every rational number and show that none of them could be squared to give us 2 somehow and thats actually quite difficult so for this particular theorem were going to attack it by contradiction were going to assume that there is such a rational number and then see if we can derive something contradictory or an absurdity from this so well assume I suppose we start by saying proof assume such a number exists and well call it P over Q exists now its a little bit tricky because rational numbers are a little bit quirky and that you can have the

same rational number being represented by different integers so for example two-thirds is also a four sixths or six ninths or negative two over negative three so what were going to do is were going to also assume that we are in the lowest possible form of that rational number and that is to say precisely that the numerator and the denominator have no common factors theyve been cancelled out and moreover that P and Q have no common factors thats about all we can say about that so lets just start working on it so we assumed that such a number exists that means that P over Q squared is equal to 2 right now were going to just work on this a little bit just to rearrange it into slightly more convenient form so I will expand out the square so P squared over Q squared is equal to 2 and then Ill just rewrite this as P squared is equal to 2q squared is a if and only its because Q is not 0 because its a rational number P over Q is a rational number okay so weve rearranged this in to a slightly suggestive form and if we think about it for a second this statement here is telling us something about what P and Q can be so P squared is equal to 2q squared what that means is that P squared is in fact an even number because theres two times something and if the

square is even that means the number itself must be even think about that for a second if you square any odd number you get another odd number back so this actually implies that P itself is an even number so this employs that P squared and hence P is even so we now has some new information about P and we should try and incorporate that down as a mathematical statement somehow so well say that p is equal to 2 times some other integer m and well substitute that back into expression just here so substituting back in we get so P squared is going to become 4 M squared equals to Q squared which can be written as 2 m squared is equal to Q squared now that looks very much like what we just had with P and Q so comparing those two boxes we can apply the same reasoning a second time and we can deduce that Q must in fact be even also so by the same reasoning q is also even so therefore P and Q both even now thats actually problematic because if P and Q are both even they share a common factor of two and right at the stars we built into our assumptions that P and Q have no common factors so eventually derived the contradiction if we have it that p and q squared must equal 2 then its necessarily true that p and q are both even numbers as

we just derived and that contradicts our assumption or our statement that P and Q have no common factors we just chose the lowest representative of the equivalence class of rational numbers there so P and Q are both even contradictory contradicting our assumption that they have no common factors so weve assumed the negation of our theorem we have derived an absurdity or a contradiction if you like and therefore our theorem is true is proven and we can finished off with a nice little skeleton white okay so now that weve established that there are numbers that cannot be represented by restaurant numbers we need to cut with the kind of working definition of what the real numbers should be now it turns out that in fact we can – we can construct the real numbers directly from the rationals but this is kind of beyond the scope of this course so were not going to what were going to do instead is were going to assume that our real numbers are an extension of the rationals Kepler make sense the rational numbers with some extras that the real numbers are an ordered field in exactly the same way that the rationals are as we talked about previously containing the rationals as a subfield okay this all makes sense and so we can use inequalities etc as we kind of cavalierly talked about in previous videos that are really talking about what the real numbers were as previously discussed so

we talked about about ordering inequalities all kind of thing carries over into the real numbers and in fact we can get to the real numbers by adding a certain axiom to to what we had previously so this is called the axiom of completeness or the least upper bound property and the statement says every non-empty set of real numbers that is bounded above has at least upper bound now we havent really talked about or defined what bounded and upper bounds and that kind of thing mean yet so well get onto that in the next video but for now that suffice us to say that this axiom characterizes the real numbers if we take the real numbers as being an ordered field containing Q as a subfield and we add on this axiom that in every set of these numbers that you take that is bounded above whatever that means has the least upper bound that is sufficient to prescribe the real numbers now turns out as I previously told you before if we were to construct the real numbers from the rational numbers this axiom of completeness actually turns out to be a theorem a consequence of our construction so it is a really valid thing to do but for the time being were just going to take it as an assumption or an axiom and then were going to see what the consequences of assuming this are a little bit later once we have established a few different

math, 160.301, analysis
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