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Nairaland Mathematics Clinic by Richiez(m): 12:54am On Jan 01, 2013 |
We diagnose and solve math problems here This thread is the meeting point for nairaland math gurus...I dare anyone to ask a question in mathematics without me having an answer to them, LETS START 8 Likes 1 Share |
Re: Nairaland Mathematics Clinic by Kaxmytex(m): 1:43am On Jan 01, 2013 |
[size=28pt]guy,i luv ur way... Hapi new year![/size] |
Re: Nairaland Mathematics Clinic by Richiez(m): 7:15am On Jan 01, 2013 |
@kaxmytex, thank alot and happy new year too, do u av any new yr maths question 4 me? 1 Like |
Re: Nairaland Mathematics Clinic by Calebsworld(m): 7:20am On Jan 01, 2013 |
@ Richiez,my question,is how can I become a mathematician,as in having distinctions in the subject.God bless u. |
Re: Nairaland Mathematics Clinic by Zikstar4me(m): 8:06am On Jan 01, 2013 |
@richiez, what level of math are u talkin abt? 4 Likes |
Re: Nairaland Mathematics Clinic by CrazyMan(m): 9:34am On Jan 01, 2013 |
Zikstar4me: @richiez, what level of math are u talkin abt?Good question. |
Re: Nairaland Mathematics Clinic by Onaolamipo: 10:12am On Jan 01, 2013 |
Zikstar4me: @richiez, what level of math are u talkin abt? the first poster is not bold enough to ask that kind of question. even those that are studying maths and applied maths in university cannot be bold enough to ask such question 1 Like 1 Share |
Re: Nairaland Mathematics Clinic by Calebsworld(m): 11:51am On Jan 01, 2013 |
Onaolamipo:.Pride goeth before a fall,if u prove to be too intelligent or wise,u will look miserable b4 a fool.Richliez said,ask him any question on maths,Dumb ass kids came here with unreasonable questions. 6 Likes |
Re: Nairaland Mathematics Clinic by Adol16: 1:41pm On Jan 01, 2013 |
Maths gurus pls help me solve the simultaneous equations x + y =5 x^x + y^y=31 10 Likes 2 Shares |
Re: Nairaland Mathematics Clinic by Youngsage: 3:34pm On Jan 01, 2013 |
Adol16: Maths gurus pls help me solve the simultaneous equationsx + y =5 -------- eqn i. x^x + y^y=31 ----- eqn ii. From equatns i & ii, take the log of both sides (xlog^x + ylog^y) = log 31------ eqn iii. log(x+y) = log 5 ------------ eqn iv. xylogxy= log 31 -------- eqn v. log xy = log 5 --------- eqn vi. (Using elimination method); xylog5 = log31 xy = log 31/log 5 =6.2 approx. 6. Recall, from eqn i, x + y = 5. So x = 6/y or y = 6/x. Therefore x=2 when y=3. ans: x=2, and y=3. 61 Likes 3 Shares |
Re: Nairaland Mathematics Clinic by Richiez(m): 4:07pm On Jan 01, 2013 |
@calebsworld, here are a fewtips; 1. you have to develop a great interest for maths, and have that strong desire to become a maths guru. 2. Realize that there's nothing magic about mathematics. 3. make friends with people who appreciate mathematics. 4. practise as many maths questions as possible, starting from the very simple ones to more complex ones. develop the urge to solve them on your own, but you can always seek for assistance whenever you get hooked. 5. do not forget any new thing you have learnt, keep reviewing them. 6. learn to memorize some basic maths formulas as this will build your confidence. GOOD LUCK 17 Likes |
Re: Nairaland Mathematics Clinic by Richiez(m): 4:55pm On Jan 01, 2013 |
Zikstar4me: @richiez, what level of math are u talkin abt?both secondary and tertiary, infact anyone 1 Like |
Re: Nairaland Mathematics Clinic by olawalebabs(m): 5:00pm On Jan 01, 2013 |
Good thread. 1 Like 1 Share |
Re: Nairaland Mathematics Clinic by Zikstar4me(m): 5:05pm On Jan 01, 2013 |
Ok then, please can sm1 explain the pigeonhole principle to me? |
Re: Nairaland Mathematics Clinic by Calebsworld(m): 9:23pm On Jan 01, 2013 |
