Chikis Posted April 19, 2016 Share Posted April 19, 2016 If I multiply [math]\frac{a-n}{a+n}[/math] by minus sign, do I really get [math]-1[/math] Link to comment Share on other sites More sharing options...

ajb Posted April 19, 2016 Share Posted April 19, 2016 Why do you think you might? Explain your reasoning and people may be able to really point out your mistake. Link to comment Share on other sites More sharing options...

Chikis Posted April 19, 2016 Author Share Posted April 19, 2016 (edited) My reason is this I wanted to simplfy [math]\frac{a-n}{a+n}[/math] completely. Looking at the numerator and denominator, these are the same terms which would have cancelled out if not for the opposite signs. Knowing that [math](-)(-)=+[/math], I decided to multiply the numerator by minus sign to make the simplification possible. Edited April 19, 2016 by Chikis Link to comment Share on other sites More sharing options...

ajb Posted April 19, 2016 Share Posted April 19, 2016 So you get [math]-1 \frac{a-n}{a+n} = \frac{-a +n}{a+n}[/math] 2 Link to comment Share on other sites More sharing options...

Chikis Posted April 19, 2016 Author Share Posted April 19, 2016 (edited) So you get[math]-1 \frac{a-n}{a+n} = \frac{-a +n}{a+n}[/math] This is the same as[math]-\left(\frac{a+n}{a+n}\right)[/math] Agreed? Edited April 19, 2016 by Chikis Link to comment Share on other sites More sharing options...

ajb Posted April 19, 2016 Share Posted April 19, 2016 No, I do not agree. You can convince yourself that this is not true but setting some values to a and n. 2 Link to comment Share on other sites More sharing options...

Sensei Posted April 19, 2016 Share Posted April 19, 2016 Chikis, when I am having objections what result will be I am opening Open Office, entering some random chosen a,n (f.e.), and math formula. f.e. in your case: A1=some a B1=some n C1=A1-B1 D1=A1+B1 E1=C1/D1 F1=-E1 (or -1*E1) Then you can make a graph: A2=A1+1 select some A fields, Edit > Fill > Down. And whole A column is filled with values incremented by 1 (for example). Then fill the rest B-F Select E column, and make line graph from it. You see how it's changing with variable a. 1 Link to comment Share on other sites More sharing options...

Chikis Posted April 21, 2016 Author Share Posted April 21, 2016 No, I do not agree. I don't know why you don't have to agree. When you have[math]-\left(\frac{a-n}{a+n}\right)[/math], remembes that the minus sign multiplies every term at the numerator to give back your [math]\frac{-a+n}{a+n}[/math] So what is the point here? Link to comment Share on other sites More sharing options...

ajb Posted April 21, 2016 Share Posted April 21, 2016 I agree with what you have written above. But this is not going to be -1 for all values of a and n. It is if n =0 and a not equal to zero. Link to comment Share on other sites More sharing options...

studiot Posted April 21, 2016 Share Posted April 21, 2016 (edited) Trying it on something simpler is good advice. Look at this, there are ways of writing the number 1.[math]1 = \left( 1 \right) = \frac{1}{1} = \frac{{\left( 1 \right)}}{{\left( 1 \right)}} = \left( {\frac{1}{1}} \right)[/math]So if we put a minus sign in front we have -1 [math] - 1 = - \left( 1 \right) = - \frac{1}{1} = - \frac{{\left( 1 \right)}}{{\left( 1 \right)}} = - \left( {\frac{1}{1}} \right)[/math] Which ones of these are correct ? [math] - \left( {\frac{1}{1}} \right) = \left( {\frac{{ - 1}}{1}} \right)[/math] [math] - \left( {\frac{1}{1}} \right) = \left( {\frac{1}{{ - 1}}} \right)[/math] [math] - \left( {\frac{1}{1}} \right) = \left( {\frac{{ - 1}}{{ - 1}}} \right)[/math] None of these Edited April 21, 2016 by studiot Link to comment Share on other sites More sharing options...

Chikis Posted April 21, 2016 Author Share Posted April 21, 2016 Number 3 is not correct. Link to comment Share on other sites More sharing options...

