mm-291 == Subject: Re: Somebody please help me.alt.math.undergrad:> Can anybody figure out this riddle? What has two eyes and lives in> one? I've been trying for a week now and it's driving me crazy.Does> anybody have any idea. Please help.An iris: it's spelled with two s and 'lives' in your eye.Brian ==== Subject: Re: very interesting differentiation problem> The derivative of an o(x^n) function is an o(x^(n-1)) function.Not true. Consider x^(n-1)*sin(e^x). ==== Subject: Re: very interesting differentiation problem> Problem.Let f(x) be infinitely many differentable on R and> f(x)=o(x^n) when x tends to infty.(x->infty).Prove that for any k>n > (k)> function f (x) have at least on zero. (k) > I notice that f (x) means k-th derivative f at x.That's false as sta. Let f(x) = e^(-x). Then f(x) = o(x^n) as x -> oo for any n, but no derivative of f has a zero.Perhaps you meant to write f(x) = o(x^n) as |x| -> oo. Take n = 1. We want to show f''(x) = 0 somewhere. Suppose not. Then f'' > 0 everywhere or < 0 everywhere. This implies f' has a nonzero limit - possibly infinite - either at +oo or -oo. Now use the fundamental theorem of calculus to show f(x)/x does not tend to 0 as |x| -> oo. ==== Subject: The problematic point is this: by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id i0VCeR821954; === Virgil worte:...For any natural number, n, let p_1,...,p_n be the first n primes and consider N = 1 + (product of p_1 through p_n).N is larger than any of the lis primes, and is not divisible by anyof them, so must be divisible by (or even equal to) a prime largerthan any in the list.Here is my prblem - why can't we just say:N is larger than any of the lis primes, and is not divisible by anyof them - hence N is a PRIME not in the list.Why do we have to state that N is either a Prime -or- divisible by (oreven equal to) a prime larger than any in the list ?-Ben ==== Subject: Re: The problematic point is this:> Virgil worte:> ...> For any natural number, n, let p_1,...,p_n be the first n primes and > consider N = 1 + (product of p_1 through p_n).N is larger than any of the lis primes, and is not divisible by any> of them, so must be divisible by (or even equal to) a prime larger> than any in the list.> Here is my prblem - why can't we just say:> N is larger than any of the lis primes, and is not divisible by any> of them - hence N is a PRIME not in the list.Here are some more:2*3*5*7*11*13 + 1 = 30031 = 59 * 509 2*3*5*7*11*13*17 + 1 = 510511 = 19 * 97 * 277 2*3*5*7*11*13*17*19 + 1 = 9699691 = 347 * 27953 2*3*5*7*11*13*17*19*23 + 1 = 223092871 = 317 * 703763 2*3*5*7*11*13*17*19*23*29 + 1 = 6469693231 = 331 * 5712*3*5*7*11*13*17*19*23*29*31*37 + 1 = 7420738134811 = 181 * 60611* 676421 2*3*5*7*11*13*17*19*23*29*31*37*41 + 1 = 304250263527211 = 61 * 450451 * 11072701 Far from being a rarity, these example seem to be quite common. Is there any known theory here? For example, we see that among the first 13 primes, there are 7 Euclid pseudo-counterexamples, for a ratio of 7/13. Does this ratio have a limit as the number of primes goes to infinity? ==== Subject: Re: The problematic point is this:[snip]> Here are some more:> 2*3*5*7*11*13 + 1> = 30031 = 59 * 509> 2*3*5*7*11*13*17 + 1> = 510511 = 19 * 97 * 277> 2*3*5*7*11*13*17*19 + 1> = 9699691 = 347 * 27953> 2*3*5*7*11*13*17*19*23 + 1> = 223092871 = 317 * 703763> 2*3*5*7*11*13*17*19*23*29 + 1> = 6469693231 = 331 * 571> 2*3*5*7*11*13*17*19*23*29*31*37 + 1> = 7420738134811 = 181 * 60611* 676421> 2*3*5*7*11*13*17*19*23*29*31*37*41 + 1> = 304250263527211 = 61 * 450451 * 11072701> Far from being a rarity, these example seem to be quite common. Is there> any known theory here?I don't know. But you might be interes in. David ==== Subject: Re: The problematic point is this:> Virgil worte:> ...> For any natural number, n, let p_1,...,p_n be the first n primes and > consider N = 1 + (product of p_1 through p_n).N is larger than any of the lis primes, and is not divisible by any> of them, so must be divisible by (or even equal to) a prime larger> than any in the list.