Richiez: @calebsworld, here are a fewtips;thanks boss,that's all I need,God bless u d more. |
Re: Nairaland Mathematics Clinic by Richiez(m): 10:04pm On Jan 01, 2013 |
Youngsage: nice try bro, although you made some some mistakes especially at the point where you took log of both sides. e.g it ought to be log(x+y)= log5 and not logx + logy = log5. but i must admit, you did a very good job. here's another approach; x + y = 5.......(1) x^x + y^y =31.......(2) from eqn(1), x=5-y, hence we substitute this value for x in eqn(2). (5-y)^(5-y) + y^y = 31.......(3) now, we have reduced the problem to what value of y that must be inserted in eqn (3) to obtain 31. a thorough look at eqns (1) & (2) shows that the values of x & y must be positive integers within the range 0<x<5 and 0<y<5 clearly, y=3 i.e (5-3)^(5-3) + 3^3 2^2 + 27 4+27=31 now that we have established that y=3, we shall substitute the value of y in eqn(1) to get x recall that x+y=5 therefore; x+3=5 or x=5-3=2 hence, x=2 when y=3 38 Likes 1 Share |
Re: Nairaland Mathematics Clinic by Richiez(m): 10:29pm On Jan 01, 2013 |
Zikstar4me: Ok then, please can sm1 explain the pigeonhole principle to me? well funny as the name sounds, the pigeonhole principle is a very important principle in mathematics. it states that if there are x<y and there are x pigeonholes and y pigeons, then at least one pigeonhole must contain more than one pigeon. this principle is used in solving basic mathematical arguments, for example; if there are only 4 colours of shirts and five students are required to choose any colour of shirt that they prefer, then atleast two students must choose same colour. wow.....maths is really interesting! 11 Likes |
Re: Nairaland Mathematics Clinic by 1stknight(m): 10:32pm On Jan 01, 2013 |
Youngsage: Bro pls I don 't understand how you got the fifth equation. 3 Likes |
Re: Nairaland Mathematics Clinic by Richiez(m): 10:33pm On Jan 01, 2013 |
olawalebabs: Good thread. Calebsworld: thanks boss,that's all I need,God bless u d more. thanks! 1 Like |
Re: Nairaland Mathematics Clinic by Gboliwe: 6:36am On Jan 02, 2013 |
Youngsage:Maths is frightening with equations like this 2 Likes |
Re: Nairaland Mathematics Clinic by Zikstar4me(m): 6:40am On Jan 02, 2013 |
@richiez, was tryin to learn that theorem in a buk, though it is xplained to b easy as u jst said in the buk and d xamples are cheap, Bt the exercises are smtin else |
Re: Nairaland Mathematics Clinic by Richiez(m): 7:02am On Jan 02, 2013 |
Zikstar4me: @richiez, was tryin to learn that theorem in a buk, though it is xplained to b easy as u jst said in the buk and d xamples are cheap, Bt the exercises are smtin else maybe you should post some of the exercises here |
Re: Nairaland Mathematics Clinic by Zikstar4me(m): 8:09am On Jan 02, 2013 |
17 people corespond by mail with one another- each one with all d rest. In their letters only 3 diff topics are discussed. Each pair of corespondents deal with only one of these topics. Prove that there are at least three people who write to each other about the same topic. 1 Like |
Re: Nairaland Mathematics Clinic by Richiez(m): 1:17pm On Jan 02, 2013 |
Zikstar4me: 17 people corespond by mail with one another- each one with all d rest. In their letters only 3 diff topics are discussed. Each pair of corespondents deal with only one of these topics. Prove that there are at least three people who write to each other about the same topic. you should note that most pigeon hole problems involves logical reasoning and not calculations. SOLUTION no of mail correspondents = 17 no of topics = 3 since the correspondents are in pairs, no of possible pairs = 8 (i.e only 8 pairs from 17) let pigeons = no of pairs =8 and pigeonholes = no of topics = 3 if we assume that the letters wud be distributed evenly i.e pigeons wud be distributed evenly into the pigeon holes, then atleast all 3 pigeon holes will an extra pigeon this implies that some pairs will have same topic as other pairs 2 Likes |
