[|2x-1| - |x+5| = 3] && [Show that 7^n -1 is divisible by 6]

[Pontifex]

Short n' Sweet
I'm not sure this is the correct forum. I found the problems in a calculus book, so I assume this is the right place...

So the work:

|2x-1| - |x+5| = 3
|2x-1| - |x+5| -3 = 0

For x >= 0:

2x-1 - (x +5) -3 = 0
2x - 1 -x -5 -3 = 0
x - 9 = 0
x = 9
(This matches up with the book thus far)

For x < 0:

-(2x -1) - -(x+5) - 3 = 0
1 - 2x + x + 5 - 3 = 0
3 - x = 0
x = 3

Completely wrong according to the book. Should be x = -7/8

This makes sense as my answer should be something less than zero, but I can't see my error. I'm almost certain that the error has to do with the second term [- -(x+5)], but I fail to see where I'm messing up.

Show that 7^n -1 is divisible by 6

I remember something similar to this in a proofs class I took, but for the life of me I can't remember the equation denoting a quotient. I believe that would be the first step. I can visualize taking out a number of sixes from the product of "7^n -1" but I'm not sure how to attack it algebraically so that I can prove it.

Thanks for your help.

Longer Version

I tend to be verbose, more than others would usually expect. I usually cause people to skim over my content, thus missing the context or some subtleties of the point I'm trying to make. Hence why I'm splitting up in this way.

I'm re-learning math from the ground up. I was never very good at Math throughout most of my life and I'm marrying a math major. I've been all the way through calculus and can ace most anything I put my mind to. But her influence has lead me to believe that I'm missing out. Most ironically I seem to have a better grasp of the higher level abstract theory than I do the basics. I only got as high as 4th year calculus (Calculus 4d in the nomenclature of my college), but I audited a higher level stats class and was doing quite well, despite complications.

When I was working through the basic introductions in the book I have (Calculus Concepts and Contexts by "James Stewart", if you're interested) I found that I'd had no exposure to the properties of the absolute value function or the properties of inequalities (yay American School system). Working through the problems in the book has been very helpful in making sense of some of the things I'd had to learn by rote before.

So, because of that, I'm not looking to solely solve problems; But to actually think about, understand and visualize what's going on with whichever problem I'm working with.

Thanks in advance.

--Pontifex

Edit: Accidentally hit the submit button before I was through. ><

 

[BigGlenntheHeavy]

\(\displaystyle (7^{n}-1) \ = \ (7-1)(7^{n-1}+7^{n-2}+7^{n-3}+...+7+1) \ = \ 6(7^{n-1}+7^{n-2}+7^{n-3}+...+7+1)\)

\(\displaystyle Hence, \ 6 \ is \ a \ factor \ of \ (7^{n}-1)\)

 

[Pontifex]

\(\displaystyle (7^{n}-1) \ = \ (7-1)(7^{n-1}+7^{n-2}+7^{n-3}+...+1) \ = \ 6(7^{n-1}+7^{n-2}+7^{n-3}+...+1)\)

\(\displaystyle Hence, \ 6 \ is \ a \ factor \ of \ (7^{n}-1)\)

\(\displaystyle (7^{n-1}+7^{n-2}+7^{n-3}+...+1)\)
This is 7^n? Where 1 = 7^0, I suppose.

Why is this a summation? Is this how you usually express it?

Usually when I think of powers it's 7*7*7*7...

No that doesn't seem right, say you have 7^4, as I have above. By your expansion it should be something like: 7^3 + 7^2 + 7^1 + 7^0 = 343 + 49 + 7 + 1 = 400 (where n=4); Where as 7^4 = 2401.

Agree with this:
\(\displaystyle (7-1)(7^{n-1}+7^{n-2}+7^{n-3}+...+1) \ = \ 6(7^{n-1}+7^{n-2}+7^{n-3}+...+1)\)

But where do you get the (7-1) from? Are you pulling it out of the expansion? If so, shouldn't you be noting it?

