Please help with exam prep! - Proofs

**omgmath** · June 17th 2007, 04:56 PM

1. Some natural numbers can be expressed as a difference of two squares, but others cannot. For example, 12=4^2 - 2^2, but it is impossible to write 10 as a difference of two squares. Find a way to determine whether or not a given natural number can be expressed as the difference of two perfect squares.

2. In rectangle ABCD, M is the midpoint of side AD and N is the midpoint of side DC. Segments AN and CM intersect at E
a) Prove that <AEM=<MBN
b)Determine how the angles in part a are related to the length: width ratio of the rectangle.

3. A sequence is defined by the equation below:
a) Prove that every term of the sequence is less than 3.
b) Prove that every term of the sequence is greater than the preceding term.

Any help you can give me on any/all of these questions would be great, thanks.

**Jhevon** · June 17th 2007, 04:58 PM

Originally Posted by omgmath

3. A sequence is defined by the equation below:
a) Prove that every term of the sequence is less than 3.
b) Prove that every term of the sequence is greater than the preceding term.

The sequence equation is missing

**omgmath** · June 17th 2007, 05:00 PM

Sorry I forgot the last equation, here it is:

**Jhevon** · June 17th 2007, 05:54 PM

Originally Posted by omgmath

3. A sequence is defined by the equation below:

$t_1 = 1$ , $t_{n + 1} = \sqrt {2 t_n + 1}$

a) Prove that every term of the sequence is less than 3.

Now i'm one to always be off by some technicality, so use these answers as a general outline and fix it up to your liking.

Let the sequence $\left( t_n \right)$ be defined recursively as $t_1 = 1$ , $t_{n + 1} = \sqrt { 2 t_n + 1}$ for all integers $n \geq 1$

We show that $t_n < 3$ for all integers $n \geq 1$

We proceed by induction.

Let P(n): " $t_n < 3$ for all integers $n \geq 1$ "

For P(1) we have $t_1 = 1 < 3$ , so P(1) is true.

Assume P(k) is true for some integer $k > 1$

Since P(k) is true, we have

$t_k < 3$

$\sqrt {2 t_{k - 1} + 1} < 3$

$\Rightarrow 2 \sqrt {2 t_{k - 1} + 1} < 2 \cdot 3 = 6$

$\Rightarrow 2 \sqrt {2 t_{k - 1} + 1} + 1 < 6 + 1 = 7$

$\Rightarrow \sqrt { 2 \sqrt {2 t_{k - 1} + 1} + 1} < \sqrt {7}$

$\Rightarrow \sqrt {2 t_k + 1} < \sqrt {7}$

$\Rightarrow t_{k + 1} < \sqrt {7} < 3$

So we have $t_{k + 1} < 3$ , which is P(k + 1)

Thus the induction proof is complete, and we have $t_n < 3$ for all integers $n \geq 1$

b) Prove that every term of the sequence is greater than the preceding term.

Let $\left( t_n \right)$ be the sequence defined above. We show that $t_{n + 1} > t_n$ for all integers $n \geq 1$

We proceed by induction.

Let P(n): " $t_{n + 1} > t_n$ for all integers $n \geq 1$ "

We have for P(1): $t_{1 + 1} = t_2 = \sqrt {3} > 1 = t_1$ . So P(1) is true.

Assume P(k) is true for some integer $k \geq 1$

Then we have:

$t_{k + 1} > t_k$

$\Rightarrow 2 t_{k + 1} > 2 t_k$

$\Rightarrow 2 t_{k + 1} + 1 > 2 t_k + 1$

$\Rightarrow \sqrt {2 t_{k + 1} + 1} > \sqrt {2 t_k + 1}$

$\Rightarrow t_{k + 2} > t_{k + 1}$ , which is P(k + 1)

Thus the induction proof is complete and we have $t_{n + 1} > t_n$ for all integers $n \geq 1$

**Soroban** · June 17th 2007, 06:27 PM

Hello, omgmath!

3. A sequence is defined by: . $t_1 = 1,\;t_{n+1} \:=\:\sqrt{2t_n + 1}$

a) Prove that every term of the sequence is less than 3.

b) Prove that every term of the sequence is greater than the preceding term.

How about an inductive proof?

(a) $S(n)\!: \;t_n < 3$ for all $n.$

Verify $S(1)\!:\;t_1 \,=\,1 \:<\:3$ . . . true

Assume $S(k)$ is true: . $t_k \:< \:3$

Multiply both sides by 2: . $2\!\cdot\!t_k \:<\:6$

Add 1 to both sides: . $2\!\cdot\!t_k + 1 \:<\:7$

Take the square root of both sides: . $\sqrt{2\!\cdot\!t_k + 1} \:< \:\sqrt{7}$

The left side is: $t_{k+1}$
The right side is: $\sqrt{7} \:=\:2.645... \:<\:3$

Therefore: . $t_{k+1} \:<\:3$
. . The inductive proof is complete.

Part (b) is proved in a similar manner.

[Edit: Jhevon beat me to it . . .]

