Abstract Math Problems

[Loktart]

Hello,

I have a few problems I need to workout, each I hit a wall after a few steps.

1. Prove that if n is an integer greater than 4, 2n > n[sup:1jpduz6t]2[/sup:1jpduz6t] + 1.

I understand the base case, proving it, the inductive hypothesis, however, I'm not sure where to go after that with the algebraical inequality

2. Prove carefully, using the axioms for an ordered field, that:
1 + x[sup:1jpduz6t]2[/sup:1jpduz6t] < (1 + x)[sup:1jpduz6t]2[/sup:1jpduz6t] <=> x > 0:

The list of axioms are:

Axioms for an ordered field
(1) (Commutativity) For all a and b, a + b = b + a and ab = ba;
(2) (Associativity) For all a, b and c,(a+b)+c = a+(b+c) and (ab)c = a(bc);
(3) (Distributivity) For all a, b and c, a(b + c) = ab + ac;
(4) (Zero) There is an element 0 such that for all a, a + 0 = a = 0 + a;
(5) (Unity) There is an element 1 such that for all a, a1 = a; furthermore, we
assume that 1 6= 0
(6) (Subtraction) For all a, the equation a+x = 0 has a unique solution x = -a.
Similarly, the equation a + x = b has a unique solution b - a.
(7) (Division) If a 6= 0, The equation ax = 1 has a unique solution x = a[sup:1jpduz6t]-1[/sup:1jpduz6t].
Similarly the equation ax = b has a unique solution x = b=a = ba[sup:1jpduz6t]-1[/sup:1jpduz6t].
(8) (Trichotomy) Either a = b, a < b or b < a, and only one these holds.
(9) (Multiplication Law) If c > 0, then ac < bc if and only if a < b,
if c < 0, then ac < bc if and only if b < a;
(10) (Addition Law) a < b if and only if a + c < b + c;
(11) (Transitivity) If a < b and b < c, then a < c.

Really stumped on this one.


3.Is it possible to find a statement that is logically equivalent to (i.e., has the same
truth table as) (A => B), but only involves the operations "or" and "&"? (and
does not use ~, i.e., negation). Why or why not?

This question asks you to find a logically equiavlent statment to the implication A=>B with the use of ONLY the operators "and" and "or" (i.e. A and B=>B or a etc.).
Looking at the truth table for this we have

A | B | A=>B
T | T | T
T | F | F
F | T | T
F | F | T

So the rewrite has to be equivalent, i.e. the same truth table.

Thanks in advance for any help.

 

[Loktart]

1. Prove that if n is an integer greater than 4, 2[sup:th6bxlxp]n[/sup:th6bxlxp] > n[sup:th6bxlxp]2[/sup:th6bxlxp] + 1.

This question is the one giving me the most problems.

So far I have:

Base Case: 2[sup:th6bxlxp]5[/sup:th6bxlxp] > (5)[sup:th6bxlxp]2[/sup:th6bxlxp] + 1 = 32>26

Inductive Hypthesis: Assume 2[sup:th6bxlxp]n[/sup:th6bxlxp]>n[sup:th6bxlxp]2[/sup:th6bxlxp]+1

2[sup:th6bxlxp](n+1)[/sup:th6bxlxp] > n[sup:th6bxlxp]n+1[/sup:th6bxlxp] + 1


I have no idea where to go from here =( Any help would be appreciated.

 

[pka]

Here is a lemma: \(\displaystyle a^2 + b^2 \geqslant 2ab\; \Rightarrow \;K^2 + 1 \geqslant 2K\)
Suppose that: \(\displaystyle K \geqslant 5\; \Rightarrow \;2^K \geqslant K^2 + 1\)
Look at: \(\displaystyle 2^{K + 1} = 2^K + 2^K \geqslant \left( {K^2 + 1} \right) + \left( {K^2 + 1} \right) \geqslant K^2 + 2K + 1\).

 

[Loktart]

How did you get 2[sup:2fgk2pzt]k+1[/sup:2fgk2pzt]=2[sup:2fgk2pzt]k[/sup:2fgk2pzt]+2[sup:2fgk2pzt]k[/sup:2fgk2pzt]

Isn't 2[sup:2fgk2pzt]k+1[/sup:2fgk2pzt] = 2[sup:2fgk2pzt]k[/sup:2fgk2pzt] X 2[sup:2fgk2pzt]1[/sup:2fgk2pzt]

The lemma you provided is for 2K, how can that apply to 2[sup:2fgk2pzt]k[/sup:2fgk2pzt]?

 

[pka]

How did you get 2[sup:225w2u1g]k+1[/sup:225w2u1g]=2[sup:225w2u1g]k[/sup:225w2u1g]+2[sup:225w2u1g]k[/sup:225w2u1g]
Isn't 2[sup:225w2u1g]k+1[/sup:225w2u1g] = 2[sup:225w2u1g]k[/sup:225w2u1g] X 2[sup:225w2u1g]1[/sup:225w2u1g] YES, THAT IS THE SAME THING!

\(\displaystyle 2^{K + 1} = 2\left( {2^K } \right) = 2^K + 2^K\)

Sorry I just assumed that you could to basic operations.

Here is more: \(\displaystyle \left( {K^2 + 1} \right) + \underbrace {\mathop {\left( {K^2 + 1} \right)}\limits_ \uparrow \geqslant \left( {K^2 + 1} \right) + \mathop {2K}\limits_ \uparrow }_{}\)