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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 95,861

Hi;

How would you judge this answer? And why?

*Last edited by bobbym (2009-09-30 19:05:12)*

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 95,861

Hi;

How about this one? Do you like this line of reasoning?

Show that there are at least 3 real roots in [-2,2] for the function:

f(x) = x^5 - 3x -1

Here is how he does it:

Look at these values:

f(-2) = -27

f(-1) = 1

f(0) = - 1

f(1) = - 3

f(2) = 25

Look at the values of f(-2) and f(-1) they went from negative to positive (crossed the x axis). So somewhere between -2 and -1 is a root.

Look at the values of f(-1) and f(0) they went from + to - (crossed the x axis). So somewhere between -1 and 0 is a root.

Look at the values of f(1) and f(2) they went from - to + (crossed the x axis). So somewhere between 1 and 2 is a root.

So there are 3 or more real roots between -2 and 2.

*Last edited by bobbym (2009-10-02 00:35:51)*

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 95,861

Hi;

#3)

A person reduces the assertion that there is always a square between n and 2n inclusive for all n to this inequality:

She manipulates this to:

She says proved:

Someone else says that you must use induction, presumably on the inequality. Who is right?

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**JaneFairfax****Member**- Registered: 2007-02-23
- Posts: 6,868

bobbym wrote:

Hi;

How would you judge this answer? And why?

*Last edited by JaneFairfax (2010-01-01 03:44:46)*

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**JaneFairfax****Member**- Registered: 2007-02-23
- Posts: 6,868

bobbym wrote:

A person reduces the assertion that there is always a square between n and 2n inclusive for all n to this inequality:

That would be looking for a square between and . It does not imply that there is a square between and .

Correct proof of #3.

If the statement were false, there would exist a natural number

and a natural number satisfyingWith some simple algebraic manipulation, we should arrive at

It can be easily checked that there are no natural numbers

satisfying for or and so we have a contradiction.*Last edited by JaneFairfax (2010-01-01 13:22:14)*

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 95,861

OK.

Congratulations on your deep sense of knowing how to perplex me.

With some simple algebraic manipulation, we should arrive at

What did you do here?

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**mathsyperson****Moderator**- Registered: 2005-06-22
- Posts: 4,900

She removed the n's by doubling the first inequality and combining them.

2k² < (k+1)²

From there it's simple manipulation.

Why did the vector cross the road?

It wanted to be normal.

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 95,861

Hi mathsyperson;

Thanks, clear now, it is my first blind spot of the year.

Nice proof Jane. Happy New year to you.

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**If it ain't broke, fix it until it is.**

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**scientia****Member**- Registered: 2009-11-13
- Posts: 224

bobbym wrote:

Hi;

How would you judge this answer? And why?

The first step is invalid because in order to split the limit of a sum as the sum of limits, you have to make sure all the limits you are taking exist. In this case, none of the three limits exists. The rest of the proof is treating ∞ as a number, which it is not.

The second problem looks fine to me.

The third problem can also be proved this way. Note that

JaneFairfax wrote:

can be written

since *n* is positive. There are *n*+1 integers from *n* to 2*n* inclusive and so there are at least *n*+1 integers between *k*² and (*k*+1)². Hence these two integers must differ by at least *n*+2:

Combining with *k*² < *n* gives

which is another contradiction.

*Last edited by scientia (2010-01-09 01:53:46)*

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 95,861

Hi Jane and scientia;

1) I agree that the way that guy does that limit has many holes in it.

2) That is a perfectly acceptable method to look for roots, just keep in mind not all roots cross the x axis so that idea will miss some.

3) This was the trickiest one of all. Very pretty ideas were given to prove that equality.The question has not been answered and here is why.

A person reduces the assertion that there is always a square between n and 2n inclusive for all n to this inequality:

She manipulates this to:

She says proved:

The question was, is her manipulation a proof. Do you like where she stops? Do you find her proof valid? Is more required? When does algebraic manipulation constitute a proof?

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**If it ain't broke, fix it until it is.**

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**scientia****Member**- Registered: 2009-11-13
- Posts: 224

bobbym wrote:

She manipulates this to:

She says proved:

The question was, is her manipulation a proof. Do you like where she stops? Do you find her proof valid? Is more required? When does algebraic manipulation constitute a proof?

So you just want to prove

for all natural numbers *n* (even though this is the wrong way to go about solving the original question of finding a square between *n* and 2*n*)?

I dont know what the person did, but if we want to use her method, we should proceed as follows. Start by noting that for all integers *n* ≥ 1,

Hence for all natural numbers *n*,

and the proof is completed by adding *n*² throughout.

The wrong way to go about it is to start with [1] and get [2]. This would be assuming what you want to prove to be true already, and would therefore prove nothing yet this is an all too common mistake among students (and sometimes even teachers).

So what did the person do? Did she start from [1] and get [2] (which would be incorrect) or did she start with [2] and get [1]?

*Last edited by scientia (2010-01-09 05:32:58)*

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 95,861

Hi scientia;

So what did the person do? Did she start from [1] and get [2] (which would be incorrect) or did she start with [2] and get [1]?

I can't say, I don't know.

Jane wrote:

That would be looking for a square between and . It does not imply that there is a square between and .

