1) Attachment 30295

2)Attachment 30296

3)Attachment 30297

4)Attachment 30298

I will be forever in debted if you can help me with these. My first discrete math class.

Printable View

- Mar 4th 2014, 11:42 AMoutkast32Discrete Math Practice Problems (Please help confirm)
1) Attachment 30295

2)Attachment 30296

3)Attachment 30297

4)Attachment 30298

I will be forever in debted if you can help me with these. My first discrete math class. - Mar 4th 2014, 12:13 PMromsekRe: Discrete Math Practice Problems (Please help confirm)
you have to show some work. These are all straightforward problems you should be able to do after reading your textbook.

- Mar 4th 2014, 01:18 PMSorobanRe: Discrete Math Practice Problems (Please help confirm)
Hello, outkast32!

There is no formula for #2.

I doubt that your textbook will help you.

It requires some Original Thinking.

Quote:

2. Let $\displaystyle A \,=\,\{5,6,7,8,9,10,11,12,13,14,15,16,17,18\}$

How many numbers must be chosen to ensure that two of them add up to 23?

Note that there are 7 Pairs that add up to 23:

. . $\displaystyle (5,18),\:(6,17),\:(7,16),\:(8,15),\:(9,14),\:(10,1 3),\:(11,12)$

To have a sum of 23, your choices must include one of these Pairs.

Now consider the worst-case scenario: none of the Pairs.

If you select 7 numbers,*one from each Pair,*

. . you willhave a sum of 23.*not*

When you choose an 8th number, youhave a Pair.*will*

Therefore, you must draw 8 numbers. - Mar 4th 2014, 06:48 PMSorobanRe: Discrete Math Practice Problems (Please help confirm)
Hello, outkast32!

The first problem (unnumbered) is not true.

It should be stated like this:

Quote:

Using mathematical induction, prove that:

. . $\displaystyle 1^3 + 2^3 + 3^3 + \cdots + n^3 \:=\:\frac{n^2(n+1)^2}{4}$

Verify $\displaystyle S(1)\!:\;1^3 \:=\:\frac{1^2\!\cdot2^2}{4}$ . . . True!

Assume $\displaystyle S(k)\!:\;1^3+2^3+3^3+\cdots+k^3 \;=\;\frac{k^2(k+1)^2}{4}$ .[1]

We must prove $\displaystyle S(k+1)\!:\;1^3+2^3+3^3+\cdots + (k+1)^3 \;=\;\frac{(k+1)^2(k+2)^2}{4}$ .[2]

Add $\displaystyle (k+1)^3$ to both sides of [1].

.$\displaystyle 1^2+2^3+3^3+\cdots+k^2+(k+1)^3 \;=\;\frac{k^2(k+1)^2}{4} + (k+1)^3$

Simplify the right side:

. . $\displaystyle \frac{k^2(k+1)^2}{4} + (k+1)^3 \;=\(k+1)^2\left[\frac{k^2}{4} + (k+1)\right] \;=\;(k+1)^2\left[\frac{k^2+4k+4}{4}\right]$

. . $\displaystyle =\;(k+1)^2\,\frac{(k+2)^2}{4} \;=\;\frac{(k+1)^2(k+2)^2}{4}$

We have proved [2]: .$\displaystyle 1^3+2^3+3^3 + \cdots (k+1)^3 \;=\;\frac{(k+1)^2(k+2)^2}{4}$

The inductive proof is complete.