The case in which a=0 or b=0 is eas

The case in which a=0 or b=0 is easy, so assume that a≠0≠b. Let a=±pmr and b=±pns, where m and n are integers, and r and s are rationals whose numerators and denominators in lowest terms are not divisible by p. (This is the representation used by Rotman; I just collapsed all of the factors not involving p into single rationals r and s.)By definition ∥a∥p=p−m and ∥b∥p=p−n. Without loss of generality assume that m≤n. Then pm≤pn, so p−m≥p−n, and max{∥a∥p,∥b∥p}=p−m. Thus, we must show that ∥a+b∥p≤p−m.Now a+b=±pmr±pns=±pm(r±pn−ms), where n−m≥0. Now let’s write r and s as fractions in lowest terms, say as r=ij and s=kℓ, where p does not divide i,j,k, or ℓ. Thenr±pn−ms=ij±pn−mkℓ=iℓ±pn−mkjjℓ.(1)The denominator of this last fraction clearly has no factors of p. The numerator has no factors of p if n−m>0 (why?). Thus, if n−m>0, ±pm(r±pn−ms) expresses a+b as a product of a power of p and a rational whose numerator and denominator in lowest terms have no factors of p, and by definition ∥a+b∥p=p−m, which is fine.If n−m=0, the numerator in the last fraction of (1) might have factors of p, but if so, then a+b=±pm′t for some m′>m, where t in lowest terms has no factors of p in either numerator or denominator, and ∥a+b∥p=p−m′

結果 (英文) 1:

The case in which a = 0 or b = 0 is easy, so assume that a ≠ 0 ≠ b. Let a = ± pmr and b = ± pns, where m and n are integers, and r and s are rationals whose numerators and Lowest are in Terms Not denominators divisible by P (This IS Used by The representation of Rotman; All of the I Just Collapsed Factors Not Involving The P and S R & lt INTO SINGLE Rationals.).

by Definition ∥a∥p and m = P-∥b ∥p = p-n. Without loss of generality assume that m≤n. Then pm≤pn, so p-m≥p-n, and max {∥a∥p, ∥b∥p} = p-m. Thus , WE MUST Show that ∥a-m + b∥p≤p.

Now PMR A + B = ± ± ± PNS = PM (R & lt ± PN-MS), n-WHERE-m≥0. Now the let S and R & lt apos Write AS the fractions are in Lowest Terms, say AS R & lt = ij of and S = kℓ, WHERE P does Not Divide I, J, K, or ℓ. the Then

R & lt ± PN-MS = ij of ± PN-mkℓ = iℓ ± PN-mkjjℓ. (. 1 )
The denominator of this last fraction clearly has no factors of p. The numerator has no factors of p if n-m> 0 (why?). Thus, if n-m> 0, ± pm (r ± pn-ms) expresses Product AS A + B A A Power of P and A Rational Whose of numerator and denominator in Lowest Terms have have NO Factors of P, and by Definition ∥a-P = m + b∥p, Which IS Fine.

the If-n-m = 0, the numerator in the last fraction of (1) might have factors of p, but if so, then a + b = ± pm't for some m '> m, where t in lowest terms has no factors of p in either numerator or denominator, and ∥a + b∥p = p-m '

結果 (英文) 2:

The case in who a 0 or b?0 is easy, so assume it a s a 0 s b. Let a s'pmr and b'pns, where m and n are s, and r and s are rationals whose seeators and numerors in lowest t erms are not divisible by by p. (This is is the representation used by Rotman; I just collapsed all of the factors not in-the-welling p into singles r and s.)

By definition s.p.p.m. and b-p-p-n. Without loss of generality assume that m?n. Then pm s pn, so p?m?p?n, and max?a-p,?b-p-m. Thus, we must show the he's a-b-p-p-m.

Now a-b-pns-pns-pm(r-pn-ms), where n-m-0. Now let's write r and s as fractions in lowest s, say as r?ij and s-kl, where p dos not divide i, j, k, or l. Then

R?pn-ms-ij-pn-mkl-il-pn-mkjl. (1)
The boutor of this last clear clearly has no factors of p. The numerator has no factors of p if n-m 0 (why?). Thus, if n?m 0, smh.com.au a.b. a product of a power of p and a rational whose numerator and lowest terms have no factors of of p, and by-definition a-b-p-p-m, who is fine.

If n-m?0, the numerator in the last fraction of (1) might have factors of, but if so, then a-b-pm?t for some m, wher e t in lowest terms has no factors of p in either numerator or or, and a-b-b-p-m'

結果 (英文) 3:

The case in which a=0 or b=0 is easy，so assume that a≠0≠b. Let a=±pmr and b=±pns，where m and n are integers，and r and s are rationals whose numerators and denominators in lowest terms are not divisible by p.（This is the representation used by Rotman；I just collapsed all of the factors not involving p into single rationals r and s.）
By definition∥a∥p=p−m and∥b∥p=p−n. Without loss of generality assume that m≤n. Then pm≤pn，so p−m≥p−n，and max{∥a∥p，∥b∥p}=p−m. Thus，we must show that∥a+b∥p≤p−m.
Now a+b=±pmr±pns=±pm（r±pn−ms），where n−m≥0. Now let’s write r and s as fractions in lowest terms，say as r=ij and s=kℓ，where p does not divide i，j，k，orℓ. Then
r±pn−ms=ij±pn−mkℓ=iℓ±pn−mkjjℓ.（1）
The denominator of this last fraction clearly has no factors of p. The numerator has no factors of p if n−m>0（why？） . Thus，if n−m>0，±pm（r±pn−ms）expresses a+b as a product of a power of p and a rational whose numerator and denominator in lowest terms have no factors of p，and by definition∥a+b∥p=p−m，which is fine.
If n−m=0，the numerator in the last fraction of（1）might have factors of p，but if so，then a+b=±pm′t for some m′>m，where t in lowest terms has no factors of p in either numerator or denominator，and∥a+b∥p=p−m′