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Standard VII

Mathematics

Integers

Let a1≤ a2≤...

Question

Let $a_{1} \leq a_{2} \leq . . . \leq a_{n} = m$ be positive integers. Denote by $b_{k}$ the number of those $a_{i}$ for which $a_{i} \geq k .$ Prove that $a_{1} + a_{2} + . . . a_{n} = b_{1} + b_{2} + . . . + b_{m} .$

m \sum j = 1 b_{j},

if counted by columns.

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m \sum j = 1 j_{b},

if counted by rows.

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m \sum j = b_{j},

if counted by columns.

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None of these

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Solution

The correct option is B $m \sum j = 1 b_{j},$ if counted by columns.
Let us consider a $n \times m$ array of numbers $(x i j)$ defined as follows : in row i, the first $a_{i}$ entries are equal to 1 and the remaining $m - a_{i}$ entries are equal to 0. For instance, if n=3 and $a_{1} = 2, a_{2} = 4, a_{3} = 5,$ array is
1 1 0 0 0
1 1 1 1 0
1 1 1 1 1
Now , if we examine column j, we notice that the number of 1's in that column equals the number of those $a_{i}$ greater than or equal to j , hence $b_{j} .$ The desired result follows by adding up in two ways the 1's in the array. The total number of 1's is $n \sum i = 1 a_{i},$ if counted by rows, and $m \sum j = 1 b_{j},$ if counted by columns.

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Let a1≤a2≤...≤an=m be positive integers. Denote by bk the number of those ai for which ai≥k. Prove that a1+a2+...an=b1+b2+...+bm.

Let $a_{1} \leq a_{2} \leq . . . \leq a_{n} = m$ be positive integers. Denote by $b_{k}$ the number of those $a_{i}$ for which $a_{i} \geq k .$ Prove that $a_{1} + a_{2} + . . . a_{n} = b_{1} + b_{2} + . . . + b_{m} .$