譴責(zé)申明：世界十大數(shù)學(xué)難題的解答,圓球體層式的解答解讀

1、標(biāo)題：世界十大數(shù)學(xué)難題的解答.圓球體層式解答 .作者 : 百度里的昵稱“蔡於竟道”Apology statement.Condemn stated:relatively speaking, this layer ball type theory.Is true, and I in baidus nickname, CAIin registration in May 2014, at the same time.Can other will not.B: yes.Other, opposite in other forms.0, 7 years, told the diameter of the

2、 sphere, and how the score, N, anything better to do one thing at the time, including the universe at the time, and so on.Principle of outer principle, principle of the inner principle.About is like a thing well done.The condemned man: baidus nickname, CAI YU JING DAO .道歉申明。譴責(zé)申明：相對(duì)來(lái)說(shuō)，此圓球體層式理論。真正的，是和

3、本人在百度的昵稱“蔡於竟道 ”，注冊(cè)時(shí) 2014 年 5 月，同時(shí)發(fā)表的。其他的可以都將不算。是的。其他的，相對(duì)是以其他形式講過(guò)。零七年時(shí)，講過(guò)的0，圓球體的直徑，怎樣的參比數(shù)，N，任何當(dāng)時(shí)不如做一件事，包括宇宙當(dāng)時(shí)，等。原理外層有原理，原理里層有原理。講了就是當(dāng)時(shí)如同做好了一件事。譴責(zé)人：百度里的昵稱，“蔡於竟道 ”.浙江省臺(tái)州市。2015 年.10 月.3 日 . （機(jī)械翻譯，不知道對(duì)吧？）摘要：此圓球體層式理論，能解答此世界十大數(shù)學(xué)難題。The Perfect Sphere Layer expression can solve the following ten difficult ma

4、thematic problems in the world.P=NP. 原理同源 .（也叫原里同源.）（P = NP.The principle of homology.The origin is the origin.）幾何法論證中。 若，理論證回到這個(gè)原點(diǎn)， 實(shí)際證回到這個(gè)原點(diǎn)。 （是的。不能說(shuō)理論和實(shí)際都各為一個(gè)平面。那只是幾何法中時(shí)的一個(gè)演示的一道題時(shí)。當(dāng)然也不能說(shuō)成多大圓球體。）因?yàn)檎撟C最后，理論和實(shí)際都證回到原點(diǎn)，則相等。那么，原理為等值原理（原理同源 .）。引言：零六年在網(wǎng)絡(luò)上，以其它形式講過(guò)。零九年曾經(jīng)發(fā)表過(guò)。有的他們也認(rèn)得。時(shí)在百度里，去年五六月份在老師微信群的幫助下，做好

5、的這十道題。其實(shí)它就像一個(gè)講話的形式：邏輯。相對(duì)來(lái)說(shuō)正確而正確。能通過(guò)任何考驗(yàn)而相對(duì)正確而正確。And in repeated argumentation, this sphere layer expression can demons trate any existing questions (including the location of the logic).The principle seems to begin with the origin and centers on the origin. If it is a plane in theory, and it could a

6、ctually be a plane as well,（ ） then an equation is achieved.Just like the argument in the fourth layer, it is the basic mathematical geometry method, and is very important. If the original strength at that moment coul d really keep up, that is the real original strength and it is called: This is the

7、 r eal original strength.As for the sphere theory, I once demonstrated it in Peking University sspace forum in 2006, also proved that “Zero”and many other relevant questions, and published it in 2009. It wasrecognized in some places. During May/Jun, 2014, under the assistance of wechat group of the

8、teachers office, I accomplished these ten questions on Baidu sspace.“ Cai Yu Jing Dao ”。This sphere expression is like a way of talking, the logic is relatively correct so it s correct. It can pass any tests and is relatively correct, so itscorrect.世界十大數(shù)學(xué)難題的題目：難題”之一：P （多項(xiàng)式算法）問(wèn)題對(duì)NP （非多項(xiàng)式算法）問(wèn)題 .難題”之二：