--Pontifex

 

[DrMike]

He's using the result that a^n-b^n = (a-b)(a[sup:2w3do3os]n-1[/sup:2w3do3os]+a[sup:2w3do3os]n-2[/sup:2w3do3os]b + .. + a[sup:2w3do3os]2[/sup:2w3do3os]b[sup:2w3do3os]n-3[/sup:2w3do3os] + ab[sup:2w3do3os]n-2[/sup:2w3do3os]+b[sup:2w3do3os]n-1[/sub]).

You may prefer :

if 7^{n-1[/sup:2w3do3os] = 6k+1, can you prove 7[sup:2w3do3os]n[/sup:2w3do3os]=6m+1 for some integer m? (Hint : multiply both sides by 7, and write m in terms of k). Have you heard of 'mathematical induction'?}

 

[Pontifex]

He's using the result that a^n-b^n = (a-b)(a[sup:1omnahzt]n-1[/sup:1omnahzt]+a[sup:1omnahzt]n-2[/sup:1omnahzt]b + .. + a[sup:1omnahzt]2[/sup:1omnahzt]b[sup:1omnahzt]n-3[/sup:1omnahzt] + ab[sup:1omnahzt]n-2[/sup:1omnahzt]+b[sup:1omnahzt]n-1[/sub]).

You may prefer :

if 7^{n-1[/sup:1omnahzt] = 6k+1, can you prove 7[sup:1omnahzt]n[/sup:1omnahzt]=6m+1 for some integer m? (Hint : multiply both sides by 7, and write m in terms of k). Have you heard of 'mathematical induction'?}

This is fantastic:

a^n-b^n = (a-b)(a[sup:1omnahzt]n-1[/sup:1omnahzt]+a[sup:1omnahzt]n-2[/sup:1omnahzt]b + .. + a[sup:1omnahzt]2[/sup:1omnahzt]b[sup:1omnahzt]n-3[/sup:1omnahzt] + ab[sup:1omnahzt]n-2[/sup:1omnahzt]+b[sup:1omnahzt]n-1[/sup:1omnahzt])

I did some investigation. I can express powers as sums given this template. Very cool! Thank you!

**** if I know why, but this seems to work:

A^n = (a-1)(A^n-1 + A^n-2 + ... + A + A^0) +1

Your formula for A^n - B^n seems to be intuitively correct, but I can't prove it. My algebra isn't strong enough. I'd be really interested in seeing a proof if you have one handy.

So taking that as a given. Where a=7 and b=1 I suppose. Then the solution nicely falls out.

If 7[sup:1omnahzt]n-1[/sup:1omnahzt] = 6k+1, can you prove 7[sup:1omnahzt]n[/sup:1omnahzt]=6m+1 for some integer m?

I'm not entirely sure what's going on here. I've heard of mathematical induction yes, but I haven't used it in quite a few years. I believe I fixed the equation correctly, can you confirm?

If so that makes a sort of sense given what I see of mathematical induction in the book here. But I'm unsure how I'm supposed to interpret the results:

[7[sup:1omnahzt]n-1[/sup:1omnahzt] = 6k+1]*7
7^n = 42k +1

Thus

6m+1 = 42k +1
m = 7k

(I assume 'k' is supposed to be an integer here as well as 'm')

'k' has to be special integer in this case to satisfy the equality 7[sup:1omnahzt]n-1[/sup:1omnahzt] = 6k+1; e.g. k=2 doesn't work, but k=8 does. And so if we move along to the next step in the n-1, n, etc. sequence then of course 'm' must be 7 times larger to continue to satisfy the equality. As the 7[sup:1omnahzt]n-1[/sup:1omnahzt] grows by a factor of 7 as well.

Where do the expressions '6k+1' and '6m+1' come from? What do they signify?

--Pontifex

 

[galactus]

I like Glenn's approach. But if I may give mny 2 cents and use induction.