**Jhevon** · June 17th 2007, 06:32 PM

Originally Posted by Soroban

[Edit: Jhevon beat me to it . . .]

you were a lot smoother than me though. i went back to defining $t_k$ in terms of $t_{k - 1}$ , which was totally unnecessary and unaesthetic.

you can go after the other two if you want

**omgmath** · June 17th 2007, 06:33 PM

Thanks guys, any ideas on the first 2?

**ThePerfectHacker** · June 17th 2007, 06:44 PM

Originally Posted by omgmath

1. Some natural numbers can be expressed as a difference of two squares, but others cannot. For example, 12=4^2 - 2^2, but it is impossible to write 10 as a difference of two squares. Find a way to determine whether or not a given natural number can be expressed as the difference of two perfect squares.

This is a number theory question.

Say,
$n = a^2 - b^2 = (a+b)(a-b)$
A possible factorization of $n$ is $n\cdot 1$

And hence it is reasonable to search for solutions to:
$a+b=n$
$a-b=1$

The solution is given by $(a,b) = \left( \frac{n+1}{2}, \frac{n-1}{2} \right)$ .

Now the thing that we need is for $a,b$ to be integers. And that certainly happens if $n$ is an odd positive integer.

What happens if it is even? The possibilities is that $n$ has one of the four forms: $4k,4k+1,4k+2,4k+3$ . Note we covered the cases $4k+1,4k+3$ . Let us do $n=4k$ . Then since it is divisible by 4 we can write:
$n = \left( \frac{n}{4} + 1 \right)^2 - \left( \frac{n}{4} - 1 \right)^2$

And what happens if $n=4k+2$ . Well, the is impossible because a square modulo 4 must be either 1 or 3. Hence $a^2 - b^2 = 0,1,3 (\bmod 4)$ . Not 2! Hence such a representation is impossible.

**omgmath** · June 17th 2007, 07:09 PM

Is there any way you could explain that to me a bit better, I have no clue how you got the solution for (a,b), and I dont understand what you were getting at when you said they need to be integers if n i odd...Im basically completely lost

**ThePerfectHacker** · June 17th 2007, 07:16 PM

Originally Posted by omgmath

Is there any way you could explain that to me a bit better, I have no clue how you got the solution for (a,b), and I dont understand what you were getting at when you said they need to be integers if n i odd...Im basically completely lost

Add the two equations to eliminate b.

Subtract the two equation to eliminate a.

Do you know how to solve simulatenous equations?

**ThePerfectHacker** · June 17th 2007, 07:38 PM

Originally Posted by omgmath

2. In rectangle ABCD, M is the midpoint of side AD and N is the midpoint of side DC. Segments AN and CM intersect at E
a) Prove that <AEM=<MBN
b)Determine how the angles in part a are related to the length: width ratio of the rectangle.

I have an ugly way of doing it, but I think it works.

Look at picture below. We want to show the blue angles are equal to eachother.

In coordinate geometry you perhaps learn the following theorem, let $l_2,l_1$ be two lines (non-vertical). And let $l_2$ has bigger angle measurement then $l_1$ with the x-axis. Then if $\theta$ is the angle between them we have:
$\tan \theta = \frac{m_2 - m_1}{1+m_2m_1}$ where $m_2$ is the slope of $l_2$ and $m_1$ is the slope of $l_1$ .

Here is how we use it.

Let $\alpha$ be the angle <AEM. And $\beta$ be the angle <MBN. We will show that $\tan \alpha = \tan \beta$ and deduce $\alpha = \beta$ .

Now we find $\tan \alpha$ and $\tan \beta$ by using the theorem above.

Let me do $\alpha$ and leave $\beta$ for you to do.

In the picture $\alpha$ is formed by lines AN and CM with side CM creating a bigger angle than AN with x-axis (in this case x-axis is AD).

Thus,
$\tan \alpha = \frac{m_{CM} - m_{AN}}{1 + m_{CM}\cdot m_{AN}}$ .

To compute the slope we just find the ratio between the the height and width.

$m_{CM} = \frac{CD}{DM} = \frac{2DN}{DM}$

$m_{AN} = \frac{DN}{AD} = \frac{DN}{2DM}$

Thus,
$\tan \alpha = \frac{\frac{2DN}{DM} - \frac{DN}{2DM}}{1+ \frac{2DN}{DM}\cdot \frac{DN}{2DM}} = \frac{\frac{3DN}{2DM}}{\frac{DM^2+DN^2}{DM^2}} = \frac{3DN\cdot DM}{2MN^2}$

This answers the second part of the question on how to compute the angle in terms of the sides.

Now do the same thing with $\tan \beta$ .
And show they give the same value.

**CaptainBlack** · June 17th 2007, 09:51 PM

Originally Posted by omgmath

1. Some natural numbers can be expressed as a difference of two squares, but others cannot. For example, 12=4^2 - 2^2, but it is impossible to write 10 as a difference of two squares. Find a way to determine whether or not a given natural number can be expressed as the difference of two perfect squares.