9、霍奇 (Hodge) 猜想 .難題”之三：龐加萊 (Poincare) 猜想 .難題”之四：黎曼 (Riemann) 假設(shè) .難題”之五：楊米爾斯 (Yang-Mills)存在性和質(zhì)量缺口 .難題”之六：納維葉斯托克斯 (Navier-Stokes) 方程的存在性與光滑性 .難題”之七：貝赫 (Birch) 和斯維訥通戴爾 (Swinnerton-Dyer) 猜想 .難題 之八：幾何尺規(guī)作圖問(wèn)題 .難題”之九：哥德巴赫猜想 .難題之十四色猜想 .The Perfect Sphere Layer expression can solve the following ten difficultma

10、thematic problems in the world.Problem 1: P (Polynomial Algorithm) Problem vs. NP (Non Polynomial Algorithm) ProblemProblem 2: Hodge ConjectureProblem 3: Poincare ConjectureProblem 4: Riemann HypothesisProblem 5: YangMills Existence and Mass gapProblem 6: The Existence and Smoothness of Navier - Sto

11、kes EquationsProblem 7: Birch and Swinnerton-Dyer ConjectureProblem 8: Geometric Ruler Gauge Construction ProblemProblem 9: Goldbach ConjectureProblem 10: Four Color Conjecture附加：此球體層式，能解決此世界十道數(shù)學(xué)難題。先講：物，物與數(shù)算起 （正圓球體層式），（正圓球體層式，在此以后就叫： 此圓球體或圓球體. ）原里，原來(lái)的里面。宇宙萬(wàn)物長(zhǎng)河中的任何什么，而物（相對(duì)來(lái)說(shuō)，也就是任何什么，而叫物。而在中文中，就能直接用上一

12、個(gè)名詞，“而物”。）。之，當(dāng)時(shí)什么物。之，當(dāng)時(shí)什么。This sphere layer expression can solve the world sten difficult problems in mathematics. Firstly, we talk about the objects, which can be counted by the numbers (Perfect Sphere LayerExpression). The Perfect Sphere Layer Expression, is hereinafter referred to as this Sphere

13、or the Sphere. The original interior is the original interior part.Anythi ng in the universe is the object at a certain time and being under certain circumstances. Erwu, relatively speaking, is Any Thing and being called the object. In Chinese, we use Erwu, a noun to describe it.零“ 0”，相對(duì)來(lái)說(shuō)沒(méi)有動(dòng)用任何，而任何

14、著。（此句的翻譯，應(yīng)是這樣理解。就像力學(xué)，相對(duì)來(lái)說(shuō)是平恒力學(xué)。似左邊等于右邊。為“ 0”時(shí)，則也構(gòu)成了平恒時(shí)。啥都沒(méi)有反應(yīng)！）（）似任何沒(méi)有動(dòng)用，沒(méi)有反應(yīng)！一旦有什么了。（時(shí)，先參照物一樣，參比一個(gè)數(shù)，或自然數(shù) N 中。）的似，（翻譯理解：似當(dāng)時(shí)怎么樣了。就似，就像。叫做“的似”。），動(dòng)用了任何于任何 （此句， 翻譯理解成： 任何為了任何。 “似電影片名中：勇敢的心。中， 說(shuō)：為了自由！”），任何本質(zhì)于任何本質(zhì)，而怎么了著的。（翻譯理解成：怎么樣了著的。相對(duì)來(lái)說(shuō)開(kāi)始反應(yīng)了一樣。開(kāi)始忙碌著。）。的似，動(dòng)用了，任何于任何，任何本質(zhì)于任何本質(zhì)的怎式，而怎么了著的。 （翻譯。這里時(shí)再闡明，也就是他還在

15、那講那方式。又一次地闡明著。）。而似（“而似”兩字。此時(shí)可翻譯理解成：就像似.，什么什么 .。）任何于任何，任何本質(zhì)于任何本質(zhì)的等式，而怎么了著。（翻譯此句理解。又，重新闡明了一下。相對(duì)來(lái)說(shuō)講完了。）。Zero 0, does not occupy anything relatively, and is being anything.（此處原來(lái)中文為不用翻譯的紅字，現(xiàn)在按綠字加上，其中“平恒”疑為“平衡”，按此翻譯，如有其它意思，請(qǐng)說(shuō)明）Just like mechanics, it is relatively a balance mechanics. It appears that the

16、left side equals the right side. When each side is 0, they also constitute a balance. There is not any response!It appears that it does not occupy anything, there s no response. At that time, we firstly referred to a number, or natural number N, just like a reference object. Once it does havesomethi