We must show that \(\displaystyle 7^{n}-1=6p\)

Show for n=1: \(\displaystyle 7^{1}-1=6..........\text{True}\)

Assume true for \(\displaystyle 7^{k}-1=6p\) and show the the \(\displaystyle P_{k+1}\) case is true.

\(\displaystyle 7^{k+1}-1=6p\)

We have \(\displaystyle 7^{k+1}-1=7(\underbrace{7^{k}-1}_{\text{This is 6p}})+6\)

\(\displaystyle 7\cdot 6p+6=6(7p+1)\)

and it is shown that \(\displaystyle 7^{k+1}-1\) is divisible by 6 and \(\displaystyle P_{k+1}\) has been shown. QED

 

[BigGlenntheHeavy]

But where do you get the (7-1) from? Are you pulling it out of the expansion? If so, shouldn't you be noting it?

No, I'm not pulling it out of the expansion, I"m pulling it out of my ass.

Note: \(\displaystyle a^{n}-1 \ = \ (a-1)(a^{n-1}+a^{n-2}+a^{n-3}+...+a+1) \ for \ all \ n \ greater \ or \ equal \ to \ 1.\)

\(\displaystyle Proof: \ Second \ Principle \ of \ Finite \ Induction\)

\(\displaystyle Hint \ for \ the \ proof: \ a^{n+1}-1 \ = \ (a+1)(a^{n}-1)-a(a^{n-1}-1).\)

\(\displaystyle (7^{6}-1) \ = \ (7-1)(7^{5}+7^{4}+7^{3}+7^{2}+7^{1}+1) \ = \ 117,648 \ = \ (6)(19,608), \ OK?\)

 

[Pontifex]

I like Glenn's approach. But if I may give mny 2 cents and use induction.

We must show that \(\displaystyle 7^{n}-1=6p\)

Show for n=1: \(\displaystyle 7^{1}-1=6..........\text{True}\)

Assume true for \(\displaystyle 7^{k}-1=6p\) and show the the \(\displaystyle P_{k+1}\) case is true.

\(\displaystyle 7^{k+1}-1=6p\)

We have \(\displaystyle 7^{k+1}-1=7(\underbrace{7^{k}-1}_{\text{This is 6p}})+6\)

\(\displaystyle 7\cdot 6p+6=6(7p+1)\)

and it is shown that \(\displaystyle 7^{k+1}-1\) is divisible by 6 and \(\displaystyle P_{k+1}\) has been shown. QED

Perfect. Makes good sense.

I'll have to do some more background work on mathematical induction.

It looks fantastic too. Is there a style guide and / or instructions for the LaTeX you're using for this forum?

--Pontifex

 

[Pontifex]

But where do you get the (7-1) from? Are you pulling it out of the expansion? If so, shouldn't you be noting it?

No, I'm not pulling it out of the expansion, I"m pulling it out of my ***.

Note: \(\displaystyle a^{n}-1 \ = \ (a-1)(a^{n-1}+a^{n-2}+a^{n-3}+...+a+1) \ for \ all \ n \ greater \ or \ equal \ to \ 1.\)

\(\displaystyle Proof: \ Second \ Principle \ of \ Finite \ Induction\)

\(\displaystyle Hint \ for \ the \ proof: \ a^{n+1}-1 \ = \ (a+1)(a^{n}-1)-a(a^{n-1}-1).\)

\(\displaystyle (7^{6}-1) \ = \ (7-1)(7^{5}+7^{4}+7^{3}+7^{2}+7^{1}+1) \ = \ 117,648 \ = \ (6)(19,608), \ OK?\)

Now I'm not familiar with all of the mathematical notation you may be familiar with. This ***, is this an authoritative source of mathematical knowledge? If so can you provide a link? Thanks!
(I am aware of the meaning, just a little joke. =)

I'll have to take your word for the proof, as I said above I'm not familiar with the a^n - b^n expansion as you show it, so I'll have to see a proof before I'm completely satisfied. I'll work through it a bit later, for now the simple inductive proof will have to do.

