Now here is the idea. These relationships, or rules, are correct for integers,since they follow from the definitions of addition, multiplication, and raisingto a power. We are going to discuss whether or not we can broaden the classof objects which a , b , and c represent so that they will obey these same rules, although theprocesses for a+b , and so on, will not be definable in terms of the direct action ofadding 1 , for instance, or successive multiplications by integers.
现在的想法，是这样的，这些关系或规则，对于整数，是正确的，由于它们遵循加法、乘法、和次方的定义。我们要继续讨论的是，对于a , b , 和c所代表的对象，我们是否能够扩展其类型，这样，它们将遵守同样的规则，虽然，对于这些过程来说，它们的定义，与前面并不完全相同；例如，对于a+b，并不是直接地加1，对于乘法，也不一定是通过整数的连乘。{？}

So we have increased the range of objectsover which the rules work, but the meaning of the symbols is different.
规则，是对一定范围内的对象，其作用，现在，我们已经增加了这个范围，但是，符号的意义，有所不同。

One cannot say, for instance, that −2 times 5 really means to add 5 together successively −2 times. That means nothing. But nevertheless everything will workout all right according to the rules.
例如，人们不能说-2乘以5，确实就意味着，连续加5，-2次。这没有任何意义。但尽管如此，依据这些规则，所有事情，都能正常运转。

An interesting problem comes up in takingpowers. Suppose that we wish to discover what a(3−5) means. We know only that 3−5 is a solution of the problem, (3−5)+5=3 . Knowing that, we know that a(3−5)a5=a3. Therefore a(3−5)=a3/a5, by the definition of division. With a little more work, this can bereduced to 1/a2 . So we find that the negative powers are the reciprocals of thepositive powers, but 1/a2 is a meaningless symbol, because if a is a positive or negative integer, the square of it can be greater than 1, and we do not yet know what we mean by 1 divided by a number greater than 1 !
求次方的时候，一个有趣的问题来了。假设我们希望发现，a(3−5)，意味着什么。我们只知道，3-5，是问题(3−5)+5=3，的一个解。知道了这点，我们就可以知道a(3−5)a5=a3。因此，通过除法的定义，就可知道a(3−5)=a3/a5。稍加努力，它就可被简化为1/a2。于是，我们发现，负的次方，是正的次方的导数，但是，1/a2是一个无意义的符号，因为，如果a是一个正整数或负整数，那么，其平方，大于1，现在，我们还不知道，1除以一个大于1 的数，意味着什么。

Onward! The great plan is to continue theprocess of generalization; whenever we find another problem that we cannotsolve we extend our realm of numbers. Consider division: we cannot find anumber which is an integer, even a negative integer, which is equal to theresult of dividing 3 by 5 . But if we suppose that all fractional numbers also satisfy therules, then we can talk about multiplying and adding fractions, and everythingworks as well as it did before.
向前！伟大的计划，就是继续普遍化的过程；当我们发现另外一个我们无法解决的问题时，我们就扩展我们的数字领域。考虑除法，我们找不到一个整数，哪怕是一个负整数，让它等于3除以5的结果。但是，如果我们假设，所有分数，也都满足规则，那么，我们就可以谈论分数的乘和加了，这样，所有的事情，就都会像以前一样，正常运行。

Take another example of powers: whatis a3/5 ? We know only that (3/5)5=3 , since that was the definition of 3/5 . So we know also that (a(3/5))5= a(3/5)(5)= a3 , because this is one of the rules. Then by the definition of roots wefind that
再举一个次方的例子：什么a3/5呢？我们只知道(3/5)5=3，由于这就是3/5的定义。所以，我们就也知道了 (a(3/5))5=a(3/5)(5)= a3，因为，这是规则之一。因此，通过根的定义，我们发现。

In this way, then, we can define what wemean by putting fractions in the various symbols, by using the rules themselvesto help us determine the definition—it is not arbitrary. It is a remarkablefact that all the rules still work for positive and negative integers, as wellas for fractions!
因此，以这种方式，对于我们的意思，我们就能定义了，即通过把分数，放到不同的符号中，通过利用规则本身，来帮助我们决定定义—它不是任意的。这是一个值得注意的事实，即所有的规则，对于正整数和负整数，仍能正常运行，对分数也一样

