Related questions
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.
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.
Step by stepSolved in 3 steps
- Please, I'm confused on how to do this problem. If it's possible for you to give a some detail on doing it, that'll be great! Thanks!arrow_forwardProve that 2^(27) - 1 is not a prime number.arrow_forward1. Prove that for all numbers of k, (5*2^(2k+1) - 1)/3 is an odd number. 2. Prove that for all natural numbers k, a (4^k - 1)/3 + 2*a*4^k is of the form 1 + 6a 3. Prove that for all natural numbers k, a (5*2^(2k+1) - 1)/3 + a*4^k is of the form 5 + 6a. I need help in all of those three questions please. Proof by induction.arrow_forward
- why/where/how does the 1 and the 0 about from at the end of the last few steps?arrow_forward(a) Compute 7^35 (mod 3) .. Please explain! (b) Find the last three digits of 9^30 (Please use Euler's Theorem AND the Chinese Reminder Theorem)arrow_forwardFor each of the statements below, indicate if it is true or false. You donot have to provide an explanation.(a) (∀n ∈ Z) (∃m ∈ Z) (n = (m^2) + 5)(b) (∀n ∈ Z) (∃m ∈ Z) (m = (n^2) +5)(c) (∃a ∈ Z) (∀b ∈ Z) (a != b + 10) ( != not equal to)(d) (∃a ∈ Z) (∀b ∈ Z) (a < b^2)(e) (∀n ∈ Z) (∃m ∈ Z) (∃p ∈ Z) (mp + 3 = n)arrow_forward
- Prove that n^2 is an even number if and only if 3n + 1 is an odd number by proving If n^2 is even then 3n + 1 is odd.arrow_forwardFor the previous problem how did you calculate for 1,5,17,and 37 for proving that n^2 + 1 is not divisible by 4. Thanks sorry i'm still confused on how to check for divisibility of a proofs.arrow_forwardFind all solutions for x, up to congruence. x ≡ 3^2017 (mod 100)arrow_forward
- Advanced Engineering MathematicsAdvanced MathISBN:9780470458365Author:Erwin KreyszigPublisher:Wiley, John & Sons, Incorporated
Numerical Methods for EngineersAdvanced MathISBN:9780073397924Author:Steven C. Chapra Dr., Raymond P. CanalePublisher:McGraw-Hill Education
Introductory Mathematics for Engineering Applicat...Advanced MathISBN:9781118141809Author:Nathan KlingbeilPublisher:WILEY 
Mathematics For Machine TechnologyAdvanced MathISBN:9781337798310Author:Peterson, John.Publisher:Cengage Learning,
