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#101 Re: Exercises » Proof verification for neighbor fractions » 2014-01-08 04:20:31
By the way, am I right by saying that the exception (the equality with 1 or -1) is telling us that such factorization exists?
because ovisously we could have 1*1/1*1-0*x/1*1=1/1*1*1*1
Which is one example of such a factorization that exists.
#102 Re: Exercises » Proof verification for neighbor fractions » 2014-01-08 04:18:19
Oh I really need to put it in the numerator and show it ? I didn't know that !
#103 Re: Exercises » Proof verification for neighbor fractions » 2014-01-08 04:01:48
By the way, I forgot to treat the case where there might exist one fraction which is factorizable and the other not.
So :
a/b-c/d=+-1/bd
xi/xj-c/d = +-1/((xj)*d)
(i/j-c/d=+-1/xj*d) * jd
id-jc not =l +-1/x
except for 1 and -1
The same is done for the second fractions on the left side.
#104 Re: Exercises » Proof verification for neighbor fractions » 2014-01-08 03:56:29
The reason why the left hand side cannot be equal to the right hand side is because we have considered our variables (a=zy,etc.) to be integer whose factorization were also integers.(We are dealing with fractions, this is the reason why we need them to be integers.) So, when we arive at this result : yn-lm(not equal to) +-1/zp, the left hand side is a substraction of integers. But the right hand side is a fraction with 1/zp(zp also being integers). So, the result shall be inferior to 1. The only exception where it won't be inferior is the case were the left-right hand side shall be equal to +-1.
For the *****, I'm not sure what you're referring to... I shall add more later on.
#105 Exercises » Proof verification for neighbor fractions » 2014-01-04 07:15:59
- Al-Allo
- Replies: 8
*If I turn out to have a wrong answer, please no hints or showing an valid proof. I want to do it on my own !
http://postimg.org/image/buax1y1sd/
ad/bd-bc/bd=+-1/bd is neighbor fraction
Now, reduce the common numbers :
a/b-c/d=+-1/bd
We must now prove that the left hand side has irreductible fractions. Lets see what would happen if these fraction were reductible.
let a=z*y , b=z*l , c=p*m d=p*n
z*y/z*l-p*m/p*n equality to be determined +-1/(z*l)*(p*n)
Reduce:
*ln (y/l-m/n) equality to be determined (+-1/(z*l)*(p*n))*ln
yn-lm equality to be determined +-1/z*p
yn-lm not= +-1/z*p
We have considered the initial fractions to be reductible and have arrived at a false result. An integer cannot be equal to (+-1/z*p)
So, the initial fractions must be irreductible.
*Of course, I'm considering the variables to represents integers.
#106 Re: Help Me ! » Vector question » 2013-12-01 05:19:48
mmmmmhh...... not sure....
#107 Help Me ! » Vector question » 2013-12-01 04:37:32
- Al-Allo
- Replies: 4
Hi, I'm learning vectors and there's two notions which I don't distinguish :
Is the orientation of a vector (determined by the angle it has) the same thing as the direction of a vector ???
thank you
#108 Help Me ! » trigonometry question » 2013-11-30 11:55:39
- Al-Allo
- Replies: 1
Hi, I have the following equation :
2*sin3a=sqrt(2)
sin3a=sqrt(2)/2
I was able to solve it in a geometrical way(giving 6 solutions.) but I have no idea on how to solve it in a symbolic way...(which would take clearly less time to do ) Is there any way ?
thank you
#109 Help Me ! » Log » 2013-11-12 08:49:16
- Al-Allo
- Replies: 1
Hi, quick question.
Is there a difference in notation when I say :
(log8(x))^2
and
log8(x)^2
Is it different ??? Thank you !
#110 Re: Help Me ! » Reduction ad Absurdum » 2013-10-14 06:31:10
(What ever manipulations you do with it )
Say: "What ever correct manipulations you do with it " and I'm happy.
In place of "correct" you could also say "valid".
Bob
AH yes, sorry for the mistake. So yes, any valid manipulation !