17、ng, just like using anything to anything, or any nature to any nature, it might begin to respond and do something. It appears that it uses anything to anything, any nature to any nature, so its what it is. It appears that it is an equation of anything to anything, any nature to nature, soit swhat it

18、 is.在這時(shí)講一下平衡。都知道數(shù)算中不是等式，就像力學(xué)中不平衡一樣，沒(méi)法做事。之，（時(shí)“之”字，時(shí)總結(jié)了一下。翻譯可理解成：則，怎么樣了.。）相對(duì)來(lái)說(shuō)，怎么樣的等式被動(dòng)用了，而怎么了著的。（翻譯理解時(shí)。這里，先好比一個(gè)講好了，總結(jié)一下。下面還有。）We will talk about the balance now. It is well known that nothing could be achieved without equations in calculations, just like the imbalance in mechanics. So, relatively spe

19、aking, when equations are used, then things are achieved.時(shí)，而怎么了著的。（先用自然數(shù)中（N ）參比用之，直接用上了?。┳匀粩?shù)N 中，第一個(gè)參數(shù)為“ 1”（翻譯理解：“參數(shù)”，參照物一樣，而叫它“參數(shù)”。）。時(shí)又怎么了，時(shí)又似任何于任何，任何本質(zhì)于任何本質(zhì)的怎樣的等式，而怎么了。時(shí)，第二個(gè)為 “ 2”。接著這樣， 又怎么了， 為第三個(gè)， 為“ 3”。4，5，6，7，8，9，.。N 中。時(shí)，動(dòng)用的都是怎樣的等式。到那圓球體層式時(shí)，每一層的原來(lái)的量，依舊都是相等的。而每一層只是怎么的等式來(lái)回而已。What was it like at th

20、at time?When natural number N was used as a reference, it was in directuse. Within the natural number N, the first parameter is 1.What was it like again at that time? It was like an equation of anything to anything, any nature to any nature, and it is what it is.At that time, the second was the numb

21、er “ 2” , then what, the third one is the number 3, and then 4,5,6,7,8,9 respectively among N.At that time, what kinds of equations were used? When it came to the sphere layers, the original amount of each layer was still equal. It is just an issue of certain equations for each layer.(此正圓球體層式似的構(gòu)成：)零

22、“ 0”，起數(shù)算起。（翻譯理解為：開(kāi)始數(shù)學(xué)中的演算了。）的似（翻譯理解：就似，就像 .。）從任何的角度方位來(lái)一樣。任何角度方位，那就是一空間方位角度。而先在平面上，“ 0”開(kāi)始起動(dòng)了任何于任何，任何本質(zhì)于任何本質(zhì)的等比等式，而怎么了的。N 中，時(shí)第一個(gè)參比之，為“1”。時(shí)為第一層似。又怎么了。接著“2”，為第二層。接著第三層為“3”。 4， 5， 6， 7 ,8 , 9， .，N 層中。而每一層中N 都平均平鋪開(kāi)。The Structure of the Perisphere Layer ExpressionFrom zero 0, the calculation starts. Just l

23、ike from the perspective of any orientation, any angleorientationis a spatial orientationangle. While firstly on the plane, 0 starts the geometricequations of anything to anything, and any nature to any nature, so that is the case.Among N, the one that came first to compare at that time was“1”.At th

24、at time it was the firstlayer. Then what sthe next? The number 2 is the second layer, then followed by the third layer3, and 4,5,6,7,8,9.in layer N.In each layer, N tiles averagely.時(shí)每一層都動(dòng)用怎樣的等比等式到下層。 相對(duì)來(lái)說(shuō)， 每一層的量都還相等的。 高度多一樣，為一個(gè)參數(shù) “1”。似時(shí)又一個(gè)平面，與他( 疑為“它” )一樣。并同樣的，原點(diǎn)與原點(diǎn)“0”，垂直相交。又有幾何法可構(gòu)成圓球體層式。為正圓球體層式。At

25、that time, each layer commands a certain geometric equation to the lower layer. Relatively speaking, the amount of each layer is still equal. It has the same height of a parameter 1.It is like that at that time there is another plane as well, the same as it, the original point and the original point

