• February 22nd 2013, 03:57 AM
James7361539
Hi I'm stuck on this Q

its about arithmetic sequences
I have attached a photo of the Q

I have written out all the general terms but I don't know how to prove
the sum of any term in B plus any term in C is a term in E?

Thanks in advance for replies.
• February 22nd 2013, 06:40 AM
emakarov
Quote:

Originally Posted by James7361539
I have written out all the general terms but I don't know how to prove
the sum of any term in B plus any term in C is a term in E?

Let's think about how someone can help you. Since you have not written your versions of the general terms, we cannot check if they are correct. And to prove the claim about the sum we obviously need the general terms. Therefore, the expectations is that one of the helpers will not only show how to prove the sum claim, but will also write the general terms that are used in it. Once such solution is written, there is no way to find out if you understood it. I am not even suggesting the possibility that you had not done what you've written and that you hope to just copy the solution.

On this forum, we prefer to work in a different way. Please show us the work you've done and explain precisely what your difficulty is. (It can't be that you don't know how to add two general terms, can it?) Then we all can be sure that you benefit from this conversation because it increases your understanding.
• February 22nd 2013, 07:15 AM
James7361539
Quote:

Originally Posted by emakarov
Let's think about how someone can help you. Since you have not written your versions of the general terms, we cannot check if they are correct. And to prove the claim about the sum we obviously need the general terms. Therefore, the expectations is that one of the helpers will not only show how to prove the sum claim, but will also write the general terms that are used in it. Once such solution is written, there is no way to find out if you understood it. I am not even suggesting the possibility that you had not done what you've written and that you hope to just copy the solution.

On this forum, we prefer to work in a different way. Please show us the work you've done and explain precisely what your difficulty is. (It can't be that you don't know how to add two general terms, can it?) Then we all can be sure that you benefit from this conversation because it increases your understanding.

Hey yeah sorry I'm just not very good with latex. Just a question on this form how do you wrap latex language?

Un for B = 2+(n-1)5
Un for C= 3+(n-1)5
Un for E =5+(n-1)5

Un for B + Un for C = 5+2(n-1)5

Does that do it?
• February 22nd 2013, 07:35 AM
emakarov
Quote:

Originally Posted by James7361539
Un for B = 2+(n-1)5
Un for C= 3+(n-1)5
Un for E =5+(n-1)5

Un for B + Un for C = 5+2(n-1)5

Does that do it?

You need to show that for any term b in B and any term c in C there exists a term e in E such that b + c = e. So, it is not enough to consider the situation when both b and c are the nth terms, i.e., when b = 2 + 5(n - 1) and c = 3 + 5(n - 1) for the same n.

An arbitrary term b in B has the form 2 + 5(m - 1) for some m ≥ 1 and an arbitrary term c in C has the form 3 + 5(n - 1) for some n ≥ 1. Then b + c = 5 + 5(m + n - 2). To prove that b + c is in E, we need to find some k ≥ 1 such that b + c = 5 + 5(k - 1), so we need 5 + 5(m + n - 2) = 5 + 5(k - 1). This equation is equivalent to k = m + n - 1. Since m + n - 1 ≥ 1, we found the required k.

Quote:

Originally Posted by James7361539
Hey yeah sorry I'm just not very good with latex. Just a question on this form how do you wrap latex language?

See this post and Wikibooks. In plain text, it is acceptable to write B_n for $B_n$ and x^2 for $x^2$.
• February 22nd 2013, 07:56 AM
James7361539
Quote:

Originally Posted by emakarov
You need to show that for any term b in B and any term c in C there exists a term e in E such that b + c = e. So, it is not enough to consider the situation when both b and c are the nth terms, i.e., when b = 2 + 5(n - 1) and c = 3 + 5(n - 1) for the same n.

An arbitrary term b in B has the form 2 + 5(m - 1) for some m ≥ 1 and an arbitrary term c in C has the form 3 + 5(n - 1) for some n ≥ 1. Then b + c = 5 + 5(m + n - 2). To prove that b + c is in E, we need to find some k ≥ 1 such that b + c = 5 + 5(k - 1), so we need 5 + 5(m + n - 2) = 5 + 5(k - 1). This equation is equivalent to k = m + n - 1. Since m + n - 1 ≥ 1, we found the required k.

