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# Trouble with Proofs/Inductive reasoning watch

1. Got a few questions I want to run by you guys, would love for some tips and pointers.

First:

What I did:

If n is odd it can be written as:

where a is some Integer.

This shows that any odd number can be written as where is some Integer. Therefore when dividing by 4 we are always left with a remainder of one.

Is my reasoning solid? Can it be written more formally?
--------------------------------------

Second:

My working:

Conjecture: Works for 1 and n >= 5

Assume it works for n=k

For n=k+1

And now I'm stuck. I've seen an example of this kind but it was brief and I couldn't really follow it, any pointers?

-----------------------------------

I'm concerned about my proof here because its very short and the question is 10 marks:

The perimeter of rectangles with sides and is:

.

The diagonal from one corner to the opposite corner is given by:

Given that is the shortest side on the rectangle, for c to be smallest, . Therefore:

Which is the equation of each diagonal in a square with side length , therefore rectangles with their diagonal minimized are squares..

Fin. Any good? Any errors? It did say 'by considering rectangles with Perimeter A.. which I'm not sure if I did'
----------------------------------------
Finally:

I'm pretty stuck with part b. I got as far as:

But that's about it. Would love some pointers.

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2. (Original post by SamKeene)
Got a few questions I want to run by you guys, would love for some tips and pointers.

First:

What I did:

If n is odd it can be written as:

where a is some Integer.

This shows that any odd number can be written as where is some Integer. Therefore when dividing by 4 we are always left with a remainder of one.

Is my reasoning solid? Can it be written more formally?
--------------------------------------

Second:

My working:

Conjecture: Works for 1 and n >= 5

Assume it works for n=k

For n=k+1

And now I'm stuck. I've seen an example of this kind but it was brief and I couldn't really follow it, any pointers?

-----------------------------------

I'm concerned about my proof here because its very short and the question is 10 marks:

The perimeter of rectangles with sides and is:

.

The diagonal from one corner to the opposite corner is given by:

Given that is the shortest side on the rectangle, for c to be smallest, . Therefore:

Which is the equation of each diagonal in a square with side length , therefore rectangles with their diagonal minimized are squares..

Fin. Any good? Any errors? It did say 'by considering rectangles with Perimeter A.. which I'm not sure if I did'
----------------------------------------
Finally:

I'm pretty stuck with part b. I got as far as:

But that's about it. Would love some pointers.

7 SOLID AS A ROCK
12 I am looking at the manipulation now
11 clueless in Seattle
13 I would love to see this proof too
3. (Original post by TeeEm)
13 I would love to see this proof too
I'm sure it would involve:

Then saying since P's factors can only be 1 and P either:

or

and going from there, probably considering that (a+b) and (a-b) must both be odd.
4. (Original post by SamKeene)
Got a few questions I want to run by you guys, would love for some tips and pointers.

First:

What I did:

If n is odd it can be written as:

where a is some Integer.

This shows that any odd number can be written as where is some Integer. Therefore when dividing by 4 we are always left with a remainder of one.

Is my reasoning solid? Can it be written more formally?
--------------------------------------

Second:

My working:

Conjecture: Works for 1 and n >= 5

Assume it works for n=k

For n=k+1

And now I'm stuck. I've seen an example of this kind but it was brief and I couldn't really follow it, any pointers?

-----------------------------------

I'm concerned about my proof here because its very short and the question is 10 marks:

The perimeter of rectangles with sides and is:

.

The diagonal from one corner to the opposite corner is given by:

Given that is the shortest side on the rectangle, for c to be smallest, . Therefore:

Which is the equation of each diagonal in a square with side length , therefore rectangles with their diagonal minimized are squares..

Fin. Any good? Any errors? It did say 'by considering rectangles with Perimeter A.. which I'm not sure if I did'
----------------------------------------
Finally:

I'm pretty stuck with part b. I got as far as:

But that's about it. Would love some pointers.

coming back on the induction

it works with n=5

assume 2k > k2

2 . 2k >2 . k2

2k+1 > k2+k2

2k+1 > k2+(2k +1)

(k2>2k+1 for k>=5)

2 k+1 > ...

and you can finish off
5. (Original post by TeeEm)
2k+1 > k2+k2

2k+1 > k2+(2k +1)
Could you explain how you got from the first line to the second?
6. (Original post by SamKeene)
Could you explain how you got from the first line to the second?
I replaced k2 by 2k +1 because it suits me

the inequality still holds since by doing so I am making the RHS even smaller
7. Re that last question. Consider the general difference between two consecutive squares. Notice anything?

(Once you spot the pattern, I believe it will be fairly simple to sort out the uniqueness of the solution using prime properties - admittedly I haven't tried it yet myself. I might try it in my free tomorrow)

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8. (Original post by Krollo)
Re that last question. Consider the general difference between two consecutive squares. Notice anything?

(Once you spot the pattern, I believe it will be fairly simple to sort out the uniqueness of the solution using prime properties - admittedly I haven't tried it yet myself. I might try it in my free tomorrow)

Posted from TSR Mobile
very nice
9. (Original post by TeeEm)
very nice

Posted from TSR Mobile
10. For 11, you've basically gone wrong from the start by assuming that you have a free choice for a and b. If the perimeter is A and the sides are a, b, then P = 2a+2b, and so b = (P-2a)/2. In other words, b is completely determined by a. You assume this not to be the case when you say "if a is the smallest side, we can minimize ... by setting b = a." You can't, because b is determined by a.

Instead, to simplify things, define H = P / 2, then H = a + b, and so b = H - a.

Then the length of the diagonal is ,

Note that we minimize by minimizing f(a), so you can simplify to finding a that minimizes .

You can either finish by using the AM>=GM inequaltiy, or completing the square (as quadratic in a) or differentiating.
11. (Original post by SamKeene)
Got a few questions I want to run by you guys, would love for some tips and pointers.
...
Hey, where did you get these questions they're pretty good
12. (Original post by MathMeister)
Hey, where did you get these questions they're pretty good
Here

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