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第六百一十章 埃尔德什格雷厄姆问题(数论)(1 / 2)

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公元前1650年左右的古埃及数学典籍《莱因德数学纸草书》,其中记录了古埃及人如何将有理数表示为单位分数之和。

这里有{2,3,7,12,15,18,21,29,32,36} 10个数字组成的一个数集,我们可以选择其中的2、3、12、18、36,就能得到1/2+1/3+1/12+1/18+1/36=1。

单位分数就是分子是1的分数,或者也可以说是正整数的倒数,它们是当时古埃及数字系统中唯一一类分数,他们需要用单位分数来表示其他更复杂的分数,比如将3/4写作1/2和1/4的和。

到了20世纪70年代,有关这类分数的问题再次引起了一些数学家的兴趣。当时,数学家埃尔德什(Paul Erd?s)和格雷厄姆(Ronald Graham)在探索想要设计出不满足条件的整数集有多难,也就是说,一个整数集中不能有任何子集,其倒数之和等于1。

如果A是N的子集,A具有正密度,那么存在有限的S是A的子集,使得其中数的倒数和为1。在此,数集A是自然数集的子集,无论你怎么数下去,都存在一种非零的概率,会遇到集合A中的一个数字,那么A就具有正密度。

猜想提出约半个世纪后,牛津大学数学家Thomas Bloom证明了它。

举个简单的例子,A是一个包含所有大于1的奇数的集合,它属于自然数集的子集,并满足正密度的条件,因为无论你数到10亿还是100亿,也一定会遇到奇数。然后,我们可以在A中找到有限子集S ={3,5,7,9,11,33,35,45,55,77,105},而所有这些数的倒数相加恰好等于1。

这理解起来并没有那么困难,但证明它显然就变成另一回事了。那就变成了一个大得多、复杂得多的问题。对不少数学家来说,似乎找不到什么显而易见的数学工具来解决它。

数学家Ernie Croot,他解决了所谓的埃尔德什-格雷厄姆问题的着色版本。

这是一种更弱的证明。可以这么理解,在着色版本中,整数被随机地分类,指定放到不同颜色的桶中。猜想预测,无论这种分类中用到了多少个桶,至少会有一个桶包含一个倒数之和等于1的整数子集。

Croot这篇发表于2003年的论文引入了来自调和分析的强大的新方法,那是一个与微积分密切相关的数学分支。

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