Some+Proofs

1.Suppose it is rational so it is= p/q where p and q are integers with no common divisor. 2. raise to exponent of 3 and we have 2 = (p/q)^3 or 2q^3=p^3 3. in LHS we have an even number, so in RHS q must be even. let say p= 2 r where r is an integer. 4. Substitute in 2 q^3=p^3 we have 2 q^3=(2r)^3 or 2 q^3=8 (r^3) or if we divide by 2, q^3=4 (r^3). 5. Now in the RHS we have an even number, so the LHS must be even or q = 2 s, where s is an integer. 6. From the last relation (q = 2 s) and p= 2 r (obtained above), we conclude that q and p have 2 as a common divisor. 7. The steps 1 and 6 are contradictory.
 * Cube root of two is irrationa **l: in brief

 Start with a non-empty pile of N counters. A move consists in splitting a pile into two non-empty parts. Repeat the process by applying the moves to thus obtained piles. The process ends when there is no pile left with more than a single counter.
 * Pile Splitting **: (http://www.cut-the-knot.org/ctk/chocoproofs.shtml)

Theorem
Every process described above that starts with a pile of //n// counters ends in //n// - 1 steps. 

Proof #1 (by induction)

 * 1) If there are just two counters we clearly need just one split.
 * 2) Assume that for numbers 1<//m//<//N// we have already shown that it takes exactly (//m// - 1) splits to separate a pile consisting of //m// counters. Let there be a pile of //N// counters. Split it into two with //m//1 and //m//2 counters, respectively. Of course, //m//1 + //m//2 = //N//. By the induction hypothesis it will take (//m//1 - 1) breaks to split the first pile and (//m//2 - 1) to split the second one. The total will be 1 + (//m//1 - 1) + (//m//2 - 1) = //N// - 1.

Remark
In the proof by induction the quantity //N// - 1 might appear to have come out of the blue. However, there is a justification. The realization that the number of splits only depends on the size of a pile leads to the equation f(//m// + //n//) = f(//m//) + f(//n//) + 1 which I rewrite as g(//m// + //n//) = g(//m//) + g(//n//), where g(//n//) = f(//n//) + 1. g is an additive integer function and thus is bound to be linear.


 * BREAKING CHOCOLATE **

Answer
As many as there are small squares minus 1.

Proof (by induction)

 * 1) If there are just two squares we clearly need just one break.
 * 2) Assume that for numbers 2<m<N we have already shown that it takes exactly m-1 breaks to split a bar consisting of m squares. Let there be a bar of N squares. Split it into two with m1and m2 squares, respectively. Of course, m1 + m2 = N. By the induction hypothesis it will take (m1-1) breaks to split the first bar and (m2-1) to split the second one. The total will be1 + (m1-1)+(m2-1) = N-1.