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单词 ArchimedeanProperty
释义

Archimedean property


Let x be any real number. Then there exists a natural numberMathworldPlanetmath n such that n>x.

This theorem is known as the Archimedean property of real numbers. It is also sometimes called the axiom of Archimedes, although this name is doubly deceptive: it is neither an axiom (it is rather a consequence of the least upper bound property) nor attributed to Archimedes (in fact, Archimedes credits it to Eudoxus).

Proof.

Let x be a real number, and let S={a:ax}. If S is empty, let n=1; note that x<n (otherwise 1S).

Assume S is nonempty. Since S has an upper bound, S must have a least upper bound; call it b. Now consider b-1. Since b is the least upper bound, b-1 cannot be an upper bound of S; therefore, there exists some yS such that y>b-1. Let n=y+1; then n>b. But y is a natural, so n must also be a natural. Since n>b, we know nS; since nS, we know n>x. Thus we have a natural greater than x.∎

Corollary 1.

If x and y are real numbers with x>0, there exists a natural n such that nx>y.

Proof.

Since x and y are reals, and x0, y/x is a real. By the Archimedean property, we can choose an n such that n>y/x. Then nx>y.∎

Corollary 2.

If w is a real number greater than 0, there exists a natural n such that 0<1/n<w.

Proof.

Using Corollary 1, choose n satisfying nw>1. Then 0<1/n<w.∎

Corollary 3.

If x and y are real numbers with x<y, there exists a rational numberPlanetmathPlanetmathPlanetmath a such that x<a<y.

Proof.

First examine the case where 0x. Using Corollary 2, find a natural n satisfying 0<1/n<(y-x). Let S={m:m/ny}. By Corollary 1 S is non-empty, so let m0 be the least element of S and let a=(m0-1)/n. Then a<y. Furthermore, since ym0/n, we have y-1/n<a; and x<y-1/n<a. Thus a satisfies x<a<y.

Now examine the case where x<0<y. Take a=0.

Finally consider the case where x<y0. Using the first case, let b be a rational satisfying -y<b<-x. Then let a=-b.∎

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更新时间:2025/5/4 12:14:02