Beginning Discrete math, help plz.

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January 21st 2008, 01:48 PM
p00ndawg

Beginning Discrete math, help plz.

I have a couple of questions 1 of which I more or less know what should be the proof, I just dont really know how to prove it.

1. My first question is how would I go about proving that m^2 = n^2
if and only if m = n or m = -n.

I know I have to prove it both ways, but am not quite sure how to construct it.

2. My next problem is proving that either 2x10^500 + 15 or 2x10^500 + 16
is not a perfect square.

Now I am fairly certain that perfect squares do not appear sequentially, thus proving that one of them is not square, but I am not sure how to exactly go about it.

The first question should probably be a proof that is fairly simple, these questions are both coming from chapters 1.6 and 1.7 of my book, so they shouldnt be too difficult.

help appreciated. Extra points to those who can help me understand a little bit more. :D

:confused:
January 22nd 2008, 09:54 AM
Soroban

Hello, p00ndawg!

Here is half of #2 . . .

Quote:

$\text{2. Prove that neither }\,2\cdot10^{500} + 15\,\text{ nor }\,2\cdot10^{500} + 16\,\text{ is a perfect square.}$

We observe the following . . .

$\text{If }n\text{ is an even number, its square is a multiple of 4.}$
. . $m \,= \,2k\quad\Rightarrow\quad m^2 \,=\,4k^2$

$\text{If }n\text{ is odd, its square is one more than a mutliple of 4.}$
. . $n^2 \:= \:(2k+1)^2\:=\:4k^2 + 4k + 1 \:=\:4k(k+1) + 1$

We have: . $M \;=\;2\cdot10^{500} + 15 \;=\;2\cdot10\!\cdot\!10^{499} + 12 + 3 \;=\;2\!\cdot\!2\cdot5\!\cdot\!10^{499} + 4\cdot 3 + 3$

. . Hence: . $M \;=\;4\left(5\!\cdot\!10^{499} +3\right) + 3$

Therefore, $M$ is three more than a mutiple of 4 . . . It cannot be a square.
January 22nd 2008, 09:57 AM
Jhevon

Quote:

Originally Posted by p00ndawg

I have a couple of questions 1 of which I more or less know what should be the proof, I just dont really know how to prove it.

1. My first question is how would I go about proving that m^2 = n^2
if and only if m = n or m = -n.

show that $m^2 = n^2 \implies m = n \mbox{ or } m = -n$ . we can get this by square rooting both sides of the equation

then show that $m = n \mbox { or } m = -n \implies m^2 = n^2$ . we have to deal with each case separately. we get the desired result if we square both sides of the equation
January 23rd 2008, 02:08 PM
p00ndawg

thank you guys for your help, what I am wondering although, is how in the HECK can you come up with some of these things?

Because I just started this class, should It be this kind of awkward for me to do, or come up with some of these proofs?

or rather, will it get easier as I practice?

Because I must tell you, some of these things that are coming up in these proofs are kind of ridiculoid, and while they make PERFECT sense, I am just not quite sure how I could come up with them myself.

And to tell you the truth, I am by no means bad at math in anyway, but this type of thinking really makes my head hurt.:confused:
January 23rd 2008, 02:52 PM
Plato

Quote:

Originally Posted by p00ndawg

And to tell you the truth, I am by no means bad at math in anyway, but this type of thinking really makes my head hurt.

You may be dismayed to learn that “this type of thinking” is really what mathematics is all about. What you may have thought was mathematics was just preliminary to the real heart of mathematics.
January 23rd 2008, 03:53 PM
p00ndawg

Quote:

Originally Posted by Plato

You may be dismayed to learn that “this type of thinking” is really what mathematics is all about. What you may have thought was mathematics was just preliminary to the real heart of mathematics.

Yea I totally understand what you're saying, so If I clarify what I would mean to say is that I am rather good at standard mathematics.

Of course, I only started learning this, so one would assume I wont be very good at this, Im pretty sure its safe to also assume that not many people were very good at the start of this type of mathematics as well.

of course, I have been wrong before.
January 23rd 2008, 05:28 PM
JaneBennet

(1)

$\begin{array}{lrcl} {} & m^2 &=& n^2\\ \Leftrightarrow & m^2-n^2 &=& 0\\ \Leftrightarrow & (m-n)(m+n) &=& 0\\ \Leftrightarrow & m-n=0 &\mbox{or}& m+n=0\\ \Leftrightarrow & m=n &\mbox{or}& m=-n \end{array}$
January 24th 2008, 12:41 PM
p00ndawg

thank you for the reply.