If a ) testimonies conflict one another, b ) there are a small number of witnesses, c ) the speaker has no integrity, d ) the speaker is overly hesitant or bold, or e ) the speaker is known to have motives for lying, then the epistemologist has reason to be skeptical of the speaker's claims.

* If the operation is associative, ( ab ) c = a ( bc ), then the value depends only on the tuple ( a, b, c ).

If c is another common divisor of a and b, then c also divides as + bt

If your side has two aces and a void, then you are not at risk of losing the first two tricks, so long as ( a ) your void is useful ( i. e., does not duplicate the function of an ace that your side holds ) and ( b ) you are not vulnerable to the loss of the first two tricks in the fourth suit ( because, for instance, one of the partnership hands holds a singleton in that suit or the protected king, giving your side second round control ).

If vectors a and b are orthogonal, then and:

If X is a topological space and M is a complete metric space, then the set C < sub > b </ sub >( X, M ) consisting of all continuous bounded functions ƒ from X to M is a closed subspace of B ( X, M ) and hence also complete.

Linear Diophantine equations take the form ax + by = c. If c is the greatest common divisor of a and b then this is Bézout's identity, and the equation has an infinite number of solutions.

It follows that there are also infinitely many solutions if c is a multiple of the greatest common divisor of a and b. If c is not a multiple of the greatest common divisor of a and b, then the Diophantine equation ax + by = c has no solutions.

:“ If an integer n is greater than 2, then has no solutions in non-zero integers a, b, and c. I have a truly marvelous proof of this proposition which this margin is too narrow to contain .”

If f is a surjection and a ~ b ↔ f ( a ) = f ( b ), then g is a bijection.

If ~ and ≈ are two equivalence relations on the same set S, and a ~ b implies a ≈ b for all a, b ∈ S, then ≈ is said to be a coarser relation than ~, and ~ is a finer relation than ≈.

Since mathematics is related to logic, he cites an example from mathematics: If we have a formula like ( a + b )( a-b )= a²-b² it does not tell us how to think mathematically.

* If < math > a < b </ math > and < math > b < c </ math > then < math > a < c </ math >;

* If < math > a < b </ math > and < math > c < d </ math > then < math > a + c < b + d </ math >;

* If < math > a < b </ math > and then < math > ac < bc </ math >;

* If < math > a < b </ math > and < math > c < 0 </ math > then < math > bc < ac </ math >.

*( EF1 ) If a and b are in R and b is nonzero, then there are q and r in R such that and either r = 0 or.

If it were not for an old professor who made me read the classics I would have been stymied on what to do, and now I understand why they are classics ; ;

If I even hint at it do you think it will matter that you are his nephew -- and not even a blood nephew ''??

If the circumstances are faced frankly it is not reasonable to expect this to be true.

If his dancers are sometimes made to look as if they might be creatures from Mars, this is consistent with his intention of placing them in the orbit of another world, a world in which they are freed of their pedestrian identities.

If love reflects the nature of man, as Ortega Y Gasset believes, if the person in love betrays decisively what he is by his behavior in love, then the writers of the beat generation are creating a new literary genre.

If to be innocent is to be helpless, then I had been -- as are we all -- helpless at the start.

Defoe then commented, `` If they Could Draw that young Gentleman into Their Measures They would show themselves quickly, for they are not asham'd to Say They want only a head to Make a beginning ''.

If Jews are identified as a religious body in a controversy that comes before a national or international tribunal, it is obviously compatible with the goal of human dignity to protect freedom of worship.

If they are right, they will prevail of and by themselves.

Without preliminaries, Esther asked him, `` If you are a world citizen, will you take Garry Davis' place in his tent while he goes to the hospital ''??

If we are to believe the list of titles printed in Malraux's latest book, La Metamorphose Des Dieux, Vol. 1 ( ( 1957 ), he is still engaged in writing a large novel under his original title.

If the would-be joiner asks these questions he is not likely to be duped by extremists who are seeking to capitalize on the confusions and the patriotic apprehensions of Americans in a troubled time.

If we break the minister to our bit, we are buying back our own sins.

If the record buyer's tastes are somewhat eclectic or even the slightest bit esoteric, he will find them satisfied on educational records.

If Daddy's books are out of bounds his own picture books are not.

If it will simply delay the debates until the qualifications are closed next spring, and then carry all the candidates on a tour of debates, it can provide a service to the state.