See this post and Wikibooks. In plain text, it is acceptable to write B_n for $B_n$ and x^2 for $x^2$.

Oh so you use $.........$
Okay thanks
• February 22nd 2013, 09:25 AM
James7361539
Quote:

Originally Posted by emakarov
You need to show that for any term b in B and any term c in C there exists a term e in E such that b + c = e. So, it is not enough to consider the situation when both b and c are the nth terms, i.e., when b = 2 + 5(n - 1) and c = 3 + 5(n - 1) for the same n.

An arbitrary term b in B has the form 2 + 5(m - 1) for some m ≥ 1 and an arbitrary term c in C has the form 3 + 5(n - 1) for some n ≥ 1. Then b + c = 5 + 5(m + n - 2). To prove that b + c is in E, we need to find some k ≥ 1 such that b + c = 5 + 5(k - 1), so we need 5 + 5(m + n - 2) = 5 + 5(k - 1). This equation is equivalent to k = m + n - 1. Since m + n - 1 ≥ 1, we found the required k.

See this post and Wikibooks. In plain text, it is acceptable to write B_n for $B_n$ and x^2 for $x^2$.

Hey Mr

Thanks for showing me how to do that
but now I am stuck on prooving that the square of every term in B is a term in D.

I have d=4+(l-1)5 (NB thats an l like the letter 'el') for some l>=1

b=2+(m-1)5

And I got to 4+20(m-1)+25(m-1)^2=4+(l-1)

I don't know what to do next can u please help?
Thanks from James
• February 22nd 2013, 09:43 AM
emakarov
Quote:

Originally Posted by James7361539
And I got to 4+20(m-1)+25(m-1)^2=4+(l-1)

The right-hand side should be 4 + 5(l - 1). Solve this equation for l. It is convenient not to multiply (m - 1) through. Then we get l = 4(m - 1) + 5(m - 1)^2 + 1. Obviously, this is an integer, and since m ≥ 1, we have l ≥ 1, as required.
• February 22nd 2013, 09:50 AM
MINOANMAN
James7361539

Did you notice that all the terms of the last sequence are multiples of 5?
get the general term of B: 2+(n-1)5 and add the general term of C to get A+B= 5+2(n-1)5 or A+B=5[2n-1] which is of course a multiple of 5 and therefore a term of E.
as simple as such. try the remaining alone....
• February 22nd 2013, 09:59 AM
emakarov
Quote:

Originally Posted by MINOANMAN
Did you notice that all the terms of the last sequence are multiples of 5?
get the general term of B: 2+(n-1)5 and add the general term of C to get A+B= 5+2(n-1)5 or A+B=5[2n-1] which is of course a multiple of 5 and therefore a term of E.

As described in post #4, there are two flaws in this solution. First, you don't have to add two nth terms; they may have different position in the sequences. Second, in addition to the fact that the sum is divisible by 5, we have to show that the ratio when divided by 5 is nonnegative.
• February 22nd 2013, 01:51 PM
James7361539
Quote:

Originally Posted by emakarov
You need to show that for any term b in B and any term c in C there exists a term e in E such that b + c = e. So, it is not enough to consider the situation when both b and c are the nth terms, i.e., when b = 2 + 5(n - 1) and c = 3 + 5(n - 1) for the same n.

An arbitrary term b in B has the form 2 + 5(m - 1) for some m ≥ 1 and an arbitrary term c in C has the form 3 + 5(n - 1) for some n ≥ 1. Then b + c = 5 + 5(m + n - 2). To prove that b + c is in E, we need to find some k ≥ 1 such that b + c = 5 + 5(k - 1), so we need 5 + 5(m + n - 2) = 5 + 5(k - 1). This equation is equivalent to k = m + n - 1. Since m + n - 1 ≥ 1, we found the required k.

See this post and Wikibooks. In plain text, it is acceptable to write B_n for $B_n$ and x^2 for $x^2$.

