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

]]>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.

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

]]>I would say not.

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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 .]]>

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 .]]>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.

]]>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!

]]>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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