MATH225 QUESTIONS WITH THE
CORRECT ANSWERS 2025
N= - correct answers-{1, 2, 3,...}
Z= - correct answers-{... , -2, -1, 0, 1, 2, ...}
Q= - correct answers-{ m/n | m,n ∈Z and n≠0}
Let A be a subset of Q. An elt b of Q is called an upper bound for A if - correct answers-for every a in A, a
≤b.
c is called a least upper bound of A if: - correct answers-1) it is an upper bound for A and 2) for all upper
bounds b of A, c≤b. Call a least upper bound for A, sup(A), or the supremum of A.
A field is a set (F) with 2 operations, called addition and multiplication) s.t. the following properties and
axioms hold - correct answers-8 of them (in notes) assoc, commut, identy, inverse for each
An ordering on F (also called a linear or total ordering) is a relation between pairs of elts of F, (written
x≤y) such that: - correct answers-4 of them (in notes)
F is an ordered field if - correct answers-it is a field, has an ordering, and the following conditions hold:
(2 of them in notes)
Axiom of Completeness - correct answers-Every nonempty A subset of R which has an upper bound, has
a least upper bound (i.e., a supremum)
Nested interval theorem - correct answers-For every nested decreasing sequence of closed bounded
intervals in R [a1, b1] ⊇ [a2,b2] ⊇ [a3, b3]... , there exists a real number that is an element of every one
of those intervals (i.e., the intersection of these intervals is nonempty)
Archimedean property theorem - correct answers-For all x in R, there is some n in N such that n>x. Given
any real number y > 0, there exists an n ∈ N satisfying 1/n < y.
Corollary to Archimedean property: For all y in R (y>0), - correct answers-there exists a n in N such that
1/n < y.
Thm (The rationals are dense in R) - correct answers-For all a<b in R, there is a rational strictly between
them.
Thm: Let X be a set. - correct answers-Then there is a set Y s.t. |X|<|Y|. In fact, if we let P(X) denote the
set of all subsets of X, then |X|<|P(X)|
Let A be a nonempty subset of R and suppose s is an upper bound of A. - correct answers-s=sup A if and
only if for all ε>0, there exists an a in A such that a>s-ε
a≤b if and only if - correct answers-for all ε>0, a<b+ε
for all ε>0, a<b+ε if and only if - correct answers-a≤b
, A sequence (an) converges to a real number a if and only if - correct answers-for every positive number
ε, there exists an N ∈ N such that for all n ≥ N it follows that |an − a| < ε.
(for every positive number ε), (there exists an N ∈ N) such that (for all n ≥ N) it follows that |an − a| < ε if
and only if - correct answers-A sequence (an) converges
A sequence (xn) is bounded if - correct answers-there exists a number M > 0 such that|xn|≤M for all
n∈N.
Let (an) be a convergent sequence. - correct answers-Then (an) is bounded.
(Algebraic Limit Theorem). Let lim an = a, and lim bn = b. Then, - correct answers-(i) lim(can) = ca, for all
c ∈ R; (ii) lim(an +bn)=a+b;
(iii) lim(anbn) = ab;
(iv) lim(an/bn) = a/b, provided b ̸= 0.
(Order Limit Theorem). Assume lim an = a and lim bn = b. - correct answers-(i) If an ≥0 for all n∈N, then
a≥0. (ii) If an ≤bn for all n∈N, then a≤b.
(iii) If there exists c∈R for which c≤bn for all n∈N, then c≤b. Similarly, if an ≤c for all n∈N, then a≤c.
If lim an=a and lim bn=b and for all an≥b, then - correct answers-a>b
Subsequences of a convergent sequence converge to - correct answers-the same limit as the original
sequence.
If (a_n) is monotonic and bounded - correct answers-then it's convergent.
If a sequence an has 2 subsequences which converge, but to different limits, - correct answers-then an
has no limit
A sequence (an) is called a Cauchy sequence if, - correct answers-for every ε > 0, there exists an N ∈ N
such that whenever m,n ≥ N it follows that |an − am| < ε.
A sequence (an) converges to a real number a if - correct answers-for every ε > 0, there exists an N ∈ N
such that whenever n ≥ N it follows that |an − a| < ε.
A sequence is Cauchy if and only if - correct answers-it is convergent
A sequence is convergent if and only if - correct answers-it is Cauchy
the sequence (an) does not converge to a if: - correct answers-(There exists an ε>0) (for all N in N) (there
is an n≥N) |a_n-a|≥ ε
Schröder-Bernstein Thm - correct answers-(Schroder-Bernstein Theorem). Assume there exists a 1- 1
function f : X → Y and another 1-1 function g : Y → X. Follow the steps to show that there exists a 1-1,
onto function h:X→Y and hence X∼Y.
Two real numbers a and b are equal if and only if - correct answers-for every real number ε>0 it follows
that |a−b|<ε.
