9.6 Mathematical Induction
... (Inductive hypothesis) Assume that any gathering of k people must all have the same blood type. (Inductive step) Suppose k + 1 people are gathered. Send one of them out of the room. The remaining k people must all have the same blood type (by the inductive hypothesis). Now bring the first person bac ...
... (Inductive hypothesis) Assume that any gathering of k people must all have the same blood type. (Inductive step) Suppose k + 1 people are gathered. Send one of them out of the room. The remaining k people must all have the same blood type (by the inductive hypothesis). Now bring the first person bac ...
Fuchsian groups, coverings of Riemann surfaces, subgroup growth
... Our proofs show that in Theorems 1.5, 1.6 and 1.7, any constant c satisfying 1 < c < 2µ(Γ) will do. Theorem 1.7 obviously implies the following. Corollary 1.8 Every Fuchsian group surjects to all but finitely many alternating groups. In other words, Higman’s conjecture holds for all Fuchsian groups ...
... Our proofs show that in Theorems 1.5, 1.6 and 1.7, any constant c satisfying 1 < c < 2µ(Γ) will do. Theorem 1.7 obviously implies the following. Corollary 1.8 Every Fuchsian group surjects to all but finitely many alternating groups. In other words, Higman’s conjecture holds for all Fuchsian groups ...
the strong law of large numbers when the mean is undefined
... since Xx++ ■• ■+Xn+> Xx+ • • ■+Xn = S„.lt follows from (4.6H4.8) that EXX+ > E\XX\ which, since we are assuming P(XX< 0) > 0, is impossible unless EXX+= E\XX\= oo. From (4.6) and (4.7) it now follows that lim sup(S„/n) = oo with probability 1. ...
... since Xx++ ■• ■+Xn+> Xx+ • • ■+Xn = S„.lt follows from (4.6H4.8) that EXX+ > E\XX\ which, since we are assuming P(XX< 0) > 0, is impossible unless EXX+= E\XX\= oo. From (4.6) and (4.7) it now follows that lim sup(S„/n) = oo with probability 1. ...
Pythagoras and the Pythagoreans
... a single number, the Unit, (i.e. one). They treated the unit, which is a point without position, as a point, and a point as a unit having position. The unit was not originally considered a number, because a measure is not the things measured, but the measure of the One is the beginning of number.6 T ...
... a single number, the Unit, (i.e. one). They treated the unit, which is a point without position, as a point, and a point as a unit having position. The unit was not originally considered a number, because a measure is not the things measured, but the measure of the One is the beginning of number.6 T ...
Exercises
... (a) Show that all the odd primes appearing in the prime power factorization of a number of the form n2 + 1 are of the form 4k + 1. (Hint: Find the order of n modulo an odd prime divisor of n2 + 1.) (b) Conclude from the preceding part that there exist infinitely many primes of the form 4k+1. (Hint: ...
... (a) Show that all the odd primes appearing in the prime power factorization of a number of the form n2 + 1 are of the form 4k + 1. (Hint: Find the order of n modulo an odd prime divisor of n2 + 1.) (b) Conclude from the preceding part that there exist infinitely many primes of the form 4k+1. (Hint: ...
Author`s preface
... The famous German mathematician Karl Friedrich Gauss said that mathematics is the queen of the sciences and that number theory is the queen of mathematics. His no less famous colleague Kronecker claimed that natural numbers are from God and everything else is human creation. It remains a fact that w ...
... The famous German mathematician Karl Friedrich Gauss said that mathematics is the queen of the sciences and that number theory is the queen of mathematics. His no less famous colleague Kronecker claimed that natural numbers are from God and everything else is human creation. It remains a fact that w ...
Chapter 08: Divisibility and Prime Numbers
... Given any natural number n > 1, there is a prime factorization of n. This is a consequence of the following general argument. It is based on a fundamental property of natural numbers (Section 4.7): if there is a natural number with a certain property, then there is a least one. Suppose there is a ‘b ...
... Given any natural number n > 1, there is a prime factorization of n. This is a consequence of the following general argument. It is based on a fundamental property of natural numbers (Section 4.7): if there is a natural number with a certain property, then there is a least one. Suppose there is a ‘b ...
Primes and Greatest Common Divisors
... Let S be the set of positive integers of the form ax + by (where a or b may be a negative integer); clearly, S is non-empty as it includes x + y . By the well-ordering principle S has a least element c. So c = ax + by for some a and b. If d|x and d|y then d|ax and d|by and so d|(ax + by ), that is d ...
... Let S be the set of positive integers of the form ax + by (where a or b may be a negative integer); clearly, S is non-empty as it includes x + y . By the well-ordering principle S has a least element c. So c = ax + by for some a and b. If d|x and d|y then d|ax and d|by and so d|(ax + by ), that is d ...
Fermat's Last Theorem
In number theory, Fermat's Last Theorem (sometimes called Fermat's conjecture, especially in older texts) states that no three positive integers a, b, and c can satisfy the equation an + bn = cn for any integer value of n greater than two. The cases n = 1 and n = 2 were known to have infinitely many solutions. This theorem was first conjectured by Pierre de Fermat in 1637 in the margin of a copy of Arithmetica where he claimed he had a proof that was too large to fit in the margin. The first successful proof was released in 1994 by Andrew Wiles, and formally published in 1995, after 358 years of effort by mathematicians. The theretofore unsolved problem stimulated the development of algebraic number theory in the 19th century and the proof of the modularity theorem in the 20th century. It is among the most notable theorems in the history of mathematics and prior to its proof it was in the Guinness Book of World Records for ""most difficult mathematical problems"".