Let x &gt; 0 be real. Let n 0 be the largest integer such that n 0 ⩽ x (Note that the existance of n 0 depends on the archimedian property of R).1) Why he write about natural number but he did not defines them?2) How he uses archimedian property in existance of n 0 ?

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Let   x  &amp;gt;  0 be real. Let       n    0   be the largest integer such that       n    0    ⩽  x (Note that the existance of       n    0   depends on the archimedian property of R).1) Why he write about natural number but he did not defines them?2) How he uses archimedian property in existance of       n    0   ?

ivice7u · Accepted Answer

Step 11) Isn&#039;t really answerable by anyone other than Rudin unless you can clarify what you want to know exactly (or what kind of answer you&#039;d expect).For 2) can you write down the Archimedian property and then first try to prove that for all   x  &amp;gt;  0 there is some   n  ∈      N   such that   n  &amp;gt;  x ?This proves that the set   {  m  ∈      Z    ∣  m  ≤  x  } is bounded above and thus has a maximal element. This element is the n0 from your text. (It&#039;s also called the floor of x, denoted by   ⌊  x  ⌋ )

Gauge Odom · Accepted Answer

Step 1For me, it is acceptable to take as an axiom the following statement:Well-ordering principle: Every non-empty subset of       N   has a minimum element.You are correct that Rudin never stated this property explicitly. To be fair to Mr. Rudin though, page 1 states:&quot;We shall not, however, enter into any discussion of the axioms that govern the arithmetic of the integers, but assume familiarity with the rational numbers....&quot;Using the axiom above, we can return to the decimal representation issue. For   x  ∈                R        +   we have the set   S  :=  {  z  ∈          Z          ∧            x  &amp;lt;  z  } . This set is non-empty by the archimedean property* therefore it has a minimum element m0 by the well-ordering principle. Then       m    0    −  1  =      n    0   .*The archimedean property furnishes a natural number n such that   1  ⋅      n  &amp;gt;  x . Thus   n  ∈      S and   S  ≠      ∅     .

Let x>0 be real. Let n_0 be the largest integer such that n_0 le x (Note that the existance of n_0 depends on the archimedian property of R).

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