> Here is my prblem - why can't we just say:> N is larger than any of the lis primes, and is not divisible by any> of them - hence N is a PRIME not in the list.How do you know it's not composite, yet divisible by some prime larger than any of the lis primes? For example: 2*3*5*7*11*13 + 1 = 30031 = 59 * 509.That's your counterexample. ==== Subject: Re: The problematic point is this:> Virgil worte:> ...> For any natural number, n, let p_1,...,p_n be the first n primes and > consider N = 1 + (product of p_1 through p_n).N is larger than any of the lis primes, and is not divisible by any> of them, so must be divisible by (or even equal to) a prime larger> than any in the list.> Here is my prblem - why can't we just say:> N is larger than any of the lis primes, and is not divisible by any> of them - hence N is a PRIME not in the list.Because, according to our assumptions, N need not be prime. Note that we have not assumed that the list exhausts all primes, merely that it contains all the primes up to some given prime, p_n. Thus we have not excluded the possibility of primes between p_n and N.Consider, 2*3*5*7*11*13 + 1 = 30031 = 59*509.Our conclusion merely states that given any prime, there is a larger prime. No mention of infinity. ==== Subject: Re: The problematic point is this:>Virgil worte:>For any natural number, n, let p_1,...,p_n be the first n primes and >consider N = 1 + (product of p_1 through p_n).>N is larger than any of the lis primes, and is not divisible by any>of them, so must be divisible by (or even equal to) a prime larger>than any in the list.>>Here is my prblem - why can't we just say:>N is larger than any of the lis primes, and is not divisible by any>of them - hence N is a PRIME not in the list.>Why do we have to state that N is either a Prime -or- divisible by (or>even equal to) a prime larger than any in the list ?2*3*5*7*11*13+1 is one reason :) ==== Subject: Re: The problematic point is this:> Virgil worte:> ...> For any natural number, n, let p_1,...,p_n be the first n primes and> consider N = 1 + (product of p_1 through p_n).> N is larger than any of the lis primes, and is not divisible by any> of them, so must be divisible by (or even equal to) a prime larger> than any in the list.> Here is my prblem - why can't we just say:> N is larger than any of the lis primes, and is not divisible by any> of them - hence N is a PRIME not in the list.The reason is that it's not true! Consider, for example, that1 + 2*3*5*7*11*13 = 59*509 .David> Why do we have to state that N is either a Prime -or- divisible by (or> even equal to) a prime larger than any in the list ?> -Ben ==== Subject: Took test on empty stomachI took my first college trig test this semester and ed up. I knew allthe stuff but my mind went blank on me, I even star shaking. I'm thinkingthe problem is that I didn't eat breakfast, just some coffee. She drops thelowest test score which will be this one. Anyone else experienced this. Wereyou much more relaxed on a full stomach?Matt ==== Subject: Re: Took test on empty stomach>I took my first college trig test this semester and ed up. I knew all>the stuff but my mind went blank on me, I even star shaking. I'm thinking>the problem is that I didn't eat breakfast, just some coffee. She drops the>lowest test score which will be this one. Anyone else experienced this. Were>you much more relaxed on a full stomach?Full but not so full as to make you torpid.A good night's sleep the night before is probably more effective than a lot of last-minute cramming.