If they are to be commended for foresight in their planning, what then is the judgment of a town council that compounds this problem during the planning stage??

If the Communists are sincere in wanting a united, neutral and disarmed Germany, it might well be advantageous for the German people in this nuclear age.

If only this could be done more often -- with such heartening results -- many of the earth's `` big problems '' would shrink to the insignificances they really are.

If any are left, presently, we may expect to see signs specifically prohibiting the feeding of them too.

If the raw population figures are crucially relevant, then it is idle to think of liberation, as idle as to suppose that Poland might liberate Russia.

If it is not enough that all of our internationalist One Worlders are advocating that we join this market, I refer you to an article in the New York Times' magazine section ( Nov. 12, 1961 ), by Mr. Eric Johnston, entitled `` We Must Join The Common Market ''.

If the UN troops are not mercenaries then the Hessians were not mercenaries either.

Let ( m, n ) be a pair of amicable numbers with m < n, and write m = gM and n = gN where g is the greatest common divisor of m and n. If M and N are both coprime to g and square free then the pair ( m, n ) is said to be regular, otherwise it is called irregular or exotic.

If n ≥ 1 and is an integer, the numbers coprime to n, taken modulo n, form a group with multiplication as operation ; it is written as ( Z / nZ )< sup >×</ sup > or Z < sub > n </ sub >< sup >*</ sup >.

If the ideals A and B of R are coprime, then AB = A ∩ B ; furthermore, if C is a third ideal such that A contains BC, then A contains C. The Chinese remainder theorem is an important statement about coprime ideals.

If GCD ( a, b ) = 1, then a and b are said to be coprime ( or relatively prime ).

If two numbers have no prime factors in common, their greatest common divisor is 1 ( obtained here as an instance of the empty product ), in other words they are coprime.

Fermat's little theorem states that if p is prime and a is coprime to p, then a < sup > p − 1 </ sup > − 1 is divisible by p. If a composite integer x is coprime to an integer a > 1 and x divides a < sup > x − 1 </ sup > − 1, then x is called a Fermat pseudoprime to base a.

If a and p are coprime numbers such that a < sup > p − 1 </ sup > − 1 is divisible by p, then p need not be prime.

If a is any number coprime to n then a is in one of these residue classes, and its powers a, a < sup > 2 </ sup >, ..., a < sup > k </ sup > ≡ 1 ( mod n ) are a subgroup.

If the generalized Riemann hypothesis is true, then for every coprime a and d and for every ε > 0

If GRH is true, then every proper subgroup of the multiplicative group omits a number less than 2 ( ln n )< sup > 2 </ sup >, as well as a number coprime to n less than 3 ( ln n )< sup > 2 </ sup >.

Fermat's test for compositeness, which is based on Fermat's little theorem, works as follows: given an integer n, choose some integer a coprime to n and calculate a < sup > n − 1 </ sup > modulo n. If the result is different from 1, n is composite.

If a, b, and c are coprime positive integers such that a + b = c, it turns out that

: Given an integer n, choose some integer a coprime to n and calculate a < sup > n − 1 </ sup > modulo n. If the result is different from 1, then n is composite.

If n is a positive integer, the integers between 1 and n − 1 which are coprime to n ( or equivalently, the congruence classes coprime to n ) form a group with multiplication modulo n as the operation ; it is denoted by Z < sub > n </ sub >< sup >×</ sup > and is called the group of units modulo n or the group of primitive classes modulo n. As explained in the article multiplicative group of integers modulo n, this group is cyclic if and only if n is equal to 2, 4, p < sup > k </ sup >, or 2 p < sup > k </ sup > where p < sup > k </ sup > is a power of an odd prime number.

If the tested number n is composite, the strong liars a coprime to n are contained in a proper subgroup of the group, which means that if we test all a from a set which generates, one of them must be a witness for the compositeness of n. Assuming the truth of the generalized Riemann hypothesis ( GRH ), it is known that the group is generated by its elements smaller than O (( log n )< sup > 2 </ sup >), which was already noted by Miller.

If the moduli are coprime, the Chinese remainder theorem gives a straightforward formula to obtain the solution.

If X is a smooth algebraic variety of dimension N and n is coprime to the characteristic then the there is a trace map

If is a Dirichlet character of conductor, so that is totally multiplicative and only depends on n modulo N, and if n is not coprime to N then,

If b and c are coprime, we may write s ( b, c ) as

If b, c > 0 are coprime, then

If b and c are coprime positive integers then