梳子 發問時間: 科學數學 · 1 0 年前

基礎數論,教我一下這幾題

1.Prove that if a and b are integers , with b > 0, then there exist unique integers q and r satisfying a = qb + r , where 2b &lE; r < 3b

2.Prove or disprove : If a | (b+c) ,then either a | b or a | c

3.If a | bc , show that a | gcd( a , b ) gcd( a , c )

4.Prove that if gcd(a,b)=1 , then gcd( a + b , ab ) =1

1 個解答

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  • 1 0 年前
    最佳解答

    1. Prove that if a and b are integers , with b > 0, then there exist unique integers q and r satisfying a = qb + r , where 2b ≦ r < 3b

    由Division Algorithm

    ∃!m,n∈Z such that a=mb+n where 0≤n<b

    Then a=(m-2)b+(n+2b)

    Say n+2b=r,m-2=q

    Since m,n are unique , clearly, q and r are unique

    2.Prove or disprove : If a | (b+c) ,then either a | b or a | c

    此命題不真,因為5|(2+3),但5不整除2亦不整除3

    3.If a | bc , show that a | gcd( a , b ) gcd( a , c )

    Let d1=gcd(a,b) d2=gcd(a,c)

    Then by theorem ∃r,s,t,u∈Z such that ar+bs=d1,at+cu=d2

    Then d1*d2=(ar+bs)(at+cu)=a(art+rcu+bst)+(bc)su

    Since a|bc, a|(bc)su

    So a|d1d2

    QED

    4.Prove that if gcd(a,b)=1 , then gcd( a + b , ab ) =1

    First, let gcd(a+b,b)=d

    Then d|(a+b) and d|b

    Then d|(a+b)-b=a

    So d| gcd(a,b), ie d=1

    Similarly, gcd(a+b,a)=1

    So, by theorem ∃r,s,t,u∈Z such that (a+b)r+bs=1,(a+b)t+au=1

    Then 1=[(a+b)r+bs][(a+b)t+au]=(a+b)[rt(a+b)+rau+bst]+(ab)us=1

    By thm, gcd( a + b , ab ) =1

    參考資料: 我自己+數學娘的加持
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