Hey Mr

Thanks for helping me with this and the next part
but I was reviewing this and thinking when you write 'we need to find some k>=1' would that mean hypothetically if when we equated b+c=5+5(k-1)
we got k=2 would that mean the statement is true? I've been thinking about it and I think I see that that would mean
any term in B plus any term in C would equal the 2nd term in E which of course is nonsense for these sequences but I'm talking hypothetical here.
Also hypothetically and I think this could actually happen if the sequences were slightly different what if when we
equated b+c=5+5(k-1) we got k=m+n? I think this would mean the statement is true?
In this case k would simply start at 2 ie. the 2nd term (I think if we changed the starting value of B or C to the second term this would happen?
ie. B 7 12 17) Am I right?????
• February 22nd 2013, 03:06 PM
emakarov
Quote:

Originally Posted by James7361539
when you write 'we need to find some k>=1' would that mean hypothetically if when we equated b+c=5+5(k-1)
we got k=2 would that mean the statement is true?

Yes. To show that some number x (which may depend on m and n) is in the sequence E, you need to find some k >= 1 such that x = 5 + 5(k - 1). If it happens that k = 2 satisfies this equation for every x, so be it; then x is in E.

Quote:

Originally Posted by James7361539
Also hypothetically and I think this could actually happen if the sequences were slightly different what if when we
equated b+c=5+5(k-1) we got k=m+n? I think this would mean the statement is true?

Yes. The statement to prove has the form, "For all m >= 1 and every n >= 1 there exists a k >= 1 such that 2 + 5(m - 1) + 3 + 5(n - 1) = 5 + 5(k - 1)". Since the existential quantifier "there exists" follows the universal quantifiers "for all" that introduce m and n, the expression for k is allowed to use m and n. For example, when we say, "For every person x there exists a person y such that y is the father of x", y may depend on x; we don't have to find a single y who is the father of everybody.

Quote:

Originally Posted by James7361539
In this case k would simply start at 2 ie. the 2nd term (I think if we changed the starting value of B or C to the second term this would happen?
ie. B 7 12 17)

Yes, it may happen that you find k that is always >= 2. This does not matter because 5 + 5(k - 1) is still in E for every k >= 2. The problem only asks to prove that the sum of any element of B and any element of C gives some element of E, not that all elements of E can obtained in this way.
• February 23rd 2013, 02:08 AM
James7361539
Quote:

Originally Posted by emakarov
Yes. To show that some number x (which may depend on m and n) is in the sequence E, you need to find some k >= 1 such that x = 5 + 5(k - 1). If it happens that k = 2 satisfies this equation for every x, so be it; then x is in E.

Yes. The statement to prove has the form, "For all m >= 1 and every n >= 1 there exists a k >= 1 such that 2 + 5(m - 1) + 3 + 5(n - 1) = 5 + 5(k - 1)". Since the existential quantifier "there exists" follows the universal quantifiers "for all" that introduce m and n, the expression for k is allowed to use m and n. For example, when we say, "For every person x there exists a person y such that y is the father of x", y may depend on x; we don't have to find a single y who is the father of everybody.

Yes, it may happen that you find k that is always >= 2. This does not matter because 5 + 5(k - 1) is still in E for every k >= 2. The problem only asks to prove that the sum of any element of B and any element of C gives some element of E, not that all elements of E can obtained in this way.

Yeah thanks for confirming that. I didn't really understand that whole bit about 'existential quantifiers' since I don't even know what these are but yeah I think I get it.
• February 23rd 2013, 02:37 AM
James7361539
Quote:

Originally Posted by emakarov
The right-hand side should be 4 + 5(l - 1). Solve this equation for l. It is convenient not to multiply (m - 1) through. Then we get l = 4(m - 1) + 5(m - 1)^2 + 1. Obviously, this is an integer, and since m ≥ 1, we have l ≥ 1, as required.

Hey Mr

I'm sorry if I'm bugging you but now I'm stuck on part 1
Thats the bit about prooving that there are no positive integers x and y such that etc. etc.
Could you please point me in the right direction .
Oh I've done the bit about stating and prooving a similar claim about the square of every term in C.
For that I got

g=5(f-1)^2+6(f-1)+2

Obviously g and f are arbitrary letters.

c=3+(f-1)5 for some f>=1
d=4+(g-1)5 for some g>=1

Anyway I'm pretty sure thats right but I don't know how to do part 1
I know there's probably some link with the 5 sequences but I just don't see it.
Could you please point me in the right direction?