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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 86,776

Hi 4DLiVing;

Welcome to the forum.

Can you believe that no one has noticed the above statement yet?

**In mathematics, you don't understand things. You just get used to them.Of course that result can be rigorously obtained, but who cares?Combinatorics is Algebra and Algebra is Combinatorics.**

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**4DLiVing****Member**- Registered: 2011-01-08
- Posts: 22

Hey bobbym;

I agree, -1 is not a square.

But... it is a cube. and 1 is a square. So this leaves 0 between a square and a cube.

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 86,776

Exactly! There you go. There is already a proof to this theorem as Ricky points out.

**In mathematics, you don't understand things. You just get used to them.Of course that result can be rigorously obtained, but who cares?Combinatorics is Algebra and Algebra is Combinatorics.**

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**4DLiVing****Member**- Registered: 2011-01-08
- Posts: 22

Ok.. got it... you were simply pointing out that you could not believe someone did not realize that.

This might be something intersting to prove...

For all natural numbers, there exists an infinite amount of n and m's that satisfy this condition: n^2 = m^3.

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 86,776

With the conditions you left out, there is an infinite number of solutions I believe.

**In mathematics, you don't understand things. You just get used to them.Of course that result can be rigorously obtained, but who cares?Combinatorics is Algebra and Algebra is Combinatorics.**

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**4DLiVing****Member**- Registered: 2011-01-08
- Posts: 22

a proof by induction i think would work here...

If you make the function n = (m^3)^(1/2), the solutions are

(1, 1)

(4, 8)

(9, 27)

(16, 64)

...

The independent values are all the perfect squares, the dependent values are all the perfect cubes.

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 86,776

You really do not need one because every sixth power is a square and a cube:

Since there are an infinite number of sixth powers then there is an infinite number of solutions to your equation.

Of course that result can be rigorously obtained, but who cares?

Combinatorics is Algebra and Algebra is Combinatorics.

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**4DLiVing****Member**- Registered: 2011-01-08
- Posts: 22

aha... very interesting! Expansion of the mind again!

Could an induction proof flow like this somewhat? (I haven't done higher level math in a while and I am trying to remember what I did in college.)

We are trying to prove that:

s^2 = c^3 or

s = (c^3)^(1/2).

The solution sets (c, s) are of the form (n^2, n^3) for n = 1, 2,....

Indeed, when n = 1,

we have the solution set (1, 1) and 1 = 1^2 = 1^3 = 1.

In general, substituting c = n^2 and s = n^3 into s = (c^3)^(1/2)

gives us n^3 = ((n^2)^3)^(1/2) = n^3.

Does n + 1 work?

(n + 1)^3 = (((n + 1)^2)^3)^(1/2) = (n + 1)^3

Indeed it does, so there are an infinite amount of s and c that satisfy the condition s^2 = c^3.

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 86,776

I am not working it that way. And every time I try I am coming down to n^2 * n^2 * n^2 = n^3 * n^3. It is an identity you do not need to use induction.

Of course that result can be rigorously obtained, but who cares?

Combinatorics is Algebra and Algebra is Combinatorics.

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**4DLiVing****Member**- Registered: 2011-01-08
- Posts: 22

haha now I see it...

Basically, I am trying to use a grenade launcher to kill an ant, when all I really would need to do is step on the thing!

Thanks for checking into it for me!

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 86,776

Hi 4DLiVing;

Yes, I would use the law of exponents. It is easy to see that there is an infinite amount of them.

N = { 0,1,2,3,4,5,6,7,8,9,10...}

We can map f(n) = n^6 to N. This will produce {0,1,64, 3^6, 4^6, 5^6, 6^6, ...}. The mapping produces a set that is one to one and onto. It is therefore infinite.

Of course that result can be rigorously obtained, but who cares?

Combinatorics is Algebra and Algebra is Combinatorics.

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**DrSteve****Member**- Registered: 2010-11-15
- Posts: 10

Zhylliolom wrote:

Theorem:

There are infinitely many primes.

