$Help Resolving Questions$

Question

$Help Resolving Questions$

To preface I have solved most of these and I need them checked and the others im extremely confused about and need help with.

Q) I attached my proof, i dont know if it is correct

This is my proof

Q) Is this correct?

Q) How to solve?

Q) How to solve this?

Q) Is this correct? for reference reflexive (R), irreflexive (I), symmetric (S), asymmetric (AS), antisymmetric
(ANT), transitive (T), equivalence (EQ, Is TRS), and partial order (PO, is TRA)

Discrete Mathematics

Sellymatt

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Sellymatt

0

I accidentally set the deadline too late! please please if you could give within 6 hours I would greatly appreciate it
- Walter
  
  +2
  
  I have updated the deadline for your question. Please ensure you set the correct deadline when submitting your questions in the future. Thank you!
Savionf

+3

This is a set of 5 multipart question. The offered bounty is too low, specially for such a short deadline.
- Sellymatt
  
  0
  
  I solved most of them I just needed the answers to be checked, but i can increase the bounty.
- Savionf
  
  +3
  
  Please keep in mind that in order to check your answers one need to solve them first!
- Sellymatt
  
  0
  
  I understand! I updated the bounty but unfortunately this is all I can afford, if it is too low for you then you can answer the questions that havent been solved at all, or you can decline to solve :) Sorry
Naturelover

+2

Bounty too low! Way too low…

Answer

The answer is accepted.

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Sellymatt

0

I accidentally set the deadline too late! please please if you could give within 6 hours I would greatly appreciate it

Walter

+2

I have updated the deadline for your question. Please ensure you set the correct deadline when submitting your questions in the future. Thank you!
Walter

+2

I have updated the deadline for your question. Please ensure you set the correct deadline when submitting your questions in the future. Thank you!
Savionf

+3

This is a set of 5 multipart question. The offered bounty is too low, specially for such a short deadline.

Sellymatt

0

I solved most of them I just needed the answers to be checked, but i can increase the bounty.

Savionf

+3

Please keep in mind that in order to check your answers one need to solve them first!

Sellymatt

0

I understand! I updated the bounty but unfortunately this is all I can afford, if it is too low for you then you can answer the questions that havent been solved at all, or you can decline to solve :) Sorry
Sellymatt

0

I solved most of them I just needed the answers to be checked, but i can increase the bounty.
Savionf

+3

Please keep in mind that in order to check your answers one need to solve them first!
Sellymatt

0

I understand! I updated the bounty but unfortunately this is all I can afford, if it is too low for you then you can answer the questions that havent been solved at all, or you can decline to solve :) Sorry

Answer 1

Answers can only be viewed under the following conditions:

The questioner was satisfied with and accepted the answer, or
The answer was evaluated as being 100% correct by the judge.

View the answer

Q1. In your first proof, there are mistakes in the last two lines. The last two lines should be as follows
\[(\frac{1}{n+1})(1-\frac{1}{(n+1)+1})=(\frac{1}{n+1})(\frac{1+(n+1)-1}{(n+1)+1})\]
\[=(\frac{1}{n+1})(\frac{n+1}{(n+1)+1})=\frac{1}{n+2}.\]

Q2. Is the fourth function $f=-7|1-x|-5? $. The domain should be all of $\mathbb{R}$ and the range is $(-\infty, -5]$.

***************
More detils:

The first function: Domain= $\mathbb{R}$ and the range is $[0,1)$
The second function: Domain=$\mathbb{R}$ and the range is $[-3,\infty)$
The third function: Domain=$\mathbb{R} \backslash \{-3,3\}$ and the range is $\mathbb{R}\backslash \{-1/6,0\}$
The fourth function: Domain=$\mathbb{R}$ and the range is $(-\infty, -5]$.

********

Q3. There seems to be a mistake here. I think what you want to prove is
\[1+2+\dots +n=\frac{n(n+1)}{2}.\]
Here is how I would prove it, which is similar to what you have done.
****
We aim to prove that the sum of the first $n$ natural numbers is given by:
\[1 + 2 + \dots + n = \frac{n(n+1)}{2} \]
We proceed by induction.

Base Case: Let $n = 1$.
\[ 1 = \frac{1(1+1)}{2} = \frac{1 \times 2}{2} = 1\]
Thus the base case holds.

Inductive Step: Assume the statement is true for some integer $n$, i.e.,
\[1 + 2 + \dots + k = \frac{k(k+1)}{2}\]
This is our induction hypothesis. Now, consider $n = k+1$:

\[1 + 2 + \dots + k + (k+1) = \frac{k(k+1)}{2} + (k+1) \]
\[= (k+1) \left(\frac{k}{2} + 1\right) = (k+1) \frac{k+2}{2} = \frac{(k+1)(k+2)}{2}\]
This simplifies to $\frac{(k+1)((k+1) + 1)}{2}$, which is the desired form for $n = k+1$.

Thus, by the principle of mathematical induction, the formula $1 + 2 + \dots + n = \frac{n(n+1)}{2}$ holds for all positive integers $n$.
***

Q4 We aim to prove that $2$ divides $3^n - 1$ for all positive integers $n$. That is, $3^n - 1$ is even for all $n$.

Base Case: Let $n = 1$.
\[ 3^1 - 1 = 3 - 1 = 2.\]
Since $2$ is even, the base case holds.