-- Stan Brown, Oak Road Systems, Cortland County, New York, USA http://OakRoadSystems.comAn expense does not have to be required to be considered necessary. -- IRS Form 1040 line 23 instructions ==== Subject: Re: Took test on empty stomach> I took my first college trig test this semester and ed up. I knew all> the stuff but my mind went blank on me, I even star shaking. I'm thinking> the problem is that I didn't eat breakfast, just some coffee. She drops the> lowest test score which will be this one. Anyone else experienced this. Were> you much more relaxed on a full stomach?MattMy experience is that neither a full nor an empty stomach is optimal. A light and easily digestible meal at least an hour befor the exam.Also, a good night's steep, rather than cramming all night, helps. If you must cram, do it earlier. ==== Subject: right problem by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id i0VH2KB09760; === Yes, i meant that it's need to prove that f is periodic function.Let's i give now right problem:Prove that if for function,defined on (-infty,+infty) is x+1integral f(t)dt =0 then f is periodic. xi don't anderstand you solution fully:one question:is it right that x+1d/dx integral f(t)dt =f(x+1)-f(x)??? -as i think it is right, for x x+1 x+1 xexamle because integral f(t)dt = integral f(t)dt - integral f(t)dt x a a and then taking derivative the both sides we can assume that x+1d/dx integral f(t)dt =f(x+1)-f(x) -am i right??? xThanks!!!Thanks!!! ==== Subject: Re: integral problem by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id i11CisI08322; === Refer to the Fundamental Theorem Calculus, and use the theorem tofigure out what the derivative of the integral is. Now, take thederivative on both sides just as was outlined in my other reply, andyou will arrive at the result. The main thing is to review the theorem(above), this is important.MM. ==== Subject: HELP! stuck on hamilton cycle problemFor a natural number n, define the graph G_n = (V_n, E_n) as* V_n being the set of all n-tuples consisting of 0 and 1. For example: V_2= { (0,0), (0,1), (1,0), (1,1) }* Two vertices are connec with an edge if and only if the correspondingtuples differ in exactly one position. For example there is an edge between(0,0) and (1,0) but not between (0,0) and (1,1)PROBLEM: construct a hamilton circuit in G_nI can construct a hamilton circuit for n = 1, 2 or 3. But do i do it forany n ???? ==== Subject: continuity and finite correction factors by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id i113m2i28990; === Is it appropriate to use both the finite and continuity correctionfactors together when calculating confidence intervals for smallsamples (i.e. under 50)? ==== Subject: Re: 5x5 matrix determinant by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id i11CisI08330; === I Hope i'm not wrong but I think you missed a point there.When doing Row operations on a matrix, in order for the Determinant tostay unchanged one must do it as in the following example:(1). R3 <- R3 + k*R2that way the determinant is unchanged.if instead you apply:(2). R3 <- m*R3 + k*R2then ofcourse your new determinant is m*(Det)I'm certain you know this, but maybe you got mixed up in calculatonsand that is the problem ?!.I tried it myself, applying only type (1) transforms and I got thismatrix:2, -1, 0, 4, 10, 5/2, -1, -4, -3/20, 0, 4, -1, 60, 0, 0, -4/5, -20, 0, 0, 0, -69/8now the determinant = 2*(5/2)*4*(-4/5)*(-69/8) = 138.-Ben ==== Subject: Re: complex powers> What interpretation/visualization can be given to the act of raising a> number to a complex power?> Assuming real a, and z = x + iy = Re[z] + iIm[z],> a^(x + iy) = exp(x*log(a) + iy*log(a))> This is a stretch of a^(Re[z]) and a rotation through an angle> Im[z]*log(a). said Assuming real a. However, the case a = 0 deserves specialattention. For anyone interes in investigating that case, I'd recommendusing C* (the one-point extension of the complexes) and considering threesubcases: Re(z)<0, Re(z)=0, and Re(z)>0.[The middle subcase seems to be particularly interesting.]David Cantrell ==== Subject: Re: complex powers That doesn't help because I don't know what Pye is. Could youexplain, please?On the assumption that most people reading this newsgroup have heard of theratio of circumference of a circle to diameter, and can guess what I meant,this sounds like sarcasm. Sure, pi is normally preferable in formulae. Here,trying to avoid ambiguous exponentials such as e^pii. Would prefer to usethe Greek symbol of course, but cannot make it appear in postings.Never mind all that, did you like the visualisation idea ? It is anextension of the visualisation of multiplying complex numbers together, topowers.Here is a question whilst on complex numbers; we often have problemsinvolving z^(1/n), the n-th root of complex number z, where n is a realnumber. Is there any particular significance and pattern to replacing n withcomplex number a+bi, to get z^(1/(a+bi)), what I am thinking of as a complexroot of a complex number ?Constructive ideas welcome.Ian Hutcheson ==== Subject: Re: complex powers>That doesn't help because I don't know what Pye is. Could you>explain, please?help everyone follow discussions.