Real Analysis Minitest #1

Field Axioms

Associativity: (a + b) + c = a + (b + c) / a(bc) = (ab)c

Communativity: a + b = b + a / ab = ba

Identity: a + 0 = a / b * 1 = b

Inverses: a + (-a) = 0 / b(1/b) = 1

Distributitivity: a(b + c) = ab + ac

Order axioms

Trichotomy: x e R-, x = 0, x e R+

Transitivity: If x < y and y < z, then x < z

Compatibility: If x < y, then x + z < y + z / If x < y and z > 0, then xz < yz

Other: for all a, b e R+, a + b and ab e R+

Total order

x < y, y < x, x = y

Natural, Integers, Rational, Real

N = {0, 1, 2, 3,…}

Z = {0, +-1, +-2, +-3,…}

Q = {p/q: p, q e Z, q >0}

R = constructed from Q

Multiplication Rule Proof

1. (+)(+) = (+) → given x, y > 0, 0 * y < xy

2. (+)(-) = (-) → a e R-, b e R+, then -ab e R+, ab + -ab = (a-a)b = 0, by uniqueness of solutions (-a)b = -ab, so ab e R-

3. (-)(+) = (-) → same as above

4. (-)(-) = (+) → a, b e R-, then -b e R+, so a(-b) e R-, By uniqueness of additive inverses a(-b) = -ab, hence -ab e R- and therefore ab e R+

Proposition Proof: the equation ax + b = c is unique

1. Input value for x: a(c-b/a) + b = a/a(c-b) + b = c + 0 = c

2. Find value for x: ax = c - b → 1/a * a = 1/a(c-b) = (c-b/a)

3. Conclusion: x = x, thus unique

Corollary Proof: 0 * a = 0

0 * a + a = a → (0 + 1)a = a → a = a

Corollary Proof: (-1)a = -a

(-1)a + a = 0 → (-1 + 1)a = 0 → (0)a = 0 → 0 = 0

Proposition Proof: if a <= b and c <= d, then a+c <= b+d

since b-a >= 0 and d-c >= ), (b+d) - (a+c) >= 0 → (b-a) + (d-c) >= 0

Triangle Inequality Proof: a, b e R Ia + bI <= IaI + IbI

Ia + bI = a + b or -a + -b, since +-a <= IaI and +-b <= IbI, therefore Ia + bI <= IaI + IbI

Reverse Triangle Inequality Proof: x, y e R, IIxI - IyII <= Ix-yI

IaI = Ia-b+bI <= Ia-bI + IbI → IaI - IbI <= Ia-bI → IbI - IaI <= Ia-bI → IIaI-IbII <= Ia-bI

Proposition Proof: there is no r e Q st r² = 2

1. suppose the contrary, that such r e Q exists: r = p/q p,q e Z, q>0, one is even

2. 2 = r² = p²/q² → 2q² = p², then p² is even and therefore p is even

3. write p = 2k, and obtain 2q² = (2k)² = 4k²

4. q² = 2k² → q² is even → q is even

5. Conclusion: no such r e Q exists because p and q would both be even

LUBP

x e R is an upper bound of S e R if x >= s for all s e S. Say x e R is the LUB of S e R if it is an upper bound and x >= y for any upper bound y of S. Write x = sup(S)

Supremum Theorem Proof: If x e R is an upper bound of S e R, TFAT:

1. x = sup(S)

2. for all ep >0, E s e S st Ix-sI < ep

3. for all ep > 0, E s e S st x - ep < s <= x

1) => 3): suppose contrary, s <= x - ep. Then x - ep/2 < x and is an UB of S. Thus x /= sup(S)

3) => 1): suppose x/= sup(S). Then there is an UB y of S st. y < x. Take ep = x-y, then x- ep = y, so s <= x-ep, proving the contrary of 3.

Theorem Proof: There is a unique s e R+ st s² = 2

1. Let S = {x e R: x >= 0 and x² < 2. This is bounded above by the LUBP, s = sup(S) exists

2. case 1: suppose s² < 2

   1. consider (s+h)² for small h < 1, TBC, (s+h)² = s² + 2sh + h² < s² + 2sh + h = s² + h(2s +1) < 2 <=> h < (2-s²)/(2s +1)

   2. because we can choose a positive h, this shows that s+h e S, and therefore not an UB of S, thus s² >= 2

3. case 2: suppose s² > 2

   1. given ep > 0 TBC, by the sup theorem E x e S st s-ep < x <= s. Then s² - x² = (s+x)(s-x) < 2sep < s² - 2 <=> ep < (s² - 2)/(2s)

   2. because we can choose a positive ep, this shows that E x e S st x² > 2, which is a contradiction. This s² <= 2

4. Conclusion: s² = 2

LUBP completeness

every nonempty subset of R that is bounded above has a LUB

Equivalence relation

Reflexive: m * n = n * m

Symmetric: [m, n] + [p, q] = [p, q] + [m, n]

Transitive: [m, n] ~ [p, q] ~ [r, s]