Advanced Mathematics Questions (1 Viewer)

[Loktart]

Hello, I have a couple of questions that I keep hitting a wall on.

1. Mathematical Induction

Prove that for n>4, 2^n > (n^2)+1

(I get stuck here on the algebra of the inequauility)

2. Prove Carefully using the axioms of ordered field that:

1+x^2 < (1+x)^2 <=> x>0

3. Is it possible to find a a statement that is logically equivalent to (ie has the same truth table) as (A=>B), but without the usage of the operators "or", "and" "~"(or any type of negation)

 

[Timothy.Siu]

1. Mathematical Induction

Prove that for n>4, 2^n > (n^2)+1

(I get stuck here on the algebra of the inequauility)

i.e. Prove that 2ⁿ-n²-1 >0 n>4

test for n=5 LHS=32-25-1=6 6>0 therefore n=5 is true

assume true for some particular n=k, i.e.

2^k-k²-1 >0

test for n=k+1 i.e.
that 2^k+1-(k+1)²-1 >0 is true

LHS=2.2^k-(k+1)²-1
=2(2^k-k²-1)+2k²+2 -(k+1)²-1
from our assumption we know that 2^k-k²-1 >0

then, LHS >2k²+2-(k²+2k+1)
LHS> k²+1-2k
LHS> (k-1)²
therefore LHS>0
yada yada yada, ur done

2. Prove Carefully using the axioms of ordered field that:

1+x^2 < (1+x)^2 <=> x>0

i dont really get the question?
just expand it? RHS=x^2+2x+1 and if x>0 then RHS>LHS?
3. Is it possible to find a a statement that is logically equivalent to (ie has the same truth table) as (A=>B), but without the usage of the operators "or", "and" "~"(or any type of negation)

no idea what this question means

 

[Trebla]

2. Prove Carefully using the axioms of ordered field that:

1+x^2 < (1+x)^2 <=> x>0

I'm guessing you mean prove that 1 + x² < (1+x)² is equivalent to x > 0?

The axioms of ordered fields (I think) are:
- If a ≤ b, then a + c ≤ b + c
- If a ≥ 0, and b ≥ 0, then ab ≥ 0

So
x > 0
=> 2x > 0 (using the second axiom with a = 2 (which is > 0) and b = x)
=> x² + 1 + 2x > x² + 1 (using the first axiom with c = x² + 1, a = 0 and b = 2x)
=> (x + 1)² > x² + 1

 

[Loktart]

Below are the Axioms of the Numbered and Ordered Fields

1. Axioms about Addition and Multiplication { Field axioms
(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 = a1.
Similarly the equation ax = b has a unique solution x = b=a = ba1.

2. Order Axioms
(1) (Trichotomy) Either a = b, a < b or b < a, and only one these holds.
(2) (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;
(3) (Addition Law) a < b if and only if a + c < b + c;
(4) (Transitivity) If a < b and b < c, then a < c.

As for the last question, basically you have to attempt to rewrite the implication A => B using the operators ("and" and/or "or") i.e. A and B => A or B such that the rewrite is logically equivalent to the standard implication A=>B.

However, you can't use any other operator, meaning you can't use the "not".

Here is the question directly:

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?

 

[lolokay]

what are you learning this stuff in?

 

[Loktart]

Introduction to Advanced Mathematics (i.e. Abstract Mathematics)

 

[Loktart]

Bump

 

[Affinity]

2. just what you would do usually.. they just want you to put in the precise reason for each step.


3.A -> B is just shorthand for ~A V B, the usual approach is to start with ~ V and define everything in terms of that.

If you want some other ways of writing A -> B you can have


(A->A) -> (A->B)

 

[Loktart]

1. Mathematical Induction

Prove that for n>4, 2^n > (n^2)+1

(I get stuck here on the algebra of the inequauility)

i.e. Prove that 2n-n2-1 >0 n>4

test for n=5 LHS=32-25-1=6 6>0 therefore n=5 is true

assume true for some particular n=k, i.e.

2k-k2-1 >0

test for n=k+1 i.e.
that 2k+1-(k+1)2-1 >0 is true

LHS=2.2k-(k+1)2-1
=2(2k-k2-1)+2k2+2 -(k+1)2-1
from our assumption we know that 2k-k2-1 >0

then, LHS >2k2+2-(k2+2k+1)
LHS> k2+1-2k
LHS> (k-1)2
therefore LHS>0
yada yada yada, ur done

How did you get this

"Prove that 2n-n2-1 >0 n>4"

When the question was to prove

2^n > (n^2) + 1

 

[youngminii]

What the f* is going on?

 

[Loktart]

What the f* is going on?

Basiclly i'm trying to prove by induction that

2^n > (n^2) + 1

 

[youngminii]

Basiclly i'm trying to prove by induction that

2^n > (n^2) + 1

Nono, how come I've never ever heard of anything similar to questions 2 & 3 before o.o
What course are you doing? Is it in HSC?

 

[Loktart]

I actually don't attend HSC, i googled math help forums, and found this to be of some resource. If this is forum is intended for only HSC then I apologize and thank the previous posters for help.

The only question I'm stuck on now is proving the induction question.

Thank you.

 

[3unitz]

How did you get this

"Prove that 2n-n2-1 >0 n>4"

When the question was to prove

2^n > (n^2) + 1

subtract (n^2) + 1 from both sides

 

[Iruka]

For the indution Q, I assume that you have proved the basis step.

We assum that 2^n>n^2+1

Multiply both sides by 2, then

2^(n+1)>2n^2+2.

Now since n>4, n^2 >2n, so 2n^2+2>n^2+2n+2= (n+1)^2+1

 

[jet]

Do you take secondary school education in australia? because maybe the VCE (Victoria) forums may be of more help for you in regards to the last two questions.