--Pontifex

 

[Pontifex]

Now the second issue. I assume this is an algebraic issue, but I honestly can't see it. If anyone would like to take a stab, I'd appreciate it.

|2x-1| - |x+5| = 3
|2x-1| - |x+5| -3 = 0

For x >= 0:

2x-1 - (x +5) -3 = 0
2x - 1 -x -5 -3 = 0
x - 9 = 0
x = 9
(This matches up with the book thus far)

For x < 0:

-(2x -1) - -(x+5) - 3 = 0
1 - 2x + x + 5 - 3 = 0
3 - x = 0
x = 3

Completely wrong according to the book. Should be x = -7/8

This makes sense as my answer should be something less than zero, but I can't see my error. I'm almost certain that the error has to do with the second term [- -(x+5)], but I fail to see where I'm messing up.

 

[BigGlenntheHeavy]

Bridge builder, you're great at talking the talk, why not try walking the walk, you might even impress your betroth with your erudition.

 

[Pontifex]

Bridge builder, you're great at talking the talk, why not try walking the walk, you might even impress your betroth with your erudition.

Pithy and Profound. Like it!

Will do. I'll post a proof of your version once I'm able.

Thanks again.

--Pontifex

 

[Deleted member 4993]

Short n' Sweet
I'm not sure this is the correct forum. I found the problems in a calculus book, so I assume this is the right place...

So the work:

|2x-1| - |x+5| = 3
|2x-1| - |x+5| -3 = 0

The way I tackle these problems is such;

There are four choices:

(2x-1) is positive and (x+5) is positive

(2x-1) is positive and (x+5) is negative ? not a possible choice (why?)

(2x-1) is negative and (x+5) is positive

(2x-1) is negative and (x+5) is negative

so first choice : (2x-1) > 0 and (x+5) > 0

(2x-1) - (x+5) = 5

x -4 = 5

x = 9 check conditions

2x -1 = 17 >0 checks

x + 5 = 14 > 0 checks

third choice : (2x-1) <0 and (x+5) >0

-(2x-1) - (x+5) = 3

-3x -4 = 3

x = -7/3 check conditions

2x -1 = -14/3 - 1 = -17/3 < 0 checks

x + 5 = -7/3 +5 = 8/5 > 0 checks

fourth choice: (2x-1) <0 and (x+5) <0

-(2x-1) + (x+5) = 3

-x + 6 = 3

x = 3 check conditions

2x -1 = 6 - 1 = 3 > 0 does not match assumption hence solution discarded

so two solutions to the inequality:

x = 9

and

x = -7/3

You could chek (or find) these solutions by graphing and finding points of intersection of:

y = |2x-1|

and

y = 3 + |x+5|

 

[Pontifex]

Thanks for your reply. I was going to tackle it on my own, but I don't think I would have gotten very far. My example that I was working from to try to understand the mechanics didn't mention this particular wrinkle; But it makes perfect sense in retrospect.

(2x-1) is positive and (x+5) is negative ? not a possible choice (why?)

Because this assumption results in a figure that violates the assumption. Namely x = -1/3. I worked that one out for myself, after staring at it for about 30 minutes.

I was able to duplicate the correct results for x = 9 and x = -7/3, with the guidance of examining the four possibilities. Thanks!

 

[Deleted member 4993]

Thanks for your reply. I was going to tackle it on my own, but I don't think I would have gotten very far. My example that I was working from to try to understand the mechanics didn't mention this particular wrinkle; But it makes perfect sense in retrospect.

(2x-1) is positive and (x+5) is negative ? not a possible choice (why?)

Because this assumption results in a figure that violates the assumption. Namely x = -1/3.

Correct.

However, more straight-forward logic is:

if x+5 < 0 then x < 0 (more specifically x < -5) - then (2x-1) < 0 (cannot be positive)


I worked that one out for myself, after staring at it for about 30 minutes.

I was able to duplicate the correct results for x = 9 and x = -7/3, with the guidance of examining the four possibilities. Thanks!