Re: Nairaland Mathematics Clinic by Nobody: 10:49pm On Jan 02, 2013 |
Yea |
Re: Nairaland Mathematics Clinic by Calebsworld(m): 12:21am On Jan 03, 2013 |
U are da bomb @ Richlez,pls can u pls do some explanations and exercises on this assumption topic M.I(Mathematical Induction).thanks. |
Re: Nairaland Mathematics Clinic by Richiez(m): 12:26am On Jan 03, 2013 |
oh! mathematical induction...dat very interesting topic, i'l do some explanation in d morning, let me av some sleep...i like ur spirit 1 Like |
Re: Nairaland Mathematics Clinic by Zikstar4me(m): 7:10am On Jan 03, 2013 |
I dont understand this part pls explain Richiez: 1 Like |
Re: Nairaland Mathematics Clinic by Richiez(m): 7:25am On Jan 03, 2013 |
okay what i mean is this; since there are 8 pigeons, the first 3 goes into each 3 hole, it then remains 5, in a similar way, we distribute another 3 to each 3 hole and son till all 8 pigeons get exhausted. by so doing, all 3 holes must have extra pigeons this means at least 3 pairs must have same topic since each hole represents a topic |
Re: Nairaland Mathematics Clinic by marcangelo(m): 9:08am On Jan 03, 2013 |
These maths looks weird.. Is dis university maths? 1 Like 1 Share |
Re: Nairaland Mathematics Clinic by Richiez(m): 9:18am On Jan 03, 2013 |
Calebsworld: U are da bomb @ Richlez,pls can u pls do some explanations and exercises on this assumption topic M.I(Mathematical Induction).thanks. Mathematical induction is a method of mathematically proving that a statement is true for all natural numbers i.e positive integers it usually involves 2 basic steps; 1. showing that the statement is true when the lowest natural number is used e.g n=0 or n=1 (preferably) 2. showing that if the statement is true for any natural number n=k, then it is true for the next natural number n=k+1 let us do some examples; 1. prove by mathematical induction 1+2+3+4+.......+n+(n+1) = n(n+1)/2 SOLUTION note that n represents the number of elements in the series step 1---we have to show that the statement is true for n=1 1= 1(1+1)/2 1= 2/2 1=1 therefore since L.H.S = R.H.S the statement is true for n=1 we can even go ahead to show that the statement is true for any other small value of n, say n= 4 1+2+3+4 = 4(4+1)/2 10= 4(5)/2 10= 20/2 10= 10 therefore it is also true for n=4 next is step 2 here we shall consider an arbitrary value of n, say n=k and assume it is true i.e 1+2+3+4+....+k = k(k+1)/2 ................(1) if the above statement is true, then the final step is to show that it is also true for n= k+1 1+2+3+4+....+k+(k+1) = (k+1)(k+1+1)/2 1+2+3+4+....+k+(k+1) = (k+1)(k+2)/2 ...............(2) to show that eqn 2 is true, let us add (k+1) to both sides of eqn 1 1+2+3+4+....+k+(k+1) = k(k+1)/2 + (k+1).............(3) the job here is to make the R.H.S of eqn 3 look like that of eqn 2 taking l.c.m as 2 1+2+3+4+....+k+(k+1) = [k(k+1)+2(k+1)]/2 since (k+1) is a common factor in the R.H.S we factorize 1+2+3+4+....+k+(k+1) = (k+1)(k+2)/2 This is same as eqn 2 above, hence the statement 1+2+3+4+....n+(n+1) = n(n+1)/2 is true for all natural numbers. feel free to post more exercises. 16 Likes |
Re: Nairaland Mathematics Clinic by Richiez(m): 9:22am On Jan 03, 2013 |
marcangelo: These maths looks weird.. Is dis university maths? yep, but don't be scared, any level of maths is accepted here 2 Likes |
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