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The Proof is Trivial! Watch

1. Solution 104

Imprmis, notice that . This gives .
Let . Define, , . If for some , , it follows that .
The sequence satisfies .
The set is dense in .
From Heine - Borel we conclude that is a compact Hausdorff space. Now since and agree on a dense subspace of , it follows that they agree on .

Another one?

Problem 117**

Find all continuous functions defined for each , which satisfy , where and are positive integers at least one of which is greater than .

By the way, very elegant solution to problem 112. I solved it using my beloved complex analysis.

I noticed discussion regarding mathematical books. I do not know whether or not Spivak's book is a must, but Rudin's Principles of Mathematical Analysis definitely is. For algebra - look at Lang's book - his approach is rigorous and modern, yet he does not give many examples, not to mention his proofs. In other words, his books will teach you how to run if you know how to walk. Another good reference which comes to my mind is Waerden's Algebra - it is good but a bit out of date. For topology - Munkres is good, but I prefer Kelley's approach.

(Original post by Jkn)
Come do some STEP on my thread! It's really empty
I shall check it out!
2. (Original post by james22)
Does this require spotting some complicated foruier series or is there an elementary solution?
It may or may not be...
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...unproven

(Original post by Mladenov)
I shall check it out!
3. (Original post by Mladenov)
Nice. I will give you something different.

Check your solutions to problem 86. You have missed solutions and is not a solution, for example.

Problem 87*

A group of boys and girls went to a dance party. It is known that for every pair of boys, there are exactly two girls who danced with both of them; and for every pair of girls there are exactly two boys who danced with both of them. Prove that the numbers of girls and boys are equal.
Okay, I don't know if this has already been solved but,

Solution 87

Let there be n girls; , and m boys; .

Now, for every pair of boys there exist 2 girls that danced with both of them and let these be .

If m>n, then as there are pairs of girls and pairs of boys, and , this would imply that more than 2 girls would have danced with atleast one pair of boys, by the pigeonhole principle.

So, m is less than or equal to n.
And so, by symmetry, we have n is less than or equal to m.

And so, we can assert that m = n.
4. (Original post by Jkn)
It may or may not be...
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...unproven

Ahem... Please read the terms set out in the OP.
5. (Original post by und)
Ahem... Please read the terms set out in the OP.
That's before Mladenov showed up, game changer

But I managed to find a derivation online . Izzzaaaaaall good!
6. (Original post by und)
Ahem... Please read the terms set out in the OP.
I just got warning points for that post! what's the world coming too
7. (Original post by Jkn)
I just got warning points for that post! what's the world coming too
I smell Llewellyn
8. (Original post by bananarama2)
I smell Llewellyn
Really? They docked more when I emailed them explaining that they were wrong and that it did not constitute spam.

Instead of getting rid of the points.... or even replying to my email.... they docked two more!!!!! fml

Doubt it! He seems nice. I spoke to him on the STEP thread
9. I've deleted my other question. This is quite possibly the funnest thing of all time

Problem 116*

In the early 20th century, Srinivasa Ramanujan proved that .

i) Find the approximation to given by the first term of this series.

Ramanujan's mathematics is a bit too high level for this thread (according to some people ) and so we will now deal with another type of series that approximates , those using inverse trigonometric functions.

These series are typically written in the form , where a, b, x and y are constants (not necessarily rational or positive) that do not contain and f(x) and g(y) are inverse trigonometric functions (they may or may not be the same).

ii) By considering the equation , find a series in the given form and, hence, find a series expansion for , giving your answer in sigma notation and calculating the number of decimal places of given by the first three terms.

iii) Using what you have now learnt, find a series expansion for , whereby the first three terms give an approximation to a greater degree of accuracy that the series found in part ii.
10. (Original post by Mladenov)
Problem 95** warm-up

Let be positive real numbers. Then, .
Don't think I gave this much thought when I tried it before. Looking at it now, it turns out the proof really does live up the the name of this thread

Solution 95

Using Cauchy-Schwartz, in engel form, we get...

as required.
11. (Original post by MW24595)
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Okay, I don't know if this has already been solved but,

Solution 87

Let there be n girls; , and m boys; .