#111 Re: Help Me ! » Reduction ad Absurdum » 2013-10-14 02:58:25
AH thank you ! Last question, a real verity will never make any contradictions, right ? (What ever manipulations you do with it )
#112 Help Me ! » Reduction ad Absurdum » 2013-10-13 10:21:46
- Al-Allo
- Replies: 5
Hi, I have a little question concerning contradictions :
If I have a statement "A" that I want to prove, and only have the possibility for it to be True or False.
After some manipulations, I arrive at some contradiction. (Here's where my question begins.)
How can we know that a contradiction is enough to be sure at 100 % that a statement is not correct?
Is it because in Mathematics, for a thing to be True or False, it must always be ALWAYS "working" without arriving at some contradiction ? (Mathematical ideas must always work, and not sometimes yes, sometimes no.)
I just want to be sure of thinking of it in the right way, corrections would be greatly appreciated ! Thank you !
#113 Re: Help Me ! » Question on associativity. » 2013-10-12 04:14:56
No, it is exactly the way I've put it in my first post ^^
#114 Help Me ! » Question on associativity. » 2013-10-12 03:32:44
- Al-Allo
- Replies: 3
Hey, I just have a question concerning associativity :
I have :
by associativityAnd I needed to prove each of my steps concerning an exercise. What I found weird was in the correction part, they justify the retreat of parentheses by saying "associativty" I thought that associativity was only used when you needed to add parentheses, or it can go in both ways ? Could anyone confirm it ???
#115 Re: Help Me ! » Set question » 2013-08-25 04:41:05
Well the intersection is 15, if that's what you're asking
#116 Re: Help Me ! » Set question » 2013-08-25 02:24:11
What do you mean ??? How can it be more simplified ?
#117 Help Me ! » Set question » 2013-08-24 09:23:50
- Al-Allo
- Replies: 4
If I had the following :
and I have put it in comprension notation :
{x∈R|12<equal x <equal15 and 15<equal x <equal 30 }
Is the comprehension notation good ?
*'and' means intersection in this case
#118 Re: Help Me ! » Definition question » 2013-08-24 06:29:32
yes, it was definition ^^
#119 Help Me ! » Definition question » 2013-08-24 05:32:36
- Al-Allo
- Replies: 3
Hi, can somebody tell me the definition of : and,or
Thank you
#120 Re: Help Me ! » Geometry. » 2013-08-23 12:02:31
So, A is congruent to C because A and C are both congruent to B? Isn't that what you need to prove in the first place?
Well, isn't it self evident?
Anyway, I think I should yes ^^
#121 Re: Help Me ! » Geometry. » 2013-08-23 11:47:33
Ok, here's another version :
We have the 3 geometric figures : A,B,C
With the information given: A congruent to B, B congruent to C
So, inversely, we will have :
B congruent to A, because A is congruent to B, so, its inverse must necessarely also be true, because if it wasn't we would a contradiction with the information given to us that A congruent B, but we know that it isn't the case, so the only option left is B is really congruent to A
C congruent to B
For the same reason has the above statement.
So, we have 3 geomtric figures congruent, because A is congruent to B, and having proved the inverse of B congruent C, that is, C congruent to B, we see that the two geometric figures(A and C) are also congruent to the same figure (B), and that...
A is congruent to C
#122 Re: Help Me ! » Geometry. » 2013-08-23 11:38:54
What does equal mean in terms of geometric figures?
Well, sorry for my mistake in terms, I meant congruent,xD
#123 Re: Help Me ! » Geometry. » 2013-08-23 09:51:33
Any????
#124 Re: Help Me ! » Little question » 2013-08-23 07:04:28
Ok, thank you.
#125 Help Me ! » Little question » 2013-08-23 06:19:42
- Al-Allo
- Replies: 7
Hi, : http://imageshack.us/photo/my-images/827/34f2.png/
I'm just wondering, I know that a rational expression can't be divided by 0, but what does he man by "identically" ???I feel like there's something I'm missing... any help please ! Thank you !