# non game related math question

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#1
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Posted 27 June 2001 - 06:45 AM

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#5
Anonymous Poster_Anonymous Poster_*
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Posted 28 June 2001 - 01:09 AM

and a(m+n) = a(m) + a(n) always as well, also not just smaller than, it is equal to.

Cheers,

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#6
Members - Reputation: **326**

Posted 28 June 2001 - 02:09 AM

1) Show that a base case is true. That is, your solution works when b is input.

2) Assume your theory works when x is input.

3) Given your assumption in #2 and all other attributes of your system, show that your theory works when x+1 is input.

This is only suitable for discrete (ie, integer) problems, not continuous. Let's try it here.

We know: a(m+n) <= a(m) + a(n) for m,n = 1,2,3...

Want to prove: a(kn) <= ka(n) for k,n = 1,2,3...

Step 1, Base case: k = 1

Show that: a(kn) <= ka(n)

Easy. With k=1, we get a(n) <= a(n). True.

Step 2, Assumption: k = x

Assume that: a(xn) <= xa(n)

Step 3, Proof: k = x+1

Show that a((x+1)n) <= (x+1)a(n)

Here we go... keep in mind our precondition about a(m+n)... Also, I use <?= below here for what we want to test/show, while <= is used for known truths.

We want to prove this:

a((x+1)n) <?= (x+1)a(n)

Apply distribution rule on both sides.

a(xn + n) <?= xa(n) + a(n)

Use our precondition on the left. Note how that we can move the <?= to the right, since the <= relation on left is known. This is an ordering, whereby if we prove a more restrictive case, the less restrictive case will automatically be true.

a(xn + n) <= a(xn) + a(n) <?= xa(n) + a(n)

So can we show this:

a(xn) + a(n) <?= xa(n) + a(n)

Subtract a(n) from both sides:

a(xn) <?= xa(n)

But this is our assumption in step #2 above! Proven, Q.E.D.

You have to be careful sometimes with this method; it can easily be abused to show things that aren't true. You also have to be careful that your preconditions do match your situation, otherwise you're proving the wrong theorem. I'm pretty sure I've been careful here and not missed anything.

Some may say that I'm using the theory to prove the theory, and so it's not a proof at all. But look again at the three steps. I made an assumption at x and had to prove x+1. That alone is insufficient proof, UNLESS you have a starting point. I proved it worked for k=1. Because of that, the rest of the proof shows it will work for k=2. Hey, because it works for k=2, the proof also shows it works for k=3. And so on, and so forth, so the theory is proven for k=1,2,3... Doesn't matter what n is.

---- --- -- -

Blue programmer needs food badly. Blue programmer is about to die!

Edited by - mossmoss on June 28, 2001 9:13:29 AM

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#7
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Posted 28 June 2001 - 02:14 AM

quote:Original post by Anonymous Poster

a(kn) = ka(n) always, not smaller than or equal to.

and a(m+n) = a(m) + a(n) always as well, also not just smaller than, it is equal to.

Cheers,

not true. take a(x) = 2 - x. This is a line with slope -1 and y-intercept 2. Clearly a linear .

Now,

a(1) = 1

a(2) = 0

a(3) = -1

and

a(1 + 2) = -1 < a(1) + a(2) = 1

I think you''re confusing this problem with one of linear

*transformations*i.e. if a() was a linear transformation from R->R then you can say that a(tx0 + tx1) = t((a(x0) + a(x1)), but the original problem never said that a() was a linear transformation.

As for the original question, I don''t think you can say anything about whether that is true or not. If k <= 2 then obviously you could, but if you take a(3n) or a(n + n + n) you can say that that is <= a(n + n) + a(n) but you can''t say anything about a(n) + a(n) + a(n), since a(n) + a(n) is >= a(n + n) (be definition). If you can see what I''m saying...

War Worlds - A 3D Real-Time Strategy game in development.

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#8
Anonymous Poster_Anonymous Poster_*
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Posted 28 June 2001 - 04:43 AM

Also note that a(x+y)=2-x-y != a(x)+a(y) = 4-x-y.

It is, however, an affine function.

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#9
Anonymous Poster_Anonymous Poster_*
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Posted 28 June 2001 - 05:03 AM

For example, take a(1)=2,a(2)=1,a(m)=0 m>2.

Another example: take a(x)=x^.5

It is always true sqrt(m+n)

The proof by induction was correct, but here is a simpler proof, actually this is the exact same proof but I''ll rewrite it to make it a bit clearer:

Let k,n be positive integers.

Then a(kn)=a(n+(n+...+n))<=a(n)+a(n+...+n)

(k-1 times) (k-1) times

=a(n)+a(n+(n+...n))<=a(n)+a(n)+a(n+...n)

(k-2 times) (k-2 times)

etc...

<=a(n)+...+a(n)=na(k), i.e. a(kn)<=ka(n).

(k times)

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#10
Anonymous Poster_Anonymous Poster_*
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Posted 28 June 2001 - 05:06 AM

Line 4 above should read:

sqrt(m+n)

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#12
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Posted 28 June 2001 - 07:36 AM

quote:Original post by Anonymous Poster

a(x)=2-x is not linear. Any linear function has a(0)=0.

It is indeed a linear function of x, since x is not raised to any power (except to the power of 1), and the coefficient of x is a constant (coefficient = 1).

If linear functions always satisfied a(0) = 0, then all lines would pass through the origin, which clearly isn''t true.

Graham Rhodes

Senior Scientist

Applied Research Associates, Inc.

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#13
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Posted 28 June 2001 - 03:07 PM

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#15
Anonymous Poster_Anonymous Poster_*
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Posted 29 June 2001 - 01:28 PM

Since you are a senior scientist I thought you would be smarter than to make your erroneous correction to my previous post.

A linear function is not defined as a line. A linear function, in the most common definition, is a function f:V->V, where V is a vector space, such that

f(x+y)=f(x)+f(y) for all x,y in V.

This means that f(0)=f(0+0)=f(0)+f(0), i.e. that f(0)=0.

Just because the function a(x)=2-x is not a linear function does not mean that the graph of a(x) is not a line. These are two completely different things. a(x) here is an affine function, as I said before, and the graph of all affine functions are lines.