| T O P I C R E V I E W |
| Dmitry M. |
Posted - 10/21/2007 : 14:28:28
Hello, i'm new to the forum. I'm currently majoring in Math at my local university. I'm taking a math course called Discrete Mathematical Structures and have some homework questions which i'm having trouble solving. I was hoping someone could help me in solving them and explain how they got to that point.
1. Use mathematical induction to prove that: 4^(n) - 1 is divisible by 3 for all n.
2. Use mathematical induction to prove that:
1+2+...+n<((n+1) /2)
3. Use induction to prove the following statement:
"If p is a prime and p divides a^(n) for n>1, then p must divide a itself."
4. Prove that the sum of two prime numbers, each larger than 2, is not a prime number. (Note: This does not require induction)
Any help is appreciated, and thanks to anyone and everyone in advance. |
| 6 L A T E S T R E P L I E S (Newest First) |
| badgerigar |
Posted - 11/03/2007 : 01:23:00
for #3 try looking at the contrapositive of the thing your proving. Then it gets easy. |
| pka |
Posted - 10/21/2007 : 20:04:43
quote:
Originally posted by Dmitry M.What is QED?
Don't be so lazy. Look it up!
http://en.wikipedia.org/wiki/Q.E.D. |
| Dmitry M. |
Posted - 10/21/2007 : 19:00:05
What is QED? |
| galactus |
Posted - 10/21/2007 : 18:15:19
For #2:
The base case, n=1 is true since 1<2
Assume P_k is true and show for P_(k+1):
1+2+3+........+k<((k+1)^2/2).
Show
1+2+3+..........+k+(k+1)<[(k+1)^2]/2+(k+1)
1+2+3+........+k+(k+1)<((k+1)(k+3))/2
2+4+6+..........+2k+2(k+1)<(k+1)(k+3)
But the sum of the first 2(k+1) even integers is k(k+3)
So, k(k+3)<(k+1)(k+3)....QED
As was to be shown
|
| Dmitry M. |
Posted - 10/21/2007 : 18:14:43
quote:
Originally posted by pka
Here is help with #1.
4K+1-1=4K+1+4-4-1=4(4K-1)+(3).
If 4K-1 is divisible by three then so is 4K+1-1. WHY?
Help on #4.
The sum of two odd integers is even.
Any prime greater than 2 is odd.
So for #1 it is so because 4K+1-1 = 4(4K-1)+(3) and we know that for any multiple of 4K-1 it will be divisible by 3 and 3 is divisible by 3. |
| pka |
Posted - 10/21/2007 : 16:25:34
Here is help with #1.
4K+1-1=4K+1+4-4-1=4(4K-1)+(3).
If 4K-1 is divisible by three then so is 4K+1-1. WHY?
Help on #4.
The sum of two odd integers is even.
Any prime greater than 2 is odd.
|
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