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32 Cards in this Set
- Front
- Back
Mathematical Proof |
A mathematical proof of the statement S is a sequence of logically valid statements that connect axioms, definitions, and other already validated statements into a demonstration of the correctness of S. The rules of logic and the axioms are agreed on ahead of time. At a minimum, the axioms should be independent and consistent. The amount of detail presented should be appropriated for the intended audience. |
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The Natural Numbers |
The set of natural numbers is denoted by N, and is defined by
N = {0,1,2,3,4,...} |
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The Integers |
The set of integers is denoted by Z, and is defined by
Z = {..., -4,-3,-2,-1,0,1,2,3,4,...} |
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The Rational Numbers |
The set of rational numbers is denoted by Q, and is defined by
Q={p/q | peZ, qeZ, and q != 0} |
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Irrational Numbers |
A real number that is not rational |
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Divisible |
The integer a is divisible by the nonzero integer b if a=bc for some integer c. We denote this by b|a and also say that b divides a. |
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Quotient Remainder Thm |
Let a and b be integers with b != 0. Then there exist unique integers q and r such that a = bq + r and 0 <= r < |b| |
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Even and Odd |
An integer, n, is even if there exists an integer, k, such that n = 2k. An integer is odd if it is not even. |
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Greatest Common Divisor |
Let a and b be integers that are not both 0. The gcd of a and b is a positive integer d such that
- d|a and d|b - If c divides both a and b, then c|d
Denoted gcd(a,b) |
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gcd(a,b)=as + bt |
Let a and b be integers such that at least one is not 0. Then there are integers, s and t, such that gcd(a,b) = as + bt.
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Prime, Composite |
A positive integer p, with p > 1, is said to be prime if its only positive integer divisors are 1 and p. A pos int n, n > 1, that is not prime is called composite. The int 1 is neither prime nor composite. |
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The Fundamental Thm of Arithmetic |
Every integer n, with n >= 2, can be uniquely written as the product of primes in ascending order. |
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Relatively Prime |
Positive integers a and b are relatively prime if gcd(a,b)=1 |
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Pythagorean Triple |
The set of integers {a, b, c} is called a Pyth triple if a^2 + b^2 = c^2. It is called a primitive Pyth trip if a, b, and c have no common prime factor. |
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a -= b (mod m) |
We say that a is congruent to b, mod m, if m divides a - b. This is often expressed as: a-=b(mod m) if and only if there is an integer k for which a - b = km |
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n! |
n (n-1)(n-2) |
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Floor Function; Ceiling Function |
Floor and ceiling functions are defined for all real numbers x by
floor(x) = largest int in interval (x-1, x] ceiling(x) = largest int in interval [x, x =1)
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The Inverse of a, mod m |
Let a be a nonzero integer. An integer ~a is said to be an inverse of a mod m if a(~a) -= 1 (mod m) |
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Inverses mod m |
Let a, meZ with m > 1. Then
1) if gcd(c, m) >1, a has no inverse, mod m 2) if gcd(a, m) =1 then a has an inverse mod m. 3) if ~a and ^a are both inverses for a mod m, then ~a -= ^a (mod m) |
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Euclidean Algorithm |
98 = 66 * (1) + 32 66 = 32 * (2) + 2 32 = 2 * (16) + 0
2 is gcd(66, 98) |
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Chinese Remainder Thm |
Multiply all of the mods to get m.
Then M1 = 2730/3 = 910, 910*1 -= 1 (mod 3) so on and so forth
So if x -= 2 (mod 3)
x = (2*910*1 + ~~~ + ~~~) = solution to linear congruences |
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Fermat's Little Therom |
Let a, p e Z with p a prime. Then
a^p -= a (mod p)
and
a^(p-1) -=
1 (mod p) if gcd(a, p) = 1 0 (mod p) if gcd(a, p) != 1
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Mathematical Induction |
If {P(i)} is a set of statements such that
1) P(1) is true and 2) P(i) -> P(i + 1) for i >= 1
then P(k) is true for all positive integers k. |
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Optimaility of Deferred Acceptance Algorithm |
The DAA produces an assignment that is optimal for every suitor. |
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Infinitude of the Primes |
There is an infinite number of distinct primes |
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0^0 |
1 |
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The Well-Ordering Principle |
Every nonempty set of natural numbers has a smallest element. |
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Proof of QRT |
Existence: 1) r = a-bq 2) S = {a-bq | q e Z and a - bq >= 0} 3) Nonempty: q is bq <= a, a-bq >= 0 4) Well Ordering implies smallest element 5) Let that element be a-bq0 and set r0 = a -bq0 6) Suppose r0 >= |b|. 7) Let q1 = q0 + |b|/b. (q0 +- 1 and bq1 = bq0 + |b|) 8) a = bq0 + r0 = bq0 + |b| + (r0 - |b|) = bq1 + (r0 - |b|) 9) Thus a=bq1+(r0-|b|) and (r0-|b|)>=0 10) (r0-|b|) part of S now, and r0 is greater since |b| is greater than 0, contradicting that r0 is smallest element of S 11) Instead of r0 >= |b|, we must conclude r0 < |b|, q0 and r0 meet reqs of thm, existence estabbed
Uniqueness: 1)a=bq2+r2 and bq3+r3 2)b(q3-q2)=r2-r3, 0 <= r2, r3 < |b| 3)0<=|r2-r3|<|b| 4)|b|>|r2-r3|=|b|*|q3-q2|>=0 5)Only true if q3-q2=0 and q's are ints, which implies q2=q3 and r2=r3 estabbing uniqueness |
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Direct Proof |
Typical |
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Trivial |
If Q is true, then imp true, if P is not, Q is still true, Trivial |
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Indirect |
Prove Contrapositive |
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By Cases Proof |
If multiple cases can be made out of a claim |