This is a proof by wordsPlease show all work, so I can understand !Thank You! The Archimedean Property of the real numbers is the following statement:For every x, y E R such that x > 0 and y > 0, there exists n E N such thatnx > y.(a) Prove that for every a € R, there exists n E N such that n > a. That is, N isnot bounded above. [HINT: Proceed by contradiction, and use the Least UpperBound Property of R.](b) Use part (a) to prove that the Archimedean Property is true.(c) Use the Archimedean Property to prove that for every x ER such that x > 0,there exists n EN such that < x.n(d) Prove that there is a rational number between any two real numbers. That is, forevery a, b R with a

(a) As told we will prove the first statement using the method of contradiction. If possible, let there exist an a∈ℝ such that there exists no nx∈ℕ for which x<nx i.e. n<x for all n∈ℕ. Then a is the upper bound for the non-empty set ℕ⊂ℝ. Hence by Supremum property ℕ has a supremum say u∈ℝ. Now consider u-1∈ℝ. Since u-1<u and u is the supremum of ℕ, u-1 cannot be an upper bound of ℕ. Hence there exists n1∈ℕ such that u-1<n1. But then u<n1+1 where n1+1∈ℕ. This is a contradiction to our assumption that u is an upper bound of ℕ. Hence it follows that ℕ is not bounded above. (b) Following the first part we have proved that in general if x∈ℝ, then there exists nx∈ℕ such that x<nx. Now to prove the theorem we have to show that there exists n∈ℕ such that nx<y for every x,y∈ℝ and x,y>0 Since x>0, the statement that there is an integer n so that nx<y is equivalent to finding an n with n>yx for some n But if there is no such n then n<yx for all integers n. That is, yx would be an upper bound for the integers. This contradicts the statement we proved in (a).(c) To prove this statement we have to first prove a corollary of the Archimedean Property. Corollary: If S=1n: n∈ℕ then inf S=0 Here S≠∅ and every element of S is greater than zero. Hence S is a nonempty subset of ℝ, which is bounded below. Hence by Completeness Property of ℝ, S has an infimum. Let w=inf S. Since 0 is the lower bound of S, w≥0. For any ε>0, the Archimedean property implies that there exists n∈ℕ, such that 1ε<n. Also 1ε<n implies 1n<ε. ∴ 0≤w≤1n<ε Since ε is arbitrary the above inequality implies that w=0 i.e. inf S=0. Now to prove the statement given. By the corollary, we have inf1n: n∈ℕ=0. Then if x>0, x is not a lower bound for the set 1n: n∈ℕ. Hence there exists n∈ℕ such that 0<1n<x.(d) Let us assume without loss of generality that a,b are real numbers. Since a<b, b-a>0 and so 1b-a is a positive real number. Hence by Archimedean Property there exists n∈ℕ such that n>1b-a. Since b-a>0, we get nb-na>1 i.e. na+1<nb Now na>0 and again using Archimedean property we can find m∈ℕ, such that m-1≤na<m Then nb>na+1≥m Thus we have na<m<nb and so a<mn<b. Thus there exists a rational number q=mn such that a<q<b.