Suppose integer X can be written as the difference of two squares, then:

X = a^2-b^2 = (a-b)(a+b)

then if we suppose a>b>0, then X must have two non-zero factors u, v,
u!=v, where

u=a+b

v=a-b

adding these we see that u+v=2a, and subtracting these we see that
u-v=2b.

Now suppose we are given any non zero u and v, u>v, such that their sum is even (then their difference is even as well), we can write:

a=(u+v)/2

and:

b=(u-v)/2

then uv=(a+b)(a-b).

So we see that for any X to be written as the difference of two squares
it is necessary and sufficient that X have two distinct factors whoes sum
is even.

RonL

**Soroban** · June 18th 2007, 04:33 AM

Hello, omgmath!

I have an even uglier solution for #2 . . .

2. In rectangle ABCD, M is the midpoint of side AD and N is the midpoint of side DC.
. . Segments AN and CM intersect at E
a) Prove that <AEM=<MBN
b) Determine how the angles in part a are related to the length: width ratio of the rectangle.

Let $L = AD,\;W = CD$

Let $\angle CMD = \alpha\quad\Rightarrow\quad\angle MCD \,=\, 90 -\alpha$

Let $\angle NAD = \beta\quad\Rightarrow\quad\angle AND = 90-\beta\quad\Rightarrow\quad\angle ANC = 90 + \beta$

In $\Delta CEN:\;\angle CEN \;=\;180^o - (90 - \alpha) - (90 + \beta) \;=\;\alpha - \beta$

. . Hence: . $\angle AEM \;=\;\alpha - \beta$

We have: . $\tan(\angle AEM) \;=\;\tan(\alpha - \beta) \;=\;\frac{\tan\alpha - \tan\beta}{1 + \tan\alpha\cdot\tan\beta}$

$\text{Since: }\tan\alpha \:=\:\frac{W}{\frac{L}{2}} \:=\:\frac{2W}{L}\;\text{ and }\;\tan\beta \:=\:\frac{\frac{W}{2}}{L} \:=\:\frac{W}{2L}$

. . we have: . $\tan(\angle AEM) \;=\;\frac{\frac{2W}{L} - \frac{W}{2L}}{1 + \left(\frac{2W}{L}\right)\left(\frac{W}{2L}\right) }$

. . which simplifies to: . $\boxed{\tan(\angle AEM) \:=\:\frac{3LW}{2(L^2+W^2)}}$ . [1]

Here's the uglier part . . .

In right triangle $BAM\!:\;BM^2 \:=\:\left(\frac{L}{2}\right)\!^2 + W^2\quad\Rightarrow\quad BM \:=\:\frac{\sqrt{L^2+4W^2}}{2}$
In right triangle $BCN\!:\;BN^2 \:=\:L^2 + \left(\frac{W}{2}\right)\!^2\quad\Rightarrow\quad BN \:=\:\frac{\sqrt{4L^2+W^2}}{2}$
In right triangle $NDM\!:\;MN^2 \:=\:\left(\frac{L}{2}\right)\!^2 + \left(\frac{W}{2}\right)\!^2\quad\Rightarrow\quad MN \:=\:\frac{\sqrt{L^2+W^2}}{2}$

$\text{Then: }\;\cos(\angle MBN) \:=\:\frac{BM^2 + BN^2 - MN^2}{2(BM)(BN)} \:=\:\frac{\frac{L^2+4W^2}{4} +\frac{4l^2+W^2}{4} - \frac{L^2+W^2}{4}}{2\left(\frac{\sqrt{L^2+4W^2}}{2 }\right)\left(\frac{\sqrt{4L^2+W^2}}{2}\right)}$

. . which simplifies to: . $\cos(\angle MBN) \;=\;\frac{2(L^2+W^2)}{\sqrt{L^2+4W^2}\sqrt{4L^2+W ^2}} \;=\;\frac{adj}{hyp}$

Since: $opp^2 \:=\:hyp^2 - adj^2\quad\Rightarrow\quad opp^2 \;=\;\left[\sqrt{L^2+4W^2}\sqrt{4L^2+W^2}\right]^2 - \left[2(L^2+W^2)\right]^2$

. . Then: . $opp^2 \:=\:9L^2W^2\quad\Rightarrow\quad opp \:=\:3LW$

And we have: . $\boxed{\tan(\angle MBN) \;=\;\frac{3LW}{2(L^2+W^2)}}$ . [2]

Therefore, from [1] and [2]: . $\angle AEM \:=\:\angle MBN$

~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~

Let $r \,=\,\frac{L}{W}$
We have: . $\tan\theta \;=\;\frac{3LW}{2(L^2+W^2)}$
$\text{Divide top and bottom by }W^2\!:\;\;\tan\theta \;=\;\frac{3\left(\frac{L}{W}\right)}{2\left[\left(\frac{L}{W}\right)^2 + 1\right]} \;=\;\frac{3r}{2(r^2+1)}$
Therefore: . $\boxed{\theta \;=\;\tan^{-1}\!\left(\frac{3r}{2(r^2+1)}\right)}$

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