>On the assumption that most people reading this newsgroup have heard of the>ratio of circumference of a circle to diameter, and can guess what I meant,>this sounds like sarcasm. Sure, pi is normally preferable in formulae.Oh, you meant pi. I am sorry you think it was sarcasm, but since pi is not only (as you say) preferable but also shorter, I never made the connection. >Here, trying to avoid ambiguous exponentials such as e^pii.??But if you use p for pi, then you have exp(pi*i) in your notation as e^pi, which I should think even worse. What's wrong with e^(pi*i) or exp(pi*i) ?-- Stan Brown, Oak Road Systems, Cortland County, New York, USA http://OakRoadSystems.comAn expense does not have to be required to be considered necessary. -- IRS Form 1040 line 23 instructions ==== Subject: Re: complex powersApologies for repea posting- had trouble with downloading but then itappeared. ...it was sta that i^i is real. I can show that formulaicly, but Iwanto be able to visualize it.*******************The formula I supplied wasz^(a+ib) = (r^a) (e^(-b theta)) ((e^(ia theta + b ln r))I will now try to use it on your example to assist visualisation of i^i inthe complex planeOn Argand diagram orig representation of complex number is one unit up onimaginary axis.Formula predicts rotation to new angle, (domina by a when a and b are ofsimilar size), which in general is a theta + b ln r.With this particular example new angle is 0 times pi/2 + 1 ln 1 = 0,(which you could view as a positive rotation of 3pi/2), which puts it on thepositive part of the real axis.The size predic by this formula is in general r^a times e^(-b theta)With this particular example it shrinks because it is 1^0 times e^(-1 pi/2)Thus i^i = 1/(sqrt ( e^pi) = 0.2078 approximatelyThe formula suggests that you can do much more in terms of visualisation;for example, find out what relation must exist between a and b to obtain:-i) A new complex number with no rotation, viz a theta + b ln r = oii) A new complex number with no change of size (modulus), viz r^a timese^(-b theta) = rComplex numbers are full of interesting suprises and useful shortcuts. Youget immediate trig formulae (eg tan of 3 theta), a better understanding ofroots of polynomials, the revelation that sine and cosine are justexponentials, and mind expanding challenges, such as what are the cosine orlogarithm of a complex number. Cauchy's tricks allow us to integratefunctions that seemed impossible - there is plenty to discover.I read An Imaginary Tale: the History of i - Paul J. Nahin and found itfascinating. He shows how the whole subject was full of apparent confusionsand paradoxes and how Wessel, Euler, Gauss and Cauchy broke through thesuccessive veils of mystery. Challenging but worth it.Ian Hutcheson ==== Subject: Re: complex powersIn the following p represents PyeNo doubt you already know that for n > 1, the mapping z to z^n causes thesize to be raised to the nth power and the angle to be multiplied by n.The mapping z to z^(1/n) produces n solutions equally spaced in angle aroundz and diminished in size to the n-th root of the size of z. What about z toa complex power ?With real k, e^[k ln(x)] = x^k, but ,with complex variablesln(z) = ln(r) +i(theta) + i(2np) (i.e. multivalued)Now e^[k ln(z)] = [e^k ln(r)] [e^ ik(theta)] [e^ i2nkp]luckily, any power of e^ i2p = 1 so we are always justified in usinge^[k ln(z)] = (r^k)[e^ ik(theta)] =z^kIf we now let k = a+ib, (which was not forbidden by any previousassumption), thenz^(a+ib) = (r^(a+ib))[e^ i(a+ib)(theta)] = (r^a) (e^(-b theta) (r^ib)((e^(ia theta))using (r^b)^i = (e^b ln r)^i, we obtainz^(a+ib) = (r^a) (e^(-b theta)) ((e^i(a theta + b ln r))Comparing this with z = r e^ i(theta)] indicates the resulting complexnumber in polar form, in terms of the size and sign of a and b. The new sizewill be (r^a) e^(-b theta) and angle of the result, which may amount toseveral rotations, is given by (a theta + b ln r),However, as expressions like these become the more complica, theirusefulness in supplying visual analogies diminishes. Tristan Needham's bookwith its many illustrations and his idea of Amplitwist takes visualexplanations of the complex as far as they can be used.I hope that some of you found this of some use .Ian Hutcheson ==== Subject: Re: complex powers>In the following p represents PyeThat doesn't help because I don't know what Pye is. Could you explain, please?-- Stan Brown, Oak Road Systems, Cortland County, New York, USA http://OakRoadSystems.comAn expense does not have to be required to be considered necessary. -- IRS Form 1040 line 23 instructions ==== Subject: Re: Strange resultMy graphing calculator helps me to see that y=x and y=e^x and y=- ln(x) allpass through the same point which has x co-ordinate 0.56714329. That is onlyan answer and not a method of course, but it does suggest that iterationsshould start near to x=0.5 It ended up alternating in a complex conjuga pair.sounds interesting - could you supply more details of complex pair and theworkings, and then perhaps someone who knows quite a bit about numericalmethods could help more.Ian Hutcheson ==== Subject: Re: Strange result> My graphing calculator helps me to see that y=x and y=e^x and y=- ln(x)> all pass through the same point which has x co-ordinate 0.56714329.FWIW, that's sometimes called the omega constant. See, if interes.> That is only an answer and not a method of course,The method is simple if you use the Lambert W function.David ==== Subject: fermat little theorem... by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id i11IpWO04194; === Does anyone knows how to prove fermat's little theorem that states:n^p congruent to 1 (mod p) with p prime an (n,p)=1I would apreciate any help. ==== Subject: fermat little theorem by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id i122ROI07809; === First of all, the correct statement on this result is:1. For any prime p, and integer n relatively prime to p, n^[p-1] == 1(mod p).2. Clearly this is equivalent to n^p == n (mod p), and this latterversion is true also if p is a divisor of n. That is, this secondversion should be true for ALL integers n.3. And this can easily be proved by mathematical induction on n,after we have checked that for all integers b and c, (b+c)^p == b^p +c^p (mod p).4. Finally, this last result is just an application of the binomialtheorem along with the observation that the binomial coefficientC(p,k) is a multiple of p for 0 Does anyone knows how to prove fermat's little theorem that states:n^p congruent to 1 (mod p) with p prime an (n,p)=1> If you know a little group theory there's a simple proof. Consider the ring of integers mod p. It has a multiplicative subgroup of order p-1 consisting of exactly those n for which (n,p) = 1. [This proof will go through the same for p not prime, with phi(p) replacing p-1].For any element a in the multiplicative subgroup, a^(p-1) = 1, because a is in a group of order p-1 whose identity is 1. ==== Subject: math homework Common deminatorshello please find the common deminators for these fractions. 71/177 69/190 69/170 51/158 28/92. thanks for your help this is a tough problem also i need to order them from least to greatest thanks a lot!!!!! ==== Subject: Re: math homework Common deminators>hello please find the common deminators for these fractions. 