2011 Prewinter
Tutorial #1
MA 101 : BB/KVS
1. If a, b ∈ R, prove the following.
(a) a + b = 0 ⇒ b = −a
(b) −(−a) = a
(c) (−1)a = −a
(d) (−1)(−1) = 1.
Hint:
START
0=a+b
⇒0=a+b=b+a
⇒ b is an additive inverse of a
⇒ b = −a
START
0=a+b
⇒ (−a) + 0 = (−a) + (a + b)
⇒ (−a) = ((−a) + a) + b
⇒ (−a) = (0) + b
⇒ (−a) = b
(a) Method 1
Method 2
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data
commutativity of addition
definition
uniqueness
data
equals added to equals yields equals
defn of 0 and associativity of addn.
defn of additive inverse
defn of 0
Deficient Method
What is deficient with the method: a + b = 0 ⇒ a + b − a = 0 − a ⇒ b = −a?
(b) By definition of additive inverse, (−a) + a = 0. Using (a) on this, a = −(−a).
More elaborately, definition of additive inverse a + b = b + a = 0 is symmetric in a and b. That is the definition
allows us to say a = −b and b = −a from the two equations. Consequently, the second conclusion b = (−a) in
the first yields a = −(−a).
What is terrible with −(−a) = (− · −)a = +a = a using minus × minus = +?
(c) By definition of additive inverse we have 1 + (−1) = 0. Multiplying by a, 1 · a + (−1) · a = 0 · a. By definition of
multiplicative identity, 1·a = a and Theorem 2.1.2(c) in our text BaSh, 0·a = 0. Consequently, a+(−1)·a = 0.
And, likewise, (−1) · a + a = 0. The latter two equations imply (−1)a = −a, by definition of additive inverse.
(d) Parts (b) and (c) imply (d) for a = (−1).
Remark: Solution in parts (b)–(c) are correct and written with full justification. However, it is still better to write
them in the format of (a) (Methods 1 & 2) for greater clarity.
2. Give examples for
(a) S ⊂ Q such that sup S ∈ Q
(c) U ⊂ R \ Q such that sup U ∈ R \ Q
(b) T ⊂ Q such that sup T 6∈ Q
(d) V ⊂ R \ Q such that sup V 6∈ R \ Q.
Hint:
(a) A clever answer like S = {0} ⊂ Q and sup S = 0, teaches us nothing. This clever answer does not verify that
the supremum of S is indeed 0. Likewise, all examples where S is finite, teach us nothing.
As an excellent exercise: give a definition of maximum of a finite subset of R and prove that it equals the
supremum of the set.
Take S = {− n1 |n ∈ N}. every element of S is of the form of a ratio of integers which allows us to conclude
S ⊂ Q. We next claim sup S = 0 ∈ Q. Merely saying this carries no weight. For a proof, consider :
START
0<1
⇒ 0 + (−1) < 1 + (−1)
⇒ −1 < 0
⇒ (−1) · n1 < 0 · n1
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::
⇒ − n1 < 0
::
done in class
compatibility of < with addition
defn of 0 and additive inverse.
1 < n was done in class and prove separately compatibility of <
with division by positive reals
using (−1)a = −a (question 1c) and 0 · a = 0 for all real a
Above proves that 0 is an upper bound for S. If < 0, by Archimedean, there exists a natural N such that
N > − 1 . Using compatibility of order with multiplication twice, get − N1 > . (Are you sure? Check this
carefully) This proves that anything less than 0 is not an upper bound. Thus, 0 is the least upper bound.
Finally sup S = 0 ∈ Q. Done.
√
(b) Take T = {x ∈ Q|x2 < 2} ⊂ Q. Now, sup T = 2 6∈ Q.
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2011 Prewinter
Tutorial #1
MA 101 : BB/KVS
√
√
(c) Take U = { 2 − n1 |n ∈ N} ⊂ R \ Q with sup U = 2 ∈ R \ Q.
√
(d) Take V = {−
2
n |n
∈ N} ⊂ R \ Q with sup V = 0 6∈ R \ Q.
Solution for part (a) carries full credit while (b)–(c), although correct, gets zero marks.
n2 −1
2 +3
2n
n→∞
3. Applying the definition of limit of a sequence, prove lim
= 12 .
Hint:
2
n −1
2.5
1
1.25
and a = 12 . The difference |an − a| = 2n
2 +3 − 2 = 2n2 +3 < n2 . By the archimedean property,
q
pick a natural N () such that N () > 1.25
. Then, for any n > N (), |an − a| < . So, an → a.
Call an =
n2 −1
2n2 +3
4. Complete the sentence: A sequence (xn ) in R does not converge to x ∈ R, i.e., xn 6→ x if . . .
Hint:
there exists a real 0 > 0 and a sequence of strictly increasing naturals n1 < n2 < n3 < · · · such that |xnk − x| ≥ 0
for all k.
Remark: The above conclusion is 100% correct and get 0% marks. Do you know how to logically
arrive at the conclusion?