studiot Posted April 21, 2016 Share Posted April 21, 2016 Agreed. Either we multiply the top by -1 or the bottom by -1, but not both. In fact [math]\left( {\frac{{ - 1}}{{ - 1}}} \right) = 1[/math] Not -1 So using this fact we have if [math]\frac{{a - n}}{{a + n}} = 1*\frac{{a - n}}{{a + n}} = \frac{1}{1}*\frac{{a - n}}{{a + n}} = \left( {\frac{1}{1}} \right)*\frac{{\left( {a - n} \right)}}{{\left( {a + n} \right)}}[/math] Taking the negative and working both of the correct ways 1 and 2 in my post 10 Either (1) [math] - \frac{{a - n}}{{a + n}} = - 1*\frac{{a - n}}{{a + n}} = - \frac{1}{1}*\frac{{a - n}}{{a + n}} = - \left( {\frac{1}{1}} \right)*\frac{{\left( {a - n} \right)}}{{\left( {a + n} \right)}} = \frac{{\left( { - 1} \right)\left( {a - n} \right)}}{{\left( 1 \right)\left( {a + n} \right)}} = \frac{{ - a + n}}{{a + n}} = \frac{{n - a}}{{n + a}}[/math] Or (2) [math] - \frac{{a - n}}{{a + n}} = - 1*\frac{{a - n}}{{a + n}} = - \frac{1}{1}*\frac{{a - n}}{{a + n}} = - \left( {\frac{1}{1}} \right)*\frac{{\left( {a - n} \right)}}{{\left( {a + n} \right)}} = \frac{{\left( 1 \right)*\left( {a - n} \right)}}{{\left( { - 1} \right)\left( {a + n} \right)}} = \frac{{a - n}}{{ - a - n}}[/math] To show that this is the same as (1), multiply it by 1 and use tha fact that this is the same as multiplying by [math]\left( {\frac{{ - 1}}{{ - 1}}} \right)[/math] [math]\frac{{a - n}}{{ - a - n}} = 1*\frac{{a - n}}{{ - a - n}} = \left( {\frac{{ - 1}}{{ - 1}}} \right)\frac{{a - n}}{{ - a - n}} = \frac{{\left( { - 1} \right)*\left( {a - n} \right)}}{{\left( { - 1} \right)*\left( { - a - n} \right)}} = \frac{{n - a}}{{n + a}}[/math] Are you right now? Link to comment Share on other sites More sharing options...

Chikis Posted April 22, 2016 Author Share Posted April 22, 2016 We are just complicating issue by making this topic long. What I understand and see here is that when you have [math]\frac{a-n}{a+n}[/math] and you multiply the expression by [math]-1[/math], it becomes [math]\frac{-a+n}{a+n}[/math] The numerator and denominator cancels out to give [math]-1[/math] What are we still aguing here? But this is not going to be -1 for all values of a and n. It is if n =0 and a not equal to zero. How do you mean? Give a particular example. Link to comment Share on other sites More sharing options...

ajb Posted April 22, 2016 Share Posted April 22, 2016 [math]\frac{-a+n}{a+n}[/math] Okay, this is what you get... I agree The numerator and denominator cancels out to give [math]-1[/math] What are we still aguing here? This is not true. The minus sign as you write it is 'attached to' a and not (a+n). There is no cancellation here. How do you mean? Give a particular example. Try it for yourself carefully... Let us choose a = 2 n =1. Then your opening expression is equal to 1/3, which is not 1 or -1. This is enough to show you are wrong. Link to comment Share on other sites More sharing options...

studiot Posted April 22, 2016 Share Posted April 22, 2016 We are just complicating issue by making this topic long. What I understand and see here is that when you have and you multiply the expression by , it becomes The numerator and denominator cancels out to give What are we still aguing here? Up to this point I was not arguing. In fact I congratulated you on your correct reply to my first post in this thread. I was trying to show what happens when you put a minus sign infornt of the simplest possible fraction 1/1. And you got it right. So why change when the fraction becomes more complicated? There is no difference. I even offered a simple rule to follow, but you seem to have missed it. Agreed. Either we multiply the top by -1 or the bottom by -1, but not both. Link to comment Share on other sites More sharing options...

Chikis Posted April 22, 2016 Author Share Posted April 22, 2016 (edited) This is not true. The minus sign as you write it is 'attached to' a and not (a+n). There is no cancellation here. Try it for yourself carefully... Let us choose a = 2 n =1. Then your opening expression is equal to 1/3, which is not 1 or -1. This is enough to show you are wrong. Okay, what about when [math]-\frac{a-n}{a+n}[/math]is equal to [math]-\left(\frac{a+n}{a+n}\right)[/math]. The fraction becomes -1. Am sure you agree to this one.Okay, coming to when a = 2 and n = 1, if we put the values in the expression, [math]-\left(\frac{a+n}{a+n}\right)[/math], we have [math]-\left(\frac{2+1}{2+1}\right)[/math] [math]=-\left(\frac{3}{3}\right)[/math]. This gives -(1) = -1. What do you have to say about this one? Edited April 22, 2016 by Chikis Link to comment Share on other sites More sharing options...