26、 0 also intersect vertically. Sphere layers may be constituted by geometry methods.（這里講一下，此圓球體層的厚度應(yīng)該這樣證明：）此圓球體層式中，到時(shí)候哪一層中的一個(gè)數(shù)碰到對(duì)應(yīng)的運(yùn)算。則同樣， 這個(gè)數(shù)作為一個(gè)原點(diǎn)運(yùn)作成此圓球體層式，（但，時(shí)這個(gè)數(shù)原來(lái)的本質(zhì)和量不變。）（這時(shí)也看出P=NP。）。就這樣證明出： 一定的條件下， 每次怎樣的碰到一個(gè)距離的數(shù)，就用一個(gè)此圓球體式參比之。時(shí)可看出，一定條件下為一個(gè)同等的參比值（參照物一樣，叫它“參比值”。下面的還有，叫它“參數(shù)”。理解一樣。），This Sphere Lay

27、er Expression is a perfect sphere layer Expression.We will talk about it here. The thickness of the sphere layer should be proved like this: in this sphere layer expression, a number in a certain layer will encounter the corresponding operationin due course. Similarly, this number, as an original po

28、int, could be operated to form this sphere layer expression. (However, at that time, the original nature and amount of this number remained unchanged). (At that time we could also see P = NP). So it is proved that under certain conditions, each time we encounter a distance number, we use this sphere

29、 expression as a reference. At that time we could see that, under certain conditions, it was the same reference value.只是在當(dāng)時(shí)怎么樣了。說(shuō)回來(lái)，時(shí)可為一個(gè)參數(shù)“1”。就是說(shuō)，在一定條件下，每隔一個(gè)距離。的似（就似）隔著一個(gè)“1”。那么這樣，在一定條件下，每一層圓球體層的高度先可以一樣。）The only concern is what it was like at that time.Again, at that time it could be a parameter of

30、1. That is to say, under certain conditions, it appears to be separated by a 1 for every other distance. Therefore, under certain conditions, the height of each sphere layer can firstly be the same.也再講一下每一層此圓球體層的厚度。也就是直接證明了一下： 時(shí)因?yàn)樵谝粋€(gè)一定條件下，時(shí)都是等比等式來(lái)回。時(shí)自然數(shù)中直接一個(gè)參數(shù)，可為“1”。時(shí)一個(gè)一定條件下的厚度多一樣。We also talk about

31、 the thickness of each sphere layer and it directly proves that: under a certain condition, at that time it came back and forth according to some geometric equations. At that time, a direct parameter among the natural numbers could be 1. At that time, the thickness was all the same under a certain c

32、ondition.相對(duì)來(lái)說(shuō)到時(shí)候的量似在每一層圓球?qū)永镆粯?，只是怎樣的平均平鋪著。也就是平均平鋪著。因?yàn)槊恳粚拥较乱粚?，都?dòng)用的等比等式來(lái)去。那么，每一層量多相等。來(lái)去只求有當(dāng)時(shí)情況。之 .當(dāng)時(shí)情況怎樣。之當(dāng)時(shí)可怎樣。 之，當(dāng)時(shí)怎樣。 (而 N 似，此圓球體還在運(yùn)作似。 )Relatively the amount in due course is like that in every sphere layer, the only concern is just how they tile, how they tile averagely. Because from each layer to

33、 the next, the geometric equations are applied to coming and going, so the amount of each layer is almost equal. Only the situationat that moment concerns us regarding coming and going. So what was the situation at the moment? How could they be that possibly? What was the case at that moment? (As fo

34、r N alike, the sphere appears to be still operating).相對(duì)來(lái)說(shuō)： 此圓球體內(nèi)，時(shí)又有哪個(gè)點(diǎn)要怎么樣的。時(shí)又可圓球體式起，而起怎樣對(duì)應(yīng)的數(shù)學(xué)運(yùn)算。： 又有能在一平面上的任意點(diǎn)的對(duì)應(yīng)情況。也就一平面上的怎樣對(duì)應(yīng)的情況。： 圓層表面 .可平面幾何。每一層的圓球?qū)芋w，可立體幾何。： P 與 NP 。此圓球體層式又可平面幾何來(lái)解釋。只是到時(shí)候又有怎樣當(dāng)時(shí)的情況而已。Relatively speaking, inside the sphere, at that time which point was the concern?At that time,

