Mojo asked in 科學數學 · 1 decade ago

大學數學證明題 -1

請證明:

If each m and n is a counting number then each of m+n and m*n is a counting number.

Update:

這是純粹理論證明

1 Answer

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    1 decade ago
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    Let the counting number set N0 = {0, 1, 2, ...}, and successor function is +1. m and n belong to N0.

    ----------------------

    (prove addition closure property)

    To prove m+n belongs to N0 by 數學歸納法:

    (a)

    If n=0, then m+n = m+0 = m, which belongs to N0.

    Therefore m+n belongs to N0 when n=0.

    (b)

    If n=1, then m+n = m+1, which is the successor of m in N0.

    Therefore m+n belongs to N0 when n=1.

    (c)

    Assume m+n belongs to N0 when n=k, where k belongs to N0.

    By associativity property, we know that m+(k+1) = (m+k)+1, which is successor of m+k in N0.

    Therefore m+n also belongs to N0 when n=k+1.

    Based on 數學歸納法 illustrated above, m+n is counting number.

    -----------------------------------------------

    (prove multiplication closure property)

    To prove m*n belongs to N0 by 數學歸納法:

    (a)

    If n=0, m*n = m*0 = 0, which belogns to N0.

    Therefore m*n belongs to N0 when n=0.

    (b)

    If n=1, m*n = m*1 = m, which belongs to N0.

    Therefore m*n belongs to N0 when n=1.

    (c)

    Assume m*n belongs to N0 when n=k, where k belongs to N0.

    By distributivity property, we know that m*(k+1) = m*k+m*1 = m*k+m.

    Both m*k and m belongs to N0, therefore, based on addition closure property, m*k+m belongs to N0, too.

    Based on 數學歸納法 illustrated above, m*n is counting number.

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    補充說明:

    In the above illustration, I assume that associativity and distributivity property are true for counting numbers by definition.

    If we don't have these property by definition, 那就複雜了...

    2009-04-29 16:50:43 補充:

    ----------------------------------

    剛上網看了一下 Real Number System 的定義, 對於binary operation addition and multiplication 的5個 algebraic property (associativity, commutativity, identity elements, inverse elements, distributivity), 是不用證明的.

    2009-04-29 16:50:47 補充:

    ----------------------------------

    所以我們可以以這些 property 為基礎來證明 closure.

    不過, 若你的題目不是這樣, 可能得講清楚一些才好證明.

    2009-04-30 12:25:53 補充:

    BTW, counting number 有兩種定義, 早期的不含 0, 近期在大部分領域含 0. 如果你的counting number 不含0, 就把上述證明中 n=0 的case 拿掉即可.

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