imatfaal Posted April 22, 2016 Share Posted April 22, 2016 Okay, what about when [math]-\frac{a-n}{a+n}[/math]is equal to [math]-\left(\frac{a+n}{a+n}\right)[/math]. The fraction becomes -1. Am sure you agree to this one. Okay, coming to when a = 2 and n = 1, if we put the values in the expression, [math]-\left(\frac{a+n}{a+n}\right)[/math], we have [math]-\left(\frac{2+1}{2+1}\right)[/math] [math]=-\left(\frac{3}{3}\right)[/math]. This gives -(1) = -1. What do you have to say about this one? Nope still completely wrong put 2 and 1 into your first expression and you get a different answer! The problem is that you move from a numerator of (a-n) to a numerator of (a+n) - what simple multiplication does that? Multiplying by -1 would change (a-n) to (-a+n). When moving a minus sign from applying to the whole fraction to just the top OR the bottom - you do it like studiot has explained; you need to read through his stuff again [latex]- \left( \frac{(a - n)}{(a + n)} \right) = \left( \frac{-(a - n)}{(a + n)} \right) = \left( \frac{(-a + n)}{(a + n)} \right)[/latex] Link to comment Share on other sites More sharing options...

ajb Posted April 22, 2016 Share Posted April 22, 2016 Okay, what about when [math]-\frac{a-n}{a+n}[/math]is equal to [math]-\left(\frac{a+n}{a+n}\right)[/math]. The fraction becomes -1. Am sure you agree to this one. When it is equal I agree... just that is not true for all a and n. What do you have to say about this one? You are making a mistake with the minus sign. Think about a bit more. Link to comment Share on other sites More sharing options...

Chikis Posted April 22, 2016 Author Share Posted April 22, 2016 (edited) When it is equal I agree... just that is not true for all a and n. And I say give me the value of of a and n, in which it is not true.All I can say is that [math]\frac{-a+n}{a+n}[/math] and [math]-\left(\frac{a+n}{a+n}\right)[/math] are the same and diffrent in some cases. Nope still completely wrong put 2 and 1 into your first expression and you get a different answer! The problem is that you move from a numerator of (a-n) to a numerator of (a+n) - what simple multiplication does that? Multiplying by -1 would change (a-n) to (-a+n). When moving a minus sign from applying to the whole fraction to just the top OR the bottom - you do it like studiot has explained; you need to read through his stuff again [latex]- \left( \frac{(a - n)}{(a + n)} \right) = \left( \frac{-(a - n)}{(a + n)} \right) = \left( \frac{(-a + n)}{(a + n)} \right)[/latex] The first expression is factored to get the second expression. True or force? I will only say that the expression are the same and different. Edited April 22, 2016 by Chikis Link to comment Share on other sites More sharing options...

ajb Posted April 23, 2016 Share Posted April 23, 2016 (edited) All I can say is that [math]\frac{-a+n}{a+n}[/math] and [math]-\left(\frac{a+n}{a+n}\right)[/math] are the same and diffrent in some cases. So they are not considered 'equal'. Again, you are making a mistake with the minus sign. What not try n= a... what happens to your expression? We get (a-a)/(a+a) = 0/2a = 0. Now this shows that multiplying (a-n)/(a+n) by -1 is not going to be -1 for all values of a and n. This shows that you are wrong, unless you are looking for specific values only. This only happnes when a-n =a +n which implies that n = 0. Edited April 23, 2016 by ajb Link to comment Share on other sites More sharing options...

TransientResponse Posted May 15, 2016 Share Posted May 15, 2016 (edited) If I multiply [math]\frac{a-n}{a+n}[/math] by minus sign, do I really get [math]-1[/math] distribute and multiply the -1 to every single term in the nominator. +a and -n are converted like this: -1*a= -a, and -1*-n=n. Hence the nominator is (-a+n). Note 1: -|number|*-|number|=+|number|. or shortly -*-=+ or, literally -1*-1=+1. Note 2: +*+=+ which you already know, possibly without being this closely aware of. If you assume or predict in math, geometry, you never be successful. Just learn the rules and applicate. Some details might be confusing but not. I would like to paraphrase this principle below, so it could be understood better: 1.You must first know priorities in operators/signs. Paranthesis is first>>Then power>>Then multiplying>>Then dividing>>Then Subtracting.. rely on the books, i might be forgetting something. just giving examples here. 2. "-" can be considered as subtraction operator or number sign, it won't be a problem. if you say -1*-n and (-1)*-(+n) wont matter because you know if you see * and -, you know * the operator and there cant be two operators at the same time. By the way "-(stuff)" means -1*(stuff). conclusion: if you see the sign/operator - alone, it is the short form of -1, it is not an operator. 3.Sometimes not easy to find out these even if they are written on the books. 4.So, patience, focusing, dividing the problems into small pieces if too complex are essential.. 5.Math is NOT about guessing or finding the answer.. "Rules, Analysis on(understanding) the data, How to" are essential.... 6.if you are not sure what you are doing, even if you hit the correct answer, you are doing it wrong. 7.Math, Geometriy are languages. If you see a word in math/geometry, you MUST know it has a clearer and borderly-DEFINITION. Should NOT merely rely on the meaning that the language whispers to your mind, because it often leads you to assume too much. Edited May 15, 2016 by TransientResponse Link to comment Share on other sites More sharing options...

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