Proof:

Define a topology on the set of integers by using the arithmetic progressions (from -∞ to ∞) as a basis. It is easy to verify that this yields a topological space. For each prime p let

A[sub]p[/sub] consist of all multiples of p.A[sub]p[/sub] is closed since its complement is the union of all the other arithmetic progressions with difference p. Now letAbe the union of the progressionsA[sub]p[/sub]. If the number of primes is finite, thenAis a finite union of closed sets, hence closed. But all integers except -1 and 1 are multiples of some prime, so the complement ofAis {-1, 1} which is obviously not open. This showsAis not a finite union and there are infinitely many primes.

Here is a much more elementary proof that there are infinitely many primes:

Let

be prime, and . By the Fundamental Theorem of Arithmetic M has a prime factor which clearly must be greater than .*Last edited by DrSteve (2011-01-13 02:45:59)*

If you're going to be taking the SAT, check out my book:

http://thesatmathprep.com/SAT_Sales_Page.html

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**Chalisque****Member**- Registered: 2011-07-15
- Posts: 6

Haven't read the other comments on this thread but the more interesting question is that if you require that p and q be chosen such that q is not in the extension of Q by p, is it still possible that p^q is rational.

chalisque.com/dr

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**namealreadychosen****Member**- Registered: 2011-07-23
- Posts: 16

Interesting proofs involving cubes

1. The only integer solution to (x+y+z)^3=xyz is 0,0,0

If, of x,y,z, all are ==1(mod3) or all are ==2(mod3) then x+y+z is a multiple of 3 but xyz cannot be a multiple of 3 if none if its factors is. If all are multiples of 3 then, except in the case where x,y,z=0,0,0, dividing x,y and z by 3 will also yield a solution, so a non-trivial solution with all multiples of 3 only exists if there is a solution with at least one not a multiple of 3.

If, of x,y,z, two are ==1(mod3) and one is ==2(mod3), (x+y+z)^3== (1+1+2)^3==1(mod3) but xyz==2(mod3). If one is ==1(mod3) and 2 are ==2(mod3) then (x+y+z)^3== (1+2+2)^3==2(mod3) but xyz==1(mod3).

2.

since 1=1+1/2 and and (n+2)(n+1)/2=n+1+n(n+1)/2 therefore all successive successors of 1 satisfy this sum (tounge twister ).

since 1^3= ((1+1)/2)^2 and and .

*Last edited by namealreadychosen (2011-07-31 20:18:49)*

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**radhakrishnamurty padyala****Member**- Registered: 2014-01-22
- Posts: 8

Are the sides of a triangle lenearly independent quantities?

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 86,776

Hi;

I would say not.

Of course that result can be rigorously obtained, but who cares?

Combinatorics is Algebra and Algebra is Combinatorics.

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**ShivamS****Member**- Registered: 2011-02-07
- Posts: 3,537

How's that triangle possible?

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 86,776

Hi;

Could be a (3,4,5) triangle.

Of course that result can be rigorously obtained, but who cares?

Combinatorics is Algebra and Algebra is Combinatorics.

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**ShivamS****Member**- Registered: 2011-02-07
- Posts: 3,537

When you said x, y and x+y, you meant side lengths, right?

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 86,776

Hi;

Not exactly. Those are vectors and the diagram shows that x ( one side of the triangle ) and y ( another side ) when added form the third side ( x + y ).

Now, ( x + y ) is a vector but it is a linear combination of x and y. Therefore it is not linearly dependent.

Of course that result can be rigorously obtained, but who cares?

Combinatorics is Algebra and Algebra is Combinatorics.

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**ShivamS****Member**- Registered: 2011-02-07
- Posts: 3,537

Should have read the post before and seen the arrows... For a second I thought it violated the triangle inequality.

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 86,776

Hi Shivam;

That is tentative. I am not too good with vectors. Could be a load of kaboobly doo.

Of course that result can be rigorously obtained, but who cares?

Combinatorics is Algebra and Algebra is Combinatorics.

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