testing header
Math Goodies is a free math help portal for students, teachers, and parents.
Free Math
Interactive Math Goodies Software

Buy Math Goodies Software
testing left nav
Math Forums @ Math Goodies
Math Forums @ Math Goodies
Home | Profile | Register | Active Topics | Members | Search | FAQ
 All Forums
 Homework Help Forums
 Miscellaneous Math Topics
 Mathematical Induction (Prove)

Note: You must be registered in order to post a reply.
To register, click here. Registration is FREE!

Format Mode:
Format: BoldItalicizedUnderlineStrikethrough Align LeftCenteredAlign Right Horizontal Rule Insert HyperlinkInsert EmailInsert Image Insert CodeInsert QuoteInsert List Insert Special Characters Insert Smilie
* Forum Code is ON

Math Symbols
Squared [squared] Cubed [cubed] Square Root [sqrt] Cube Root [cbrt] Pi [pi]
Alpha [alpha] Beta [beta] Gamma [gamma] Theta [theta] Angle [angle]
Degrees [degrees] Times [times] Divide [divide] Less Than or Equal To [less-than] Greater Than or Equal To [greater-than]
Plus Minus [plus-minus] Integral [integral] Sum [sum] Sub 1 [sub-1] Sub 2 [sub-2]
Element Of [element-of] Union [union] Intersect [intersect] Subset [subset] Empty Set [empty-set]

  Check here to subscribe to this topic.

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:

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
Originally posted by Dmitry M.What is QED?

Don't be so lazy. Look it up!
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):






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
Originally posted by pka

Here is help with #1.
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.
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.

Math Forums @ Math Goodies © 2000-2004 Snitz Communications Go To Top Of Page
This page was generated in 0.05 seconds. Snitz Forums 2000
testing footer
About Us | Contact Us | Advertise with Us | Facebook | Blog | Recommend This Page

Copyright © 1998-2014 Mrs. Glosser's Math Goodies. All Rights Reserved.

A Hotchalk/Glam Partner Site - Last Modified 30 Oct 2014