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I just took a test a few hours ago, and most of it was too advance for me (in my friends' terms, NOSEBLEED!.)
Still, I'm curious.
Could you tell me the answers to these questions?
Given that
Assuming that n is a natural number, prove that
"There is not a difference between an in-law and an outlaw, except maybe that an outlaw is wanted"
Nisi Quam Primum, Nequequam
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Maybe you start with the answer which is
a*a + b*b = 1, and note that plugging
in variatios equivalents of this
into the original equation will form the hypothesis.
(this is a guess)
igloo myrtilles fourmis
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Let me do the second one first. First, are you familiar with this result from elementary number theory? Two integers a and b are coprime (i.e. gcd(a,b) = 1) if and only if there exist integers r and s such that ra+sb = 1.
Well, note that
. Hence and are coprime, and the latter will never divide the former unless which is not possible for n a natural number.Last edited by JaneFairfax (2010-03-31 03:50:24)
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Well, for #1, rewrite the given equation as
and square both sides. (You can square the original equation directly but it turns out that rewriting as above and squaring will cause some terms to cancel and make life a bit easier.). Youll need to square both sides one more time to get rid of another square-root. In the end, when youve simplified everything, you should get
The LHS is just
.Ive been trying to look for a less tedious method than squaring but cant find any so far. There may well be some easier methods, but if you cant find any, then just stick to old-fashioned squaring.
Last edited by JaneFairfax (2008-08-02 00:39:27)
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Aha, heres an alternative method for #1.
Note that in order for
and to be defined, we must have and . Hence we can make the substitutions , .Last edited by JaneFairfax (2008-07-26 03:15:28)
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Very nice solution!
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Thank you.
For some reason, when I see
, I think of sin² and cos².Last edited by JaneFairfax (2008-07-26 05:09:34)
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This is to clearify my idea, just an idea
Hypothesis that
igloo myrtilles fourmis
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This is to clearify my idea, just an idea
Hypothesis that
Step with that assumption to
and
Plug a and b into original fact given, and
you get back your hypothesis that
I thought of that too, but to me it seems circular. You start with an assumption and end with the same assumption.
Last edited by Identity (2008-07-26 05:59:51)
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Identity, please just ignore JEF. I have no idea what the deuce he is doing there, but whatever hes doing hes certainly not proving the problem. We are assuming that
and trying to prove that , but it looks like hes trying to go the other direction.Offline
Jane may very well be correct that I am wrong.
I just read about types of proofs, and mine
is probably the illegal "backwards proof".
There is something called proof by example,
but how that differs from a backwards proof,
I do not know. Thanks Jane!
Suppose you choose a flawed premise,
and then prove the flawed premise using
the flawed premise as evidence. I
think this is where I am going wrong.
That is circular, and not a good idea.
Last edited by John E. Franklin (2008-07-26 07:33:52)
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Proof by example works usually to show that something is not true in all cases. For instance, prove that not all quadratic equations have two unique roots. You can prove this by taking the example of x^2 -4x + 4 = 0, 2 is the only (real) root of this equation. Even if this happened to be the only equation without two unique roots, it would still hold that all of them don't have that.
As for backwards proofs, you have to be careful with which way the implication works in your logic. All dogs have 4 legs but not everything with 4 legs is a dog. In this problem you are given a premise, equation 1 and have to show that equation 1 implies equation 2, so 1 ⇒ 2 (1 implies 2) but you can't just start with 2 and show that 2 ⇒ 1.
What you can do is proof by contrapositive. In which you say suppose 2 is not true, if you can prove that 2 is false implies 1 is false, then clearly if 1 is true, then 2 must be true.
Not however, that many times when showing two equations are equivalent, you can work in both directions, but not always. You must be careful.
A logarithm is just a misspelled algorithm.
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Actually what I did was worse than 2 ⇒ 1,
I did (2 combined with 1 ⇒ 2), which is bizarre.
igloo myrtilles fourmis
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This reminds me of a false proof that I once wrote. Luckily I managed to catch and fix it before it mattered, but it was fairly interesting.
One common false proof tries to say that if 1 ⇒ 2 then 2 ⇒ 1, which is a "converse error".
Another one says that if 1 ⇒ 2 then (not 1) ⇒ (not 2), which is an "inverse error".
One thing I did was to say that if 1 ⇒ 2 and 1 ⇒ 3, then 2 ⇒ 3.
1, 2 and 3 were actually complicated statements, which is why I didn't notice the error straight away, but once I found the error I was interested because I'd never seen one like that before.
Does anyone know if it has a special name, like the other two I mentioned?
Why did the vector cross the road?
It wanted to be normal.
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Well, for #1, rewrite the given equation as
and square both sides. (You can square the original equation directly but it turns out that rewriting as above and squaring will cause some terms to cancel and make life a bit easier.). Youll need to square both sides one more time to get rid of another square-root. In the end, when youve simplified everything, you should get
The LHS is just
.Ive been trying to look for a less tedious method than squaring but cant find any so far. There may well be some easier methods, but if you cant find any, then just stick to old-fashioned squaring.
Thanks, Jane! I actually got up to that part , but I did not know how to simplify it.
Being 13 really does have its disadvantages, don't you think?
"There is not a difference between an in-law and an outlaw, except maybe that an outlaw is wanted"
Nisi Quam Primum, Nequequam
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If you take an 8 foot pole and formed it in a circle, what is the measurement of that circle on the inside?
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The distance across the inside is about 8/3 or more exact 8/3.14159, where pi = 4/1 -4/3 + 4/5 - 4/7 + 4/9 - 4/11 ....(on and on forever)
8 / 3.14159265358979323846 = 2.54647909 feet across the circle.
Also, you might want to subtract off the width of the pipe, since 2.54647909 feet is more likely from
the inside of the pipe on one side of the circle to the outside of the pipe on the other side of the circle.
igloo myrtilles fourmis
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If you take an 8 foot pole and formed it in a circle, what is the measurement of that circle on the inside?
If you're asking for the area of the circle:
The 8 foot pole will be bent around so that it becomes the circumference of the circle. The formula for the circumference of a circle is:
So now you solve for r:
(1)The formula for the area of a cirlce is
Plugging in the r from (1),
the area of the circle isNext time you want to ask a question, please post a new topic.
Last edited by Identity (2008-08-01 14:17:19)
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JaneFairfax wrote:Well, for #1, rewrite the given equation as
and square both sides. (You can square the original equation directly but it turns out that rewriting as above and squaring will cause some terms to cancel and make life a bit easier.). Youll need to square both sides one more time to get rid of another square-root. In the end, when youve simplified everything, you should get
The LHS is just
.Ive been trying to look for a less tedious method than squaring but cant find any so far. There may well be some easier methods, but if you cant find any, then just stick to old-fashioned squaring.
Thanks, Jane! I actually got up to that part , but I did not know how to simplify it.
Being 13 really does have its disadvantages, don't you think?
Hmm.. Jane, would you mind sharing how you simplified it?
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I made a typo: it should be
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For the first one, whatever you do, dont listen to John E. Franklin. Im trying to control my temper here and not shout at anyone for posting blatantly stupid posts, and John is doing me no favours here.
John is half right. It is rather easy to see that:
If a^2 + b^2 = 1, then
What must be shown now is that if a^2 + b^2 ≠ 1, then
"In the real world, this would be a problem. But in mathematics, we can just define a place where this problem doesn't exist. So we'll go ahead and do that now..."
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Given that
Prove that
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Hold my horses!
Thats not quite right. Equality holds if and only
and are linearly dependent in .This means the condition for equality should be
.Fortunately, that still comes down to the same thing, namely
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