71/177 >69/190 69/170 51/158 28/92. thanks for your help this is a tough >problem also i need to order them from least to greatest thanks a lot!!!!!It is usually not very helpful simply to post your homework problem and tell us how hard it is. You would do better to tell us exactly what you've done and exactly where you got stuck.Actually, you might want to post in a k-12 group rather than a college undergraduate math group. It's not that your problem is beneath us, just that you're likely to get more focused and useful answers if you direct your query to the most appropriate newsgroup.But either way, the Net helps those who help themselves. Don't just post your homework, but show what you did to try to solve it. When you have the test you will not be able to post queries to Usenet, so you need to flex your mental muscles now.-- Stan Brown, Oak Road Systems, Cortland County, New York, USA http://OakRoadSystems.comAn expense does not have to be required to be considered necessary. -- IRS Form 1040 line 23 instructions ==== Subject: Re: math homework Common deminators Adjunct Assistant Professor at the University of Montana.>hello please find the common deminators for these fractions. 71/177 >69/190 69/170 51/158 28/92. thanks for your help this is a tough >problem also i need to order them from least to greatest thanks a lot!!!!!Factor the denominators:177 is a multiple of 3 (digits add up to 15), so177 = 3*(59)and 59 is prime. 190 = 10*19 = 2*5*19170 = 17*10 = 2*5*17158 = 2*7992 = 2*46 = 2*2*23So a common denominator must be a multiple of 4, 3, 5, 17, 19, 23, 59, and79. So the smallest common denominator is 2,077,594,140.-- ==== Subject: Re: math homework Common deminatorsThank you for helping me but its going to take me a long time to do thisproblem that way. so can you please help me changing these fractions intodecimals i think it would be much easier that way. I will give you thefractions again:: 71/177 69/190 69/170 51/158 and 28/92. Thanks a lot!>hello please find the common deminators for these fractions. 71/177>69/190 69/170 51/158 28/92. thanks for your help this is a tough>problem also i need to order them from least to greatest thanks a lot!!!!!> Factor the denominators:> 177 is a multiple of 3 (digits add up to 15), so> 177 = 3*(59)> and 59 is prime.> 190 = 10*19 = 2*5*19> 170 = 17*10 = 2*5*17> 158 = 2*79> 92 = 2*46 = 2*2*23> So a common denominator must be a multiple of 4, 3, 5, 17, 19, 23, 59, and> 79. So the smallest common denominator is 2,077,594,140.> --> ==== ========================================================= ==== ===> It's not denial. I'm just very selective about> what I accept as reality.> --- Calvin (Calvin and Hobbes)> ==== ========================================================= ==== ===magidin@math.berkeley.edu===== Subject: Re: math homework Common deminators>Thank you for helping me but its going to take me a long time to do this>problem that way. so can you please help me changing these fractions into>decimals i think it would be much easier that way. I will give you the>fractions again:: 71/177 69/190 69/170 51/158 and 28/92. Thanks a lot!Is it just me, or do others think we're being trolled? First these really elementary problems, then the objection to a straightforward solution -- it seems rather suspicious.-- Stan Brown, Oak Road Systems, Cortland County, New York, USA http://OakRoadSystems.comAn expense does not have to be required to be considered necessary. -- IRS Form 1040 line 23 instructions ==== Subject: Re: math homework Common deminators>Thank you for helping me but its going to take me a long time to do this>>problem that way. so can you please help me changing these fractions into>>decimals i think it would be much easier that way. I will give you the>>fractions again:: 71/177 69/190 69/170 51/158 and 28/92. Thanks a lot!> Is it just me, or do others think we're being trolled? First these > really elementary problems, then the objection to a straightforward > solution -- it seems rather suspicious.> I told her to use the internet to try to find an answer to her problems. I told her to ask how to solve it, not to have it solved for her, since she is not going to have newsgroup access during the test.For calirification purposes, she pos from her sister Lillian's computer, than off of hers. It's the first time she ever used newsgroups.Thanks for the help that you and others provided.Richard-- -------------------------------------------------------------- -------Dum spiro, spero. (Cicero) As long as I breathe, I hope. ==== Subject: Re: divisibility?Hello my name is priscilla the divisblity rule id u can divide by 5 and you would get a number without a decimal or remaniders.