It\'s about math analysis. (a) Suppose x is a positive real number. Prove that t

Question

It's about math analysis.

(a) Suppose x is a positive real number. Prove that there exists a natural number N such that 0 < 1/N < x. Hint: x^-1 is represented by a Cauchy sequence.

(b) Suppose that x, y are real numbers such that y x > 1. Prove that there exists an integer m such that x < m < y. Hint: Suppose not and find a contradiction.

(c) Given any two real numbers x and y with x < y, prove that there is a rational number m N such that x < m N < y. Hint: use (a) and (b).

Explanation / Answer

(a)  the completeness of R : Every nonempty set of real numbers that is bounded above has a supremum in R.

the Archimedean property of R : N is not bounded above in xR. That is, for each xR there exists nN such that n>x

( proof of archimedean prop. : For x<0 it is trivial. Suppose x0. Let A={nN ;nx}. By completeness we have supAR. Obviously, there is aA such that supA1/2<a, then let n=a+1 and n>x and the theorem is proved. )

now to prove : For any real number x>0, there exists a natural number N such that 0<1/N<x.

We proceed using proof by contradiction. Let 0<1/N<xfor all N N× (set of naturals), then N1/x for all NN×. Thus N× is bounded in R, contradicting the Archimedean property.

(c)  Let x,yR, and assume without loss of generality that x<y. Then yx>0, and by the Archimedean property nN such that n(yx)>1 and hence ny>nx+1. Again by the Archimedean property, mN such that m>nx and kN such that k>nx. Then m<nx<k and jN with mjk such that j1nx<j. Combining inequalities we have nx<jnx+1<ny, hence x<jn<y where j,nZ, hence j/nQ, as desired

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