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#1 Re: Help Me ! » A few questions about subsets. » 2013-10-22 06:44:24
BlitzBall wrote:I always thought an empty set is not a subset of everything "T".
Why not?
Maybe I have forgotten. It has been like 2 years. I wish my memory was long term 
#2 Re: Help Me ! » A few questions about subsets. » 2013-10-22 06:43:20
Ahh I see, so that's how it is interpret. I'll remember that. Thanks a lot 
#3 Help Me ! » A few questions about subsets. » 2013-10-20 05:06:53
- BlitzBall
- Replies: 4
Hello all, it has been a long time since. Hope you are all well
Anyway, I have a few questions about empty sets.
Lets take:
T (domain) = {a,b,c,d,e,f}
A = {a,b}
B = {c,d,e,f}
A∧B ⊆ T = is true.
But how is it true? Because A∧B is an empty set, and I always thought an empty set is not a subset of everything "T".
#4 Re: Help Me ! » Inductive step help! Again » 2012-05-25 08:09:53
Assuming that n'<n means only that you assume it for all naturals less than n.The rest is basic induction.
ahhh you are right. Maybe its best not to thing too much about T(n).
#5 Re: Help Me ! » Inductive step help! Again » 2012-05-25 07:35:46
T(n) = {1 n=1
{T(n/2)+1 Otherwise
Guess: T(n) =< 2 log n
- Assume true for all n' < n [assume T(n/2) =< 2 x log (n/2)]
T(n) =T(n/2) + 1
=<2 x log(n/2) + 1 <---by hypothesis
=2x(log n 1) + 1 <--- log(n/2) = log n log 2
<2log n
I think it's easier to understand without the T(n).
Also, what is n'? Is it primed?
#6 Re: Help Me ! » Inductive step help! Again » 2012-05-25 07:04:35
Hi, I was wondering, do you know how to do prove Time Complexity with Mathematical Induction? It's quite similar to math induction.
#7 Re: Help Me ! » Inductive step help! Again » 2012-05-19 07:08:19
Ahh i see. Thanks for your help btw 
#8 Re: Help Me ! » Inductive step help! Again » 2012-05-19 06:38:46
Yes, it is a bracketing problem.
If you have 2k and you substitute k+1 for k you would not get (2 x k + 1)
You would get 2(k+1).Also, you probably do not know this but that particular problem you have is what ignited the "Battle of Brooklyn."
Battle of Brooklyn, Very interesting lol. Didn't think that would ever end up in a war.
#9 Re: Help Me ! » Inductive step help! Again » 2012-05-19 06:21:30
Hi;
It is given to you in step 2). It is part of what you call the hypothesis.
Ahh i see.
I'm still a little puzzled..
In the inductive hypothesis: (2 × 1) + (2 × 2) + (2 × 3) + · · · + (2 × (k−1)) + (2 × k) = k × (k + 1)
Up to the point where I include n=k+1 in inductive step, I would get something like this:
(2 × 1) + (2 × 2) + (2 × 3) + · · · + (2 × (k+1−1)) + (2 × k+1) = (k+1) × (k+2)
but the actual answer to the inductive step is..
(2 × 1) + (2 × 2) + (2 × 3) + · · · + (2 × k) + (2 × (k+1)) = (k + 1) × (k + 2)
I think it something wrong with my allocation of k+1. Its something you mentioned about the bracketing earlier.
#10 Re: Help Me ! » Inductive step help! Again » 2012-05-19 06:03:13
Hi;
Ahh I get it now! By the way, do we have to simplify Right Hand Side: (k+1)(k+2) near the end?
You only do as much until it becomes clear that you have an identity.
Your bracketing is wrong on the next part of your question. If you substitute k+1 for k in (2 × (k−1)) you would get
(2 × (k−1)) ->(2 × (k+1−1)) = 2k
hmm, then where does the (2×k) come from in the inductive step?
#11 Re: Help Me ! » Inductive step help! Again » 2012-05-19 05:46:02
Hi BlitzBall;
After you have done your base case of 1 you conjecture:
If that is true then
You already believe from step 2 that
So you can substitute it into step 3 (the boxed part).
Start simplifying by algebra.
What you know now is that step 2 implies step 3. That the kth term implies the k+1 term. You already no k=1 is true, so k = 2 is true. The k = 2 implies k = 3 etcetera. Proof by induction.