> does anybody know the rule for the divisibility of 315? or could u giv> me a site to reasearch that ?> ==== Subject: Re: divisibility?Learn the basic multiplication facts for 5. Apply the rule observed in thosefacts.G C>Hello my name is priscilla the divisblity rule id u can divide by 5 and >you would get a number without a decimal or remaniders.>> does anybody know the rule for the divisibility of 315? or could u giv>> me a site to reasearch that ?>> ==== Subject: forgot the topic...., but FLUENCYI forgot the exact topic, but Darrell and Paul Tanner were discussing fluency. I have to do this way because my connection to CS was lost and so were themessage that I was reading and cannot retrieve them without pulling in a messof stuff with that one message.Here is my response about .....FLUENCY:The student recognizes which inverse operation to perform as he performs eachpreceeding step. He does not check a list of formal rules to decide whichoperation is useful for each current step. The meaning of these rules arefixed into his behavior. NOT FLUENT:The student does not know how to start a solution without reading a list offormal rules and deciding on which rule is justifiable and useful for eachstep. The rules of equality and order relation for the arithmetic operationsare only understood isola by themselves and are not integra into thestudent's behavior. G C ==== Subject: Re: Changing fractions to decimalsI think the easiest way to change these fractions to decimals is by dividingthe numerator by the denominator. For example, 71 divided by 177 is about 0.4,so 71/177 is approximately 0.4.-Christina>Can you please help me change these fractions into decimals, thankyou.>71/177 69/190 69/170 51/158 and 28/92. thanks a lot for the help!!! ==== Subject: Re: Changing fractions to decimals Adjunct Assistant Professor at the University of Montana.>Can you please help me change these fractions into decimals, >71/177 69/190 69/170 51/158 and 28/92. thanks a lot for the help!!!Do the long division until the remainders start repeating.-- ==== Subject: Re: Changing fractions to decimals>But what if i try doing the long divison and i make a mistake, this is the>hardest problem i have ever done. But my teacher said to round to the>hundreths place for the decimals i don['t get that.If you are supposed to round to the hundreths place that means thatyou only need to figure out the first couple of decimals. Figure outthe first 3; if the third one is 0,1,2,3, or 4, you drop it. It it is6, 7, 8, 9, you drop it by adding one to the previous one. if it isequal to 5, and the remainder is NOT 0, then you round up. So, for example, 28/92:The first decimal place (the tenths) is 3. 3*92 = 276, which leaves areminder of 4. For the hundreths place you are dividing 40 by 92,which gives you 0 with a remainder of 40. For the thousandths place,you are dividing 400 by 92, which gives you 4 with a remainder of400-4*92=400-368 = 32.So the decimal starts 28/92 = 0.304...so you will round down to .30.If you get .578..., then you round up to .58. If you get .295, thenyou need to see; if the remainder is not 0, then you round up to .30;if it is equal to 0, then it depends on whether you round halfs up or down.-- ==== Subject: Re: Changing fractions to decimals Adjunct Assistant Professor at the University of Montana.>I have tried that but it took a lot of time and i am afraid of making a>mistake!Yes, the larger the denominators, the longer you can expect therepetition to take. Not much you can do about it (other than trying toget someone else to do your entire homework for you, which I guess iswhat you are aiming for here....)>>Can you please help me change these fractions into decimals, >>71/177 69/190 69/170 51/158 and 28/92. thanks a lot for the help!!!>> Do the long division until the remainders start repeating.