Monthly Archives: March 2013

Lets play dominoes!

           Yesterday in class we learned about Poof by Inductions. This proof technquie allows us to prove infinitely many statements. The key to this proof is to relate it to dominoes. Proof by induction is always preformed in two steps. The first step is the base case, in relationship to dominoes it is knocking down the first domino in the series. The second step in proof by induction is the inductive step. Think of this step as all the dominoes being lines up if one falls they all fall. To conclude when in doubt remember dominoes.


A little help to get over hump day!

I saw this comic while I was surfing the internet, and I could not help to think about Casey’s jokes about other majors. I guess he actually does have a point…funny_math_cartoon4

A few things…

Last class we were introduced to something that we wished we all knew before the exam…


1) To prove something, what am I requires to do?

2) What am I allowed to use?

Since Casey shared the “BIG QUESTIONS”  I thought it would be helpful to share some of my techniques to writing proofs. The best thing technique that I found useful before writing a proof is to list out what relates to the set we were given. I go as far as writing the simple definitions that we learned in the begining of the semester to help me understand the set better.  Also if I am ever stuck on the problem I realize the best thing is to step away and try it again with fresh eyes, because most of the time you see something that you had not seen before.

I also wanted to discuss some of chapter 4.2…

statement: p =>q

Contrapositive: ~q => ~p

Converse: q => p

Inverse: ~p => ~q

The orginial staement and the contrapostive are equivalent.

When p => q and q =>p , we write that as p <=> q. Where the symbol <=> is represents the phrase  if and only if.

I hope this was helpful!



I have been up seen four the last three days so I am running low on energy, but before class starts today I want to place my POW #7 post up…

1) Getting help from my math major suitemate, Liz Mosher!



2) I think my favorite Theorem I learned thus far would have to be the Fundmental Theorem of Calculus. The reason why I like this theorem is mainly due to the fact that Calculus has been my favorite math class to date (of course FOM will probably replace it after this semester…wink, wink)

3) I have actually been working hard in this class. I think the hardest part of  FOM is keeping up with the reading, only because it is hard to follow where we are in class. Most of the times I end up reading too much of the chapter. I have done all the homework and POW as well as been keeping up with my blog weekly. To conclude I think I have been working hard.

I almost forgot?!?

Ahhh this week has been long it almost slipped my mind to post, but no need to worry I will make this post extra long!

I figured to make up for lost time I might as well go over the worksheet for this week…

#1 a) How many functions are there from (x){1,2,3} to (y){1,2,3,4,5}?

–personally speaking it is easier for me to look at functions vertically (like elementary school), it gives me the ability to draw lines between the two functions.

To find how many functions it is the number of elements in the second function powered to the number of elements in the first function, therefore the answer is 5^3.

b) How many of them are onto?

–this is where drawing the functions vertically really helps

none of them are onto because every element in y is not an output for f

c) How many of them are one-to-one

3 functions are one-to-one because the three elements in x go to a different output

#2 a) How many functions are there from (x){1,2,3,4,5} to (y){1,2,3}?
  similar to before y^x thus, 5^3

b) How many of them are onto?

there are 69 functions that are onto…getting 69 takes alot of addition

c) How many of them are one-to-one?

none because the five elements in x can’t go to different outputs in y


a) If f & g are surjective, then g0f  is surjective–True the best explanation my group could think of was that g is the co-domain of f therefore if both are surjective their composite with also be surjective

b) If g0f is surjective, then f is surjective–false

f(x)=sqrt (X), g(x)= y^2

#4 (because I have trouble using latex I am just going to write our explanations of each statement)

a) the statement states that the functions are not onto

b) If 3/(n^2), then 3/n (are group couldn’t figure out a simpler way to put it)

c) f is a decreaseing function

#5  Due to my inability to latex and the mere fact that two of the problems are due for ths weeks homework I will pass this question!


Phewwwwwww!!! If you have any questions or comments feel free to respond 🙂