Download 1. Counting (1) Let n be natural number. Prove that the product of n

1. Counting
(1) Let n be natural number. Prove that the product of n consecutive natural numbers
is divisible by n!. That is, if r is a natural number, then prove that r ¨ pr ` 1q ¨ pr `
2q ¨ ¨ ¨ pr ` n ´ 1q is divisible by n!.
` ˘
(Hint : Use the fact that for any natural numbers m and k, m
is a natural number).
k
(2) Let k, n be two natural numbers. Count the cardinality of the following set :
Sk,n “ tpx1 , x2 , . . . , xk q P Z ˆ Z ˆ . . . ˆ Z | x1 ě 1, x2 ě 2, x3 ě 3 . . . , xk ě k,
x1 ` x2 ` x3 ` . . . ` xk “ nu.
You
are
`m`k´1
˘ allowed to use the following fact. The cardinality of the set Tr,m is is
, where Tr,m is defined as follows :
k´1
Tr,m “ tpx1 , x2 , . . . , xr q P Z ˆ Z ˆ . . . ˆ Z | x1 ě 0, x2 ě 0, x3 ě 0 . . . , xr ě 0,
x1 ` x2 ` x3 ` . . . ` xr “ mu,
(Hint : Note that 1 ` 2 ` 3 ` ¨ ¨ ¨ ` k “ kpk ` 1q{2. Now, define a bijection from Sk,n
to Tk,m´kpk`1q{2 )
(3) (Problem 5.30 from D’Angelo and West)
(4) (Problem 5.37 from D’Angelo and West) Let n, k, j be natural numbers such that
n ě k ě j. Prove the following identity
ˆ ˙ ˆ ˙ ˆ ˙ ˆ
˙
n
k
n
n´j
¨
“
¨
.
k
j
j
k´j
(Hint : Just write out the formula for the binomial coefficients)
2. One more arithmetic problem
We shall find out the remainder when 278 is divided by 78. We shall make use of the following
facts
‚ 78 “ 2 ˆ 3 ˆ 13.
‚ 213 “ 2 ` 90 ˆ 91.
‚ 90 “ 2 ˆ 32 ˆ 5
‚ 91 “ 7 ˆ 13.
(1) First, prove that 78 divides 90 ˆ 91.
(2) What is the remainder when 26 “ 64 is divided by 78.
(3) Find the remainder when 278 is divided by 78. (Hint : 278 “ p213 q6 )
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3. Practice Final 3
(1) List all the elements of the following set S without repetition
S “ tf : r3s Ñ r3s | f pxq ‰ x @ x P r3su.
(2) Let us define the following subset of R.
S “ tpa, bq P R ˆ R | a ¨ b ě 0u
The set S is not an equivalence relation on R. Determine whether or not this relation
is reflexive, determine whether or not it is symmetric and determine whether or not
it is transitive. Justify your answers.
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(3) The following lemma (Lemma 0.1) is FALSE.
Lemma 0.1. For every natural number n, we have the following identity
ˆ ˙
n
“ n ¨ pn ´ 1q
2
.
(a) We shall try to provide a FALSE proof for this Lemma 0.1 by trying to count
a set in two different ways. The proof has 6 steps (Step (i), Step (ii), Step (iii),
Step (iv), Step (v) and Step (vi)).
Identity the step that has an error in reasoning. You should write your justification in less than 3 sentences.
Proof. Let n be a natural number. Let rns “ t1, 2, 3, . . . , nu
(i) Let A `be˘ the number of subsets of rns with cardinality 2. By definition,
|A| “ n2 .
(ii) Suppose we have a natural number i such that 1 ď i ď n. Let Ai be the
following subset of A
Ai “ tS P A | i P Su
In other words, elements of Ai are subsets of rns with cardinality 2 and
which has the natural number i as one of its elements. Clearly, Ai Ă A.
(iii) Each element of Ai is a set of the form ti, ju. Now j can take any value
from the set t1, 2, 3 . . . , nu ´ tiu. So, there are n ´ 1 possible values that
j can take. Consequently, |Ai | “ n ´ 1.
(iv) Let ti, ju be a subset of rns with cardinality 2, then ti, ju P Ai . This
shows that every element of A belongs to some Ai , where 1 ď i ď n.
From step (ii), we have that Ai Ă A. So, we have the following identity
A1 Y A2 Y . . . Y An “ A
(v) Hence, we can conclude that |A1 | ` |A2 | ` ¨ ¨ ¨ ` |An | “ |A|.
(vi) From Step (iii), for each i such that 1 ď i ď n, we have that |Ai | “ n ´ 1.
Using the equation in the previous step, we can conclude that n¨pn´1q “
|A|.
` ˘
From step (i) and step (vi), we can conclude that n ¨ pn ´ 1q “ n2 .
(b) Provide two examples of natural numbers n to illustrate
3
`n ˘
2
‰ n ¨ pn ´ 1q.
(4) Let a, m be positive natural numbers. Assume that m ě 2. You can assume that
we have the following identity :
2am ´ 1
1 ` 21¨a ` 22¨a ` 23¨a ` . . . ` 2pm´1q¨a “ a
2 ´1
(a) Let a, b be two natural numbers such that a divides b and a ă b. Use the
identity above to conclude that 2b ´ 1 ą 2a ´ 1.
(b) Let us define the following statements
P pnq : “2n ´ 1 is a prime number”.
Qpnq : “n is a prime number.”
Prove that the statement P pnq ùñ Qpnq is true, for all n P N.
(Hint : Look at the following definition of a prime number - Divppq “ t1, pu.
Prove the contrapositive, i.e. p Qpnq ùñ P pnqq is true. Use the identity
given earlier )
(c) The number 23 is prime. Let x be an integer such that 23 does not divide
x. Assume that x2 ” 1p mod 23q. Prove that x ” 1p mod 23q or x ” ´1p
mod 23q.
(d) The prime factorization of 211 ` 1 “ 3 ˆ 683. Also, the number 683 is prime.
Use these facts to conclude that 23 divides 211 ´ 1.
(Hint : Use Fermat’s little theorem)
(e) Give an example of a natural number n such that the statement Qpnq ùñ P pnq
is false.
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