Need help with True or False for the following... (I will put what I believe As (T) or (F)) 1. For all natural numbers a, a^2 = 0 (mod 4) or a^2 = 1 (mod 4). (T) 2. For all natural numbers a, if a^2 = 2 (mod 4), then 1 = 2. (T) 3. For all natural numbers a, If a^2 = 2 (mod 4), then 1 != 2.  (T) The negation of statement 2, which I believe is "There exists a natural number a such that a^2 ≠ 2 (mod 4) or 1 ≠ 2" (F) The negation of statement 3, which I believe is "There exists a natural number a such that a^2 ≠ 2 (mod 4) and 1 = 2" (F) The converse of statement 2, which I believe  is "If 1 = 2, then for all natural numbers a, a^2 = 2 (mod 4)" The converse of statement 3, which I believe is "If 1 ≠ 2, then for all natural numbers a, a^2 = 2 (mod 4)" (F) The contrapositive of statement 2, which is "If 1 ≠ 2, then there exists a natural number a such that a^2 ≠ 2 (mod 4)" The contrapositive of statement 3, which is "If 1 = 2, then there exists a natural number a such that a^2 ≠ 2 (mod 4)" (F) For sets A and B, A = B if and only if A is a subset of B and B is a subset of A.

Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
icon
Related questions
Question

Need help with True or False for the following... (I will put what I believe As (T) or (F))

1. For all natural numbers a, a^2 = 0 (mod 4) or a^2 = 1 (mod 4). (T)

2. For all natural numbers a, if a^2 = 2 (mod 4), then 1 = 2. (T)

3. For all natural numbers a, If a^2 = 2 (mod 4), then 1 != 2.  (T)

The negation of statement 2, which I believe is "There exists a natural number a such that a^2 ≠ 2 (mod 4) or 1 ≠ 2" (F)

The negation of statement 3, which I believe is "There exists a natural number a such that a^2 ≠ 2 (mod 4) and 1 = 2" (F)

The converse of statement 2, which I believe  is "If 1 = 2, then for all natural numbers a, a^2 = 2 (mod 4)"

The converse of statement 3, which I believe is "If 1 ≠ 2, then for all natural numbers a, a^2 = 2 (mod 4)" (F)

The contrapositive of statement 2, which is "If 1 ≠ 2, then there exists a natural number a such that a^2 ≠ 2 (mod 4)"

The contrapositive of statement 3, which is "If 1 = 2, then there exists a natural number a such that a^2 ≠ 2 (mod 4)" (F)

For sets A and B, A = B if and only if A is a subset of B and B is a subset of A.

Expert Solution
Step 1: step 1

Let's evaluate the provided statements and their negations as well as the converses and contrapositives:

1. Statement: For all natural numbers a, a^2 = 0 (mod 4) or a^2 = 1 (mod 4). (True)
   - This is true. Any natural number squared (a^2) will have a remainder of either 0 or 1 when divided by 4.

2. Statement: For all natural numbers a, if a^2 = 2 (mod 4), then 1 = 2. (False)
   - This statement is false. It implies that if a^2 is congruent to 2 modulo 4, then 1 must equal 2, which is not true.

3. Statement: For all natural numbers a, if a^2 = 2 (mod 4), then 1 ≠ 2. (True)
   - This statement is true. If a^2 is congruent to 2 modulo 4, 1 does not equal 2.


steps

Step by step

Solved in 3 steps

Blurred answer
Similar questions
  • SEE MORE QUESTIONS
Recommended textbooks for you
Advanced Engineering Mathematics
Advanced Engineering Mathematics
Advanced Math
ISBN:
9780470458365
Author:
Erwin Kreyszig
Publisher:
Wiley, John & Sons, Incorporated
Numerical Methods for Engineers
Numerical Methods for Engineers
Advanced Math
ISBN:
9780073397924
Author:
Steven C. Chapra Dr., Raymond P. Canale
Publisher:
McGraw-Hill Education
Introductory Mathematics for Engineering Applicat…
Introductory Mathematics for Engineering Applicat…
Advanced Math
ISBN:
9781118141809
Author:
Nathan Klingbeil
Publisher:
WILEY
Mathematics For Machine Technology
Mathematics For Machine Technology
Advanced Math
ISBN:
9781337798310
Author:
Peterson, John.
Publisher:
Cengage Learning,
Basic Technical Mathematics
Basic Technical Mathematics
Advanced Math
ISBN:
9780134437705
Author:
Washington
Publisher:
PEARSON
Topology
Topology
Advanced Math
ISBN:
9780134689517
Author:
Munkres, James R.
Publisher:
Pearson,