Ahh I get it now! By the way, do we have to simplify Right Hand Side: (k+1)(k+2) near the end?
Also, in the Inductive hypothesis: On the left hand side, there was (2 × (k−1)). During the Inductive step where we add k+1 to n. I had something like this ... (2 × (K+1)-1). If i expanded that at the start of inductive step, do I only expand the right hand side of (2 × (K+1)-1) to get (2 × k)?
Thanks
#12 Help Me ! » Inductive step help! Again » 2012-05-19 04:52:56
- BlitzBall
- Replies: 17
Hi guys, I'm back, it has been a while.
Look at this answer:
Question: (2×1)+(2×2)+(2×3)+· · ·+(2×(n−1))+(2×n) = n×(n+1)
Base Case: n=1
Inductive step: (2 × 1) + (2 × 2) + (2 × 3) + · · · + (2 × k) + (2 × (k+1)) = (k + 1) × (k + 2)
L.H.S: (2 × 1) + (2 × 2) + · · · + (2 × k) + (2 × (k+1))
= k × (k + 1) + (2 × (k+1)) (by hypothesis)
= (k + 1) × (k + 2)
= R.H.S.
My question is how you get k × (k + 1) + (2 × (k+1)) to (k + 1) × (k + 2)? 
Is it something to do with simplifying or finding the common term?
Thanks.
#13 Re: Help Me ! » Struggling with the Inductive Step :( » 2012-02-12 08:43:37
erm... nope.
#14 Re: Help Me ! » Struggling with the Inductive Step :( » 2012-02-12 08:37:18
Oh right, yes i get that. Thanks for your help guys!
#15 Re: Help Me ! » Struggling with the Inductive Step :( » 2012-02-12 08:24:52
Ohh right i see. In general, do you think induction is hard??
#16 Re: Help Me ! » Struggling with the Inductive Step :( » 2012-02-12 08:16:32
The funny thing is that my question is nearly the exact same as the link you posted lol! By the way with (2k-1) and 2(k+1), are we saying that (2k-1) is like the first term and 2(k+1) is the next?
#17 Re: Help Me ! » Struggling with the Inductive Step :( » 2012-02-12 08:05:55
Yup, because we're finding the proving that the next or any nth term is true.
#18 Re: Help Me ! » Struggling with the Inductive Step :( » 2012-02-12 07:59:14
Ahh I see. I get that now
#19 Re: Help Me ! » Struggling with the Inductive Step :( » 2012-02-12 07:52:31
Oh right. So when you plot K+1 on to (2k-1), you just add 2(k+1)-1 next to the formula??
#20 Re: Help Me ! » Struggling with the Inductive Step :( » 2012-02-12 07:48:51
Why is 2k-1 still exist up to that point? When you plot K+1, (2k-1) should be replaced by 2(k+1)-1.
#21 Re: Help Me ! » Struggling with the Inductive Step :( » 2012-02-12 07:43:12
Yep. Like n=k. Say you plot number 1 in both sides of k. And they are equally true. right??
#22 Re: Help Me ! » Struggling with the Inductive Step :( » 2012-02-12 07:38:07
Hi,
The thing is how do you get K^2 for the left hand side??
Is it because of the Common Factors of both sides which is K ??
#23 Help Me ! » Struggling with the Inductive Step :( » 2012-02-12 07:08:11
- BlitzBall
- Replies: 23
Hi guys, I'm back. The test was like 50/50 and have to wait for my results.
The problem i am having is math induction, I hate it! 
The question is: 1+3+5+..+(2n-1) = n^2 for all positive integers n >= 1
When n=1.
Induction step when n=k+1, I get 1+3+5+...+2(k+1)-1 = (K+1)^2
Now, the next step is to prove it. But I have no idea how to prove 1+3+5+...+2(k+1)-1 = (K+1)^2 is true 
Do I expand the left hand side 2(k+1)-1 and right hand side (K+1)^2 ???
#24 Re: Help Me ! » Algebra help! Inverse and composition functions » 2012-01-23 11:13:20
Okay. thank you once again. Thanks for all the help bobby and Mathisfun.
#25 Re: Help Me ! » Algebra help! Inverse and composition functions » 2012-01-23 11:08:36
Just checking if it's a reflective relation or not. In here: A={1,2,3} and the sets are set out as {(2,2),(1,3),(2,3),(3,3)}. Does this prove it's not Reflexive because (1,1) isn't in that set?