Inductive Step: Assume the statement is true for some integer $k$, i.e., $3^k - 1$ is even. This is our induction hypothesis. Now, consider $n = k+1$:
\[3^{k+1} - 1 = 3 \cdot 3^k - 1 = 3(3^k) - 1=3^k-1+2 \cdot 3^k.\]
Note that we have used the following calculation
\[3^(3^k)=(1+2)3^k=3^k+2\cdot 3^k.\]

Hence
\[3^{k+1} - 1 = (3^k - 1) + 2 \cdot 3^k.\]
Here, $3^k - 1$ is even by the induction hypothesis, and $2 \cdot 3^k$ is clearly even as it is a multiple of $2$.

Thus, $3^{k+1} - 1$ is also even, implying that it is divisible by $2$.

By the principle of mathematical induction, since the base case is true and the inductive step has been shown to hold, we conclude that $3^n - 1$ is divisible by $2$ for all positive integers $n$.

Q5. You have multiple mistakes in this question. I only list your mistakes:

The first relation is not symmetric (S). It is Antisymmetric (ANT). It is not EQ. It is PO.
The second relation: It is Antisymmetric.
Your answer for the third relation is fine.
The fourth relation is Antisymmetric (AS).
The fifth relation is AT, it is also PO.
The last relation is not transitive (T).

Erdos

4.7K

Erdos

0

Let me know if you have any questions.
- Sellymatt
  
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  How would Q2) be written in terms of of a set for range and domain? If you could help with that
- Erdos
  
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  I added more details.
- Sellymatt
  
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  Also just to clarify for Q5 these are just the corrections right? If i did something but you did not mention anything about it, it is correct?
- Erdos
  
  +1
  
  Yes, I only listed the ones that were incorrect. If I did not say anything, it means that what you did was correct.
- Sellymatt
  
  0
  
  if the second one is irreflexive will it be asymmetric along with antisymmetric?
- Sellymatt
  
  0
  
  same for the other parts of Q2) if they are irreflexive does that mean they are asymmetric also?
- Erdos
  
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  Yes, true for the second and fourth parts. Keep in mind that every asymmetric relation is also antisymmetric.
- Erdos
  
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  The last relation is not symmetric or antisymmetric.
- Sellymatt
  
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  does that mean that the last relation is only irreflexive?
- Erdos
  
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  It is also symmetric. It is only I and S.
- Sellymatt
  
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  I have a question for Q4) couldn't we just replace 3^k - 1 with 2k since that is how we represent a even number and then 3^(n+1) - 1 = 3^n x 3^1 - 1 (2k) x 3^1 - 1 2k x 2 = 4k -> hence and even number would this be incorrect?
- Erdos
  
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  No, your computation is incorrect.
- Sellymatt
  
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  Okay thank you
- Sellymatt
  
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  Im not sure I understand where 2⋅3^k cam from in Q4, is there a way to prove without that?. Sorry for all the questions
- Erdos
  
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  I added some more details. It should be clear now.
- Sellymatt
  
  +1
  
  Thank you so much for your help, i hope you have a great night/day !
Sellymatt

0

Thank you so much?? for Q5) did you add the ones I was missing? also for the third relation, how is it transitive because there is a possibility that x is in a class with y and y is in a class with z but x is not in a class with z. for Q2) yes the question is f=−7 |1−x |−5?
- Erdos
  
  0
  
  You are right about the third relation.
Erdos

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Please note that this question took me about 2 hours to answer. I agree with other users that the offered bounty was too low. Where I reside the minimum wage is $20! In the future, I may not accept your questions if I feel the bounty is too low.
- Sellymatt
  
  0
  
  I apologize, would 30 or 40 be appropriate?
- Erdos
  
  0
  
  It's alright. I think it should be at least $40. Note that answering the follow-up question also takes time. If you want to determine what a fair bounty is, consider how much time one needs to spend answering the question and what would be a fair hourly rate for an individual skilled enough to provide the answer. This approach will give you a good idea.
- Sellymatt
  
  0
  
  Could you tell me how I could increase the bounty, because i do not see the option
- Erdos
  
  0
  
  You can not increase the bounty after a question is accepted, but you can leave a tip. If you scroll doesn't you will see the option to leave a tip or buy coffees. This is the link for the affiliate program, in case you wanted to apply: https://matchmaticians.com/affiliates
- Sellymatt
  
  0
  
  Thank you for the link! I really appreciate your help and I sent you coffees, I hope you received it
- Erdos
  
  0
  
  Thank you for the coffees. I am glad I was able to help. Let me know if you have any further questions :)
- Sellymatt
  
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  One last question, for Q3) what is the reason the formula is wrong? like I understand the bug is in the first line meaning the formula, but why?
- Erdos
  
  0
  
  Well, you seem to replace (n+1) by 1/(n+1)(n+1) which does not make any sense at all. The last line is also does not follow from the line before, but you write the correct answer without it being justified.
- Sellymatt
  
  +1
  
  sorry i forgot to mention that this is a bogus proof question, so I only have to identify the line which is incorrect and the reasoning, but the overall answer is correct.
Erdos

+1

I understand that you are a student and it can sometimes be challenging to pay. I suggest you enroll in the website's affiliate program. You may indeed make enough money through the affiliate program to cover payments for your future questions.

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