I obviously forgot what you and Jane already know, that the first statement in her proof is wrong.

I think you have both solved the problem. Good job! Sorry to have continued to beat a dead horse.

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**If it ain't broke, fix it until it is.**

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 95,861

This prob came up somewhere else:

A coin with probability p of heads is tossed until a head appears for the first time. If the probability that the number of tosses required is {2,4,6,8,10...} is 2/5, then find p:

B solves it like this:

The chance that a head comes up on an even toss is:

Now sum that series:

Set the sum to 2/5 and solve for p.

Solve the equation any way you can, to get p = 1/3

Has B solved the problem?

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**mathsyperson****Moderator**- Registered: 2005-06-22
- Posts: 4,900

Just want to point out that to prove #2 like that, you need to observe that the function is continuous.

Otherwise you could use that proof to say that f(x) = 1/x has a root in [-1,1], for example.

Why did the vector cross the road?

It wanted to be normal.

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
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Hi;

True, but since he/she is using it on a polynomial it is OK.

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**If it ain't broke, fix it until it is.**

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 95,861

Hi;

There is a 60 percent chance of snow on Tuesday, If there is a 45 percent of chance of snow on both Tuesday and Wednesday, and a 16 percent chance that it will not snow on both Tuesday and Wednesday, what is the probability of snow on Wednesday?

A says 75%

B says 69%

C says who cares, I have boots.

Who is right? (Please don't pick C)

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**If it ain't broke, fix it until it is.**

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**mathsyperson****Moderator**- Registered: 2005-06-22
- Posts: 4,900

I agree with B.

P(T) = 0.6 and P(T n W) = 0.45, so P(T n ¬W) = 0.15.

Also, P(¬T n ¬W) = 0.16.

The remaining event is ¬T n W, and the probability of this must be 0.24.

Therefore, P(W) = 0.45 + 0.24 = 0.69 = 69%.

Why did the vector cross the road?

It wanted to be normal.

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
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Hi mathsyperson;

Good answer, B is correct!

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**If it ain't broke, fix it until it is.**

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 95,861

How about this?

How many integer values can a have?

B says just substitute into the inequalities.

Supposing a = 1 then 1 < b and b <= 1 + 3b is true.

Supposing a = 2 then 2 < b and 2b <= 2 + 3b is true.

Supposing a = 3 then 3 < b and 3b <= 3 + 3b is true.

Supposing a = 4 then 4 < b and 4b <= 4 + 3b is not true when b>4.

The same is true for a=5,6,7...

So a can take positive values of 1,2,3 and nothing greater.

Did you like this?

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**bobbym****Administrator**- From: Bumpkinland
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How many ways can you arrange 5 A's, 7 B's and 4 C's so there are 3 CA pairs?

A) Says there are:

B) Says the magic number: 87120 ways.

Who is right?

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
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Exasperating:

Let a, b, and c be real numbers such that a-7b+8c=4 and 8a+4b-c=7. What is a^2-b^2+c^2?

(a) 0

(b) 1

(c) 4

(d) 7

(e) 8

B) Says:

Looks like all you need to do is the algebra. It's tedious but...

Solve these simultaneously by multiplying the top equation by 8 and subtracting the bottom.

You get:

plug back in: You get another 2x2:

Solve for a:

Plug into the a^2-b^2+c^2

Which equals 1 so option b) 1 is the answer.

A says start by just setting a=0 and then solve from there. Because there are 3 variables and two equations, there are an infinite number of ordered triplets (a,b,c) that can satisfy the two equations. However, while the variables change, the value of a^2-b^2+c^2 stays constant. In one of these triplets, a=0. So if we set a to 0, we still get the same value of a^2-b^2+c^2 as in any other triplet.

Which method do you prefer?

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**If it ain't broke, fix it until it is.**

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 95,861

Hi;

4 checkers are randomly placed on a chessboard, what is the probability that no checker is in the same row or column as any other?

A) says:

B) says:

What do you like?

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**If it ain't broke, fix it until it is.**

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**mathsyperson****Moderator**- Registered: 2005-06-22
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I like the value of the second answer, but I like the reasoning of the first.

Even though it's wrong, at least it explained itself a bit.

Why did the vector cross the road?

It wanted to be normal.

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 95,861

Hi mathsyperson;

Maybe B) doesn't like that forum he posted on so he was purposely being enigmatic. My guess is that he used a standard formula for the rook polynomials. For an 8 x 8 board the generating function is:

Each power of x represents how many non attacking rooks. The coefficient represents the number of ways. The coefficient of x^4 is 117600. That is the number of ways to place 4 non attacking rooks on an 8x8 board.

Since the rook polys are used for derangements. I suspect that you could use the principle of inclusion and exclusion but that looks tedious.

Claude Tardif on another forum solved it in this manner:

(8*8*7*7*6*6*5*5)/4! = (117600)

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**If it ain't broke, fix it until it is.**

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 95,861

Hi;

has 4 real roots:

A says when -27 < k < 5;

B says when 5 < k < 27;

C says when -5 < k < 0.

D says never.

**In mathematics, you don't understand things. You just get used to them.**

**If it ain't broke, fix it until it is.**

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