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2011 Prewinter
Tutorial #2
MA 101 : BB/KVS
1. Establish the convergence or divergence of the following sequences.
(a) an = 3n
(b) bn =
n2
n+1
(c) cn =
(−1)n n
n+1
(d) dn = (2 + n1 )2
Hint:
(a) Given any real number B, there exists a natural M > B by the archimedean property of reals. By induction
prove that 3M > M . This implies (an ) is unbounded and hence divergent.
n
(b) By induction, n+1
≥ 12 for all natural n. Hence, bn ≥
This makes (bn ) unbounded and divergent.
n
2.
By archimedean property, the latter is unbounded.
(c) Using the odd and even subsequences which converge to distinct limits, we can prove that (cn ) diverges. Here
is a proof which uses the definition. To prove cn 6→ c for any c ∈ R, consider three cases of (i) c < 0, (ii) c = 0
and (iii) c > 0. If c = 0, take = 14 and in other two cases take = |c|
2 and argue as follows:
2k
(2k)
Case (i) All the even terms of the sequence, i.e., for n = 2k we have c − < c + = 2c < 0 < (−1)
. This
2k+1
proves that infinitely many terms of the given sequence are not found in this –neighbourhood of c. Hence
cn 6→ 0.
n
n
1
Case (ii) All terms of the sequence satisfy | (−1)
n+1 | ≥ 2 by previous part of this question. So every term of the
sequence is outside of the –neighbourhood of c = 0. So, cn 6→ 0.
Case (iii) Show that every odd term is outside of the –neighbourhood of c.
(d) The difference |dn − 4| <
6
n
< for all n > N for some N > 6/ given by the archimedean property.
2. Show that if X and Y are sequences such that X is convergent and Y is divergent, then X + Y is divergent.
Hint:
Assuming X + Y is convergent, as a difference of two convergent sequences, Y = (X + Y ) − (X) is convergent. This
contradicts the hypothesis.
Addendum 1: The logician’s converse of A ⇒ B is B ⇒ A. In our question, the logician’s converse says: If X + Y
is divergent then X is convergent and Y is divergent. This statement is false, for example, take X = (n) and
Y = (1). Attempt to fix the logician’s converse as : If X + Y is divergent then one of the sequences picked from
X and Y is convergent while the other is divergent. This fix is false, for example, take X = (n) and Y = (n). The
intelligent converse says: If X + Y is divergent then atleast one of the sequences picked from X and Y is divergent.
The intelligent converse is true. For a proof, its contrapositive is the well known proposition that the sum of two
convergent sequences is convergent.
Addendeum 2: The logician’s contrapositive of A ⇒ B is not(B) ⇒ not(A). In our question, the logician’s
contrapositive says: If X + Y is convergent then X is divergent or Y is convergent. The latter is true, as the
logician’s contrapositive of every true inference is always true. Owing to symmetry, we can get the intelligent
contrapositive: If X + Y is convergent then both the sequences X and Y are convergent or both are divergent. The
intelligent contrapositive is true and more interesting than the logician’s contrapositive.
3. Let x1 := 1 and xn+1 :=
√
2 + xn for n ∈ N. Show that (xn ) converges and find the limit.
Hint:
√
Many students just say xn → l implies xn+1 → l. Using this in the recurrence relation, l = 2 + l. Solve this
quadratic, l = 2 after rejecting the negative answer. If you want to know what is wrong with this method, apply
the same method for y1 = 1 and yn+1 = yn2 + 1.
Using a calculator or a computer program verify that the first few terms of the sequence are approximated
by 1.732051, 1.931852, 1.982890, 1.995718, 1.998929, 1.999732, 1.999933, 1.999983, 1.999996, 1.999999. We may guess
many things, but cannot rule out why (xn ) 6> 5 for n > 1030 .
√
Next, use induction to
√ show that the given sequence is monotonic. For the base case, verify x2 = 3 > 1 = x1 .
Assume √
xk+1 = 2 +√xk > xk for a natural k. By adding two and taking square roots of this inequality, get
xk+2 = 2 + xk+1 > 2 + xk = xk+1 .
Use induction to prove xn < 2 for every n.
The given sequence
being monotonically
increasing and bounded above converges to a real, say, α. Use this in
√
√
xn+1 = 2 + xn to write α = 2 + α. Solving the equation, get α = 2 after discarding α = −1.
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2011 Prewinter
Tutorial #2
MA 101 : BB/KVS
Should the above solution get full credit? There are two implicit assumptions.
are:
√ They
√
Monotonicity of the square–root function i.e., if 0, x < y, then 0 < x < y. Proof: 0 < a < b implies
0 < a2 < ab √
< b2 , by compatibility of order relation with multiplication (fill in the steps). To prove 0 < x < y
√
implies 0 < x < y assume the contrary and arrive at a contradiction using monotonicity of the squaring
function.
Another assumption is xn → α implies xn+1 → α. Prove this.
4. Let xn :=
1
12
+
1
22
+ ··· +
1
n2
for each n ∈ N. Prove that (xn ) converges.
Hint:
1
1
1
+ 2·3
+ · · · + (n−1)·n
= 1 − n1 < 1. Hence, the given sequence is bounded. It is
Verify that for n ≥ 2, xn − 1 ≤ 1·2
easily seen to be increasing. Use monotone convergence theorem.
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