please try to solve that questions.

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- Nov 14th 2006, 12:00 AMm777set, geometric sequence,fraction and arithmetic sequence
please try to solve that questions.

- Nov 14th 2006, 03:36 AMtopsquark
Show that the set 2$\displaystyle \mathbb{Z}$ of even integers is countable.

I presume you know the theorem that if two sets are isomorphic then they have the same cardinality?

The function $\displaystyle f: \mathbb{Z} \to 2\mathbb{Z}: x \mapsto 2x$ is an isomorphism between the two sets as you can easily show. Thus since $\displaystyle \mathbb{Z}$ is countably infinite so is the set 2$\displaystyle \mathbb{Z}$.

(If you don't have this theorem available, note that you can put the two sets in one to one correspondence using the above function and go from there.)

-Dan - Nov 14th 2006, 03:41 AMtopsquark
Find the 9th term of the geometric series 6, 12, 24, ...

Since it's a geometric series we know that it is of the form:

$\displaystyle a_{n+1} = ra_n$

or

$\displaystyle a_n = a_1r^{n-1}$

So $\displaystyle a_1 = 6$

$\displaystyle 12 = 6 \cdot r^{2-1}$

So $\displaystyle r = 2$

The 9th term of the series is for n = 9:

$\displaystyle a_9 = 6 \cdot 2^8 = 1536$

-Dan - Nov 14th 2006, 03:46 AMtopsquark
Find the common fraction for the recurring decimal 0.7.

We may represent the decimal by the geometric series:

$\displaystyle \sum_{n = 1}^{\infty} 7(10)^{-n} = 0.77777....$

The sum of the terms of a geometric series $\displaystyle a_1r^{-n}$ is

$\displaystyle S = \frac{a_1}{r-1}$

In this case:

$\displaystyle S = \frac{7}{10-1} = \frac{7}{9}$

-Dan - Nov 14th 2006, 03:51 AMtopsquark
Which term of the arithmatic series 5, 2, -1, ... is -85?

An arithmatic series is defined as:

$\displaystyle a_n = a_{n-1} + k$

or

$\displaystyle a_n = a_1 + (n-1)k$

We know that $\displaystyle a_1 = 5$ and $\displaystyle 2 = 5 - (2-1)k$. Thus $\displaystyle k = -3$.

$\displaystyle -85 = 5 + (n-1)(-3)$

$\displaystyle -90 = -3n + 3$

$\displaystyle -93 = -3n$

$\displaystyle n = 31$

So the 31st term will be -85.

-Dan - Nov 14th 2006, 05:36 AMSoroban
Hello, m777!

Here's alternate approach to #2 . . .

Quote:

Find the common fraction for the recurring decimal 0.7777...

We have: .$\displaystyle N \;= \;0.77777\hdots$

Multiply by 10: .$\displaystyle 10N \;=\;7.77777\hdots$

Subtract $\displaystyle N:$ . . . .$\displaystyle N \;= \;0.77777\hdots$

And we have: . . $\displaystyle 9N\:=\:7\quad\Rightarrow\quad N \,=\,\frac{7}{9}$

- Nov 14th 2006, 06:28 AMThePerfectHacker
- Nov 14th 2006, 11:37 AMtopsquark
It probably depends on the text you are using. My Topology book defines cardinality first, then introduces a proof that two sets with the same cardinality are necessarily isomorphic.

Frankly your definition might be a cleaner approach. The book rather confused me when going over this.

-Dan - Nov 14th 2006, 01:10 PMThePerfectHacker
- Nov 14th 2006, 02:08 PMPlato
I would like to see that topology text.

If this is so, that text is to be avoided. - Nov 14th 2006, 05:26 PMtopsquark
:o I owe apologies to both of you gentlemen. The word I was remembering before the bijections was "countable" not "cardinality." Yes, in my book also it defines the bijection between sets as a definition for two sets to have the same cardinality.

-Dan - Nov 15th 2006, 11:19 AMTriKri
What's a countable set and how can an infinitly big set be countable? :confused:

- Nov 15th 2006, 12:51 PMThePerfectHacker
Georg Cantor developed one of the Jewels from 20th Century mathematics. The concept of sizes of sets (even infinite sets). So some infinite sets have more elements than others. Countable is any finite set, or any set which has as many elements as the number of integers. It can be shown to be the smallest type of infinite set. The Countiuum are the real numbers and they are larger. I can explain the theory but if you are not familar with the some set theory you will not follow.