CORRECT ANSWERS 2025
N= - correct answers-{1, 2, 3,...}
Z= - correct answers-{... , -2, -1, 0, 1, 2, ...}
Q= - correct answers-{ m/n | m,n ∈Z and n≠0}
Let A be a subset of Q. An elt b of Q is called an upper bound for A if - correct answers-for every a in A, a
≤b.
c is called a least upper bound of A if: - correct answers-1) it is an upper bound for A and 2) for all upper
bounds b of A, c≤b. Call a least upper bound for A, sup(A), or the supremum of A.
A field is a set (F) with 2 operations, called addition and multiplication) s.t. the following properties and
axioms hold - correct answers-8 of them (in notes) assoc, commut, identy, inverse for each
An ordering on F (also called a linear or total ordering) is a relation between pairs of elts of F, (written
x≤y) such that: - correct answers-4 of them (in notes)
F is an ordered field if - correct answers-it is a field, has an ordering, and the following conditions hold:
(2 of them in notes)
Axiom of Completeness - correct answers-Every nonempty A subset of R which has an upper bound, has
a least upper bound (i.e., a supremum)
Nested interval theorem - correct answers-For every nested decreasing sequence of closed bounded
intervals in R [a1, b1] ⊇ [a2,b2] ⊇ [a3, b3]... , there exists a real number that is an element of every one
of those intervals (i.e., the intersection of these intervals is nonempty)
Archimedean property theorem - correct answers-For all x in R, there is some n in N such that n>x. Given
any real number y > 0, there exists an n ∈ N satisfying 1/n < y.
Corollary to Archimedean property: For all y in R (y>0), - correct answers-there exists a n in N such that
1/n < y.
Thm (The rationals are dense in R) - correct answers-For all a<b in R, there is a rational strictly between
them.
Thm: Let X be a set. - correct answers-Then there is a set Y s.t. |X|<|Y|. In fact, if we let P(X) denote the
set of all subsets of X, then |X|<|P(X)|
Let A be a nonempty subset of R and suppose s is an upper bound of A. - correct answers-s=sup A if and
only if for all ε>0, there exists an a in A such that a>s-ε
a≤b if and only if - correct answers-for all ε>0, a<b+ε
for all ε>0, a<b+ε if and only if - correct answers-a≤b
, A sequence (an) converges to a real number a if and only if - correct answers-for every positive number
ε, there exists an N ∈ N such that for all n ≥ N it follows that |an − a| < ε.
(for every positive number ε), (there exists an N ∈ N) such that (for all n ≥ N) it follows that |an − a| < ε if
and only if - correct answers-A sequence (an) converges
A sequence (xn) is bounded if - correct answers-there exists a number M > 0 such that|xn|≤M for all
n∈N.
Let (an) be a convergent sequence. - correct answers-Then (an) is bounded.
(Algebraic Limit Theorem). Let lim an = a, and lim bn = b. Then, - correct answers-(i) lim(can) = ca, for all
c ∈ R; (ii) lim(an +bn)=a+b;
(iii) lim(anbn) = ab;
(iv) lim(an/bn) = a/b, provided b ̸= 0.
(Order Limit Theorem). Assume lim an = a and lim bn = b. - correct answers-(i) If an ≥0 for all n∈N, then
a≥0. (ii) If an ≤bn for all n∈N, then a≤b.
(iii) If there exists c∈R for which c≤bn for all n∈N, then c≤b. Similarly, if an ≤c for all n∈N, then a≤c.
If lim an=a and lim bn=b and for all an≥b, then - correct answers-a>b
Subsequences of a convergent sequence converge to - correct answers-the same limit as the original
sequence.
If (a_n) is monotonic and bounded - correct answers-then it's convergent.
If a sequence an has 2 subsequences which converge, but to different limits, - correct answers-then an
has no limit
A sequence (an) is called a Cauchy sequence if, - correct answers-for every ε > 0, there exists an N ∈ N
such that whenever m,n ≥ N it follows that |an − am| < ε.
A sequence (an) converges to a real number a if - correct answers-for every ε > 0, there exists an N ∈ N
such that whenever n ≥ N it follows that |an − a| < ε.
A sequence is Cauchy if and only if - correct answers-it is convergent
A sequence is convergent if and only if - correct answers-it is Cauchy
the sequence (an) does not converge to a if: - correct answers-(There exists an ε>0) (for all N in N) (there
is an n≥N) |a_n-a|≥ ε
Schröder-Bernstein Thm - correct answers-(Schroder-Bernstein Theorem). Assume there exists a 1- 1
function f : X → Y and another 1-1 function g : Y → X. Follow the steps to show that there exists a 1-1,
onto function h:X→Y and hence X∼Y.
Two real numbers a and b are equal if and only if - correct answers-for every real number ε>0 it follows
that |a−b|<ε.