Now, for every pair of boys there exist 2 girls that danced with both of them and let these be .

If m>n, then as there are pairs of girls and pairs of boys, and , this would imply that more than 2 girls would have danced with atleast one pair of boys, by the pigeonhole principle.

So, m is less than or equal to n.
And so, by symmetry, we have n is less than or equal to m.

And so, we can assert that m = n.
Good job.

(Original post by Jkn)
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Don't think I gave this much thought when I tried it before. Looking at it now, it turns out the proof really does live up the the name of this thread

Solution 95

Using Cauchy-Schwartz, in engel form, we get...

as required.
Problem 74 remains unsolved.
12. For Problem 90 my solution is now complete, so can you remove the 'first part' bit please?
13. (Original post by Mladenov)
Problem 94 remains unsolved.
74 I think you mean! I tried it earlier today! I'll have another look tonight. I'll get it one day :')
(Original post by metaltron)
For Problem 90 my solution is now complete, so can you remove the 'first part' bit please?
Been looking on brilliant have we
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lol jk post away
14. (Original post by Jkn)
74 I think you mean! I tried it earlier today! I'll have another look tonight. I'll get it one day :')

Been looking on brilliant have we
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lol jk post away
Just want to make clear my solution has been unedited for a few days, so no I didn't look on brilliant! And stop rubbing in the fact I'm not yet on Lvl4.
15. (Original post by Jkn)
74 I think you mean! I tried it earlier today! I'll have another look tonight. I'll get it one day :')
Ops, you are right!

It is just slightly harder than 95.

(Original post by Jkn)
Hmm well in that case I suppose the interesting question here is weather or not there are an infinite number of pairs such that m and n are not both Fibonacci numbers?
The answer is no. All the solutions belong to the Fibonacci sequence.

Problem 118**

Let be a positive integer. Let be subsets of such that for any the union of any of the subsets contains at least elements. Prove that there is a permutation of such that .
16. (Original post by Mladenov)
Ops, you are right!

It is just slightly harder than 95.

The answer is no. All the solutions belong to the Fibonacci sequence.
Defend.
17. (Original post by metaltron)
Just want to make clear my solution has been unedited for a few days, so no I didn't look on brilliant! And stop rubbing in the fact I'm not yet on Lvl4.
Hahaha Tell that to "Pierre"
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He takes no ****
18. Solution 108

If we are to have integer roots then the discriminant must be positive, so we have:

The left bracket is positive so the right bracket must be positive too, giving:

Now since and are integers, their difference must be so we have giving us the quadratic:

The trivial solution is , but consider . Then it is clear that the left hand side is greater than the right hand side for all , restricting us to the possibilities and . However, the first case leads to which is a contradiction, and in the second case we have an additional root . Therefore the pair is the only solution.
19. (Original post by und)
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Solution 108

If we are to have integer roots then the discriminant must be positive, so we have:

The left bracket is positive so the right bracket must be positive too, giving:

Now since and are integers, their difference must be so we have giving us the quadratic:

The trivial solution is , but consider . Then it is clear that the left hand side is greater than the right hand side for all , restricting us to the possibilities and . However, the first case leads to which is a contradiction, and in the second case we have an additional root . Therefore the pair is the only solution.
Dayuuuuum puts my solution to shame

Loved the part with trapping the difference! Like the twist at the end of a brilliant film

Awesome bro!
20. (Original post by Jkn)
Defend.
I claim that all solutions are represented by , - up to permutation.

Suppose be a solution such that is minimal. Assume, without loss of generality, . The case leads to which is a contradiction.
Thus, we suppose . Notice that . Moreover, , and . Hence is a solution for which ; thus, we have either or ; these possibilities imply or - contradiction.
Therefore all solutions belong to the Fibonacci sequence.

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