35、starting from the sphere expression, we could also begin a certain corresponding mathematical operation.There is a corresponding situation of an arbitrary point in the plane, that is to say, how they correspond in one plane.The sphere surface can be explained by plane geometry. Each layer of the sph

36、ere body can be explained by solid geometry.P and NP. This sphere layer expression can be explained by plane geometry, the only concern is what the situation is in due course.( 時(shí)可看出 P 與 NP ：此圓球體用幾何法證明回到空間中的一原點(diǎn)。空間中的一原點(diǎn)，又幾何法證成此圓球體式。則，原點(diǎn)等于原點(diǎn)，原里本質(zhì)=原里本質(zhì)： P=NP 。）解答這圓球體層型式，這十道題基本上一樣。而幾何法中的此圓球體式能與P=NP相似。原里的

37、本質(zhì)不變與此圓球體式的等比等式的運(yùn)算，時(shí)似那數(shù)學(xué)幾何法中的原點(diǎn)“O起”動(dòng)了數(shù)學(xué)運(yùn)算。(At that time we could see P and NP: when this sphere can be proved back to one origin of thespace by the geometry method, and one origin of the space can be proved to be the sphereexpression by the geometric method. Then, the origin is equal to the origin,

38、the original nature =the original nature: P = NP). （ Answer this layer of ball type.）These ten questions are basically thesame. This sphere expression could be similar to the P = NP in the geometric method.Theoriginal nature remains unchanged and the sphere expression can be explained by geometriceq

39、uation operations, just like the origin O starts the mathematicaloperations in geometricmethod of mathematics.小言一下：而且在反復(fù)的論證中，此圓球體層式 , 能演示的任何出來(lái)的題。（包括“那個(gè)邏輯”也給定位出來(lái)。）。原理好像就圍繞這個(gè)原點(diǎn)開(kāi)始的。有，理論為一個(gè)平面,實(shí)際也可為一個(gè)平面。（糾正：不能為一個(gè)平面。錯(cuò)誤了的。只是當(dāng)時(shí)的一個(gè)題目。）則 ,相等。似在第四層論證時(shí) ,即為數(shù)學(xué)幾何法的基本 ,而且又很重要。若 ,原里有當(dāng)時(shí)的那個(gè)力量真的跟上了。 那么才是真正的原里力量。 叫它為： “

40、這才是真正的原里力量?！薄nd in repeated argumentation, this sphere layer expression can demonstrate any existing questions (including the location of the logic).The principle seems to begin with the origin and centers on the origin. If it is a plane in theory,and it could actually be a plane as well, （ Correct

41、ion: can not be said to be a plane.Wrong. ） then an equation is achieved. Just like the argument in the fourth layer, it is the basic mathematical geometry method, and is very important. If the original strength at that moment could really keep up, that is the real original strength and it is called

42、: This is the real original strength.解答題目：（ 這十道題，能用此圓球體層式， 能解釋任何一道題。 的從下說(shuō)起。）之十 . 四色猜想 .解答：此圓球體內(nèi) .一層二層三層 . 多知道不行 . 平行時(shí)就碰到 .相同的邊的著色了。（不能碰到有相同邊的，而怎樣排列著整個(gè)平面 . ）而第四層 . 好像剛剛好：幾何法展開(kāi) ,似平面上。似幅度.頻率方向和自己的周期性，并組成怎樣的正方形格子平面。Theses ten problems can be explained by this sphere layer expression, each problem can be

43、 explained. We will begin with the last problem:Problem10. Four color conjecture: There are layer one, layer two, and layer three inside the sphere body. They encounter when they parallel and the same edges are colored. (Do not touch the same edges, how would you array the whole plane?). It seems th

44、at the fourth layer is just good enough.Expand it by a geometric method, just like the amplitude, frequency direction and periodicity in a plane, and see what square grid plane can be formed.時(shí)從左到右，和從上到下，似坐標(biāo)線X,Y ，方向上，再和自己的周期性，都是“4 ”。時(shí)參比之。（翻譯理解：參照物一樣，叫做“參比”。時(shí)參比之。）時(shí)，X,Y 對(duì)應(yīng)著這一個(gè)平面。 而且無(wú)論從哪個(gè) X,Y 方向都能以相同周期性

45、的對(duì)應(yīng)參比著整個(gè)平面。那么，“ 4”個(gè)著色時(shí)，剛好且故意能排列成不碰到相同邊的四色平面著色。From the left to the right and from the top to the bottom, it is like the coordinate lines X, Y direction and its periodicity is 4. At that time it could be explained by similar reference objects., theAt that time, X, Y corresponded to the plane and it c

46、orresponded to the entire plane according to the same periodicity no matter from which X, Y direction. So, when 4 is colored, it is just fine to intentionally array to form a four colors plane that won tmeet the same edges.第 5 層 .對(duì)自己的周期個(gè)數(shù) . 到時(shí)候出現(xiàn)似有角度的怎樣平行而不平行。就是多少數(shù)出來(lái)似的。后面的都似有怎樣角度的參比之。四色著色 , 似看清了有角度在

47、平行參比中有怎樣。 到時(shí)候平面上時(shí)的怎樣等比等式, 而參比了多少。（這一層時(shí)的推論在數(shù)學(xué)幾何法中很基本，而且，很重要。）接著 ,哥德巴赫 ,黎曼等函數(shù) ,都等比等式怎樣。先講幾何尺規(guī)作圖問(wèn)題。Layer 5. Cycle Numbers. There appears to be angles that parallel and not parallel. They are just like many numbers. Please refer to this for the following similar angles.Through four color shading, It app

48、ears to see clearly what the angles are like in the parallel reference, and what the geometric equation is while on the plane in due courses, and to what extend they correspond. This layer of inference is basic and very important in geometric methodof mathematics.The followings are Goldbach and Riem

49、ann functions, all geometric equations. Firstly we will talk about geometric ruler gauge construction problem之九 .哥德巴赫猜想 .解答：（如，對(duì)應(yīng)變量，一平面圖。如圖們）右小圖們.素?cái)?shù)，（先不說(shuō) A 到 B 了的線段。）一點(diǎn)到另一點(diǎn)，而且只被這兩端點(diǎn)整除。的似看作一個(gè)量表達(dá)。 （在這里， 時(shí)要提一下這樣的一個(gè)量的表達(dá)：就是從這一點(diǎn)到那一點(diǎn)的一條直線段。看成一個(gè)量的表達(dá)。若，在哥德巴赫猜想中，就表達(dá)了一個(gè)素?cái)?shù)。）Problem 9: Goldbach Conjecture: (e.g.

50、, the Corresponding Variables, a Planar Graph as shown in figures). Small figures to the right.A prime number, (regardless of the line segment from A to B), from one point to another, could only be divided exactly by the two end points. It could be regarded as a quantity expression.( Thisquantity ex

51、pression should be mentioned here: a straight line segment from this point to that point. It slike that a prime number was expressed in the Goldbach conjecture.在此圓球體中。圓球?qū)用嫔希簳r(shí) 6 時(shí)。（如圖），（對(duì)應(yīng)變量，一平面圖。如圖們）。似每個(gè)整個(gè)圓球體層面的面積。幾何法展開(kāi)，成一平面，時(shí) XY 。時(shí)有對(duì)應(yīng)量的 X 與 Y 的對(duì)應(yīng)互換。時(shí)又有相加時(shí)， X 與 Y 相對(duì)應(yīng)的兩個(gè)奇素?cái)?shù)的和，為偶性。是對(duì)的。In this sphere,

52、on the layers of the sphere:At that time, when it is greater than or equal to 6. (as shown in the figure), (the corresponding variables, a planar graph as shown in the figures). I t slike the area of each layer of the whole sphere. Expand it by a geometric method to become a plane, at that time it w

53、as XY. At that time it had an amount corresponding the X and Y exchange. At that time when they were added, the sum of two odd prime numbers corresponding the X and Y was an even. It is correct.而9時(shí)，每層圓球體層整層的立體的體積，也可以用怎樣對(duì)應(yīng)的三個(gè)奇素?cái)?shù)相乘表達(dá)。時(shí)也又有相加時(shí)， 三個(gè)奇素?cái)?shù)的和，為奇性。 也對(duì)的。（哥德巴赫猜想的表達(dá)，證明他的本質(zhì)，依舊一樣。是幾何法中，此圓球體式的原點(diǎn)，和此圓球體式。之，原里的本質(zhì)不變。)（這里的“原里”兩字，翻譯理解是：原來(lái)的里面。）（圖解的在下面。）When it is greater than or equal