We go on in the process of generalization.Are there any other equations we cannot solve? Yes, there are. For example, itis impossible to solve this equation: . It is impossible to find a number which is rational (a fraction)whose square is equal to 2 . It is very easy for us in modern days to answer this question. Weknow the decimal system, and so we have no difficulty in appreciating the meaningof an unending decimal as a type of approximation to the square root of 2 . Historically, this idea presented great difficulty to the Greeks. Toreally define precisely what is meant here requires that we add somesubstance of continuity and ordering, and it is, in fact, quite the most difficultstep in the processes of generalization just at this point. It was made,formally and rigorously, by Dedekind. However, without worrying about the mathematicalrigor of the thing, it is quite easy to understand that what we mean is that weare going to find a whole sequence of approximate fractions, perfect fractions(because any decimal, when stopped somewhere, is of course rational), whichjust keeps on going, getting closer and closer to the desired result. That isgood enough for what we wish to discuss, and it permits us to involve ourselvesin irrational numbers, and to calculate things like the square root of 2 to any accuracy that we desire, with enough work.
我们继续这个普遍化的过程。是否有方程，是我们解不了的呢？是的，有。例如，方程，就不可解。要找到一个有理数（分数），其平方，等于2，是不可能的。现代社会，对于我们，回答这个问题，非常容易。我们知道了小数系统，所以，一个无穷的小数，可作为对2的平方根的近似，对于这种想法，我们赞赏，毫无困难。但在历史上，这个想法，曾给希腊人，造成很大困难。要在这里，真正精确地定义，它究竟意味着什么，要求我们增加一些连续和排列组合的材料{？}，事实上，在这点上，这个工作，也是普遍化的过程中最困难的一步。这个工作，是由狄德金（Dedekind），正式且严谨地完成的。然而，对于事情的数学上的严谨性，我们无需担忧，我们的意思是，我们要找一个近似分数的完整序列，完美的分数（因为任何小数，无论停在何处，当然就是有理数），会一直走下去，越来越接近所期待的结果；理解我们的意思，非常容易。对于我们希望讨论的事情，这已经足够好了，且它允许我们，与无理数打交道，在计算诸如2的平方根这种事情时，无论我们期待什么样的精度，只要工作足够，就都可以达到。

22–4Approximating irrational numbers 22-4无理数的近似
The next problem comes with what happens withthe irrational powers. Suppose that we want to define, for instance, . In principle, the answer is simple enough. If we approximate thesquare root of 2 to a certain number of decimal places, then the power is rational, andwe can take the approximate root, using the above method, and get an approximationto . Then we may run it up a few more decimal places (it is againrational), take the appropriate root, this time a much higher root because thereis a much bigger denominator in the fraction, and get a better approximation.Of course we are going to get some enormously high roots involved here, and thework is quite difficult. How can we cope with this problem?
下一个问题，随着对无理数求次方，而到来。比如，假设我们想定义。从原理上讲，答案足够简单。对于2的平方根，我们取近似，如果我们取到一定的小数位，那么，次方就是有理数，我们可以使用上面的方法，取近似的根，得到的一个近似值。那么，我们可以再多取几位小数（它又是有理数），取近似的根，这次，是一个更高级的根，因为，分数中的分母，更大了，从而，得到一个更好的近似。当然，在这里，我们可以取巨高的根，但那样，工作将会非常困难。我们该如何对付这个问题呢？

In the computations of square roots, cuberoots, and other small roots, there is an arithmetical process available bywhich we can get one decimal place after another. But the amount of laborneeded to calculate irrational powers and the logarithms that go with them (theinverse problem) is so great that there is no simple arithmetical process wecan use. Therefore tables have been built up which permit us to calculate thesepowers, and these are called the tables of logarithms, or the tables of powers,depending on which way the table is set up. It is merely a question of savingtime; if we must raise some number to an irrational power, we can look it uprather than having to compute it. Of course, such a computation is just atechnical problem, but it is an interesting one, and of great historical value.In the first place, not only do we have the problem of solving x=, but we also have the problem of solving 10x=2 , or x=log102 . This is not a problem where we have to define a new kind of numberfor the result, it is merely a computational problem. The answer is simply anirrational number, an unending decimal, not a new kind of a number.
在计算平方根、立方根、和其他小根时，有一个算术过程，可以得到，通过它，我们可以得到这些根的小数位。但是，计算无理数的次方，及随之的算法（逆向问题），所需的劳动量，非常巨大，以至于，没有简单的算法过程，可供我们使用。因此，就建立了一些表，可以让我们计算这些次方，这些表，被称为算法表、或次方表，这依赖于表，是如何建立的。这只是一个节省时间的问题；如果我们必须求某数的无理数的次方，我们可以查表，而不用去计算。当然，这种计算，只是一个技术问题，但它是一个有趣的问题，有着巨大的历史价值。最初，我们不仅有求x=的问题，而且，我们还有求10x=2, 或x=log102的问题。这种问题，并非是要为结果，定义一个新的数据种类，而只是一个计算问题。答案很简单，就是一个无理数，它是无穷小数，而不是一个新的数据种类。

Let us now discuss the problem ofcalculating solutions of such equations. The general idea is really verysimple. If we could calculate 101 , and 104/10 , and 101/100 , and 104/1000 and so on, and multiply them all together, we would get 101.414…or , and that is the general idea on which things work. But instead ofcalculating 101/10 and so on, we shall calculate 101/2 , 101/4 , and so on. Before we start, we should explain why we make so muchwork with 10 , instead of some other number. Of course, we realize that logarithmtables are of great practical utility, quite aside from the mathematicalproblem of taking roots, since with any base at all,
现在，让我们讨论，计算这种方程的解的问题。一般的想法，确实非常简单。如果我们可以计算101 、104/10 、101/100 、和104/1000，等等，那么，把它们全乘起来，我们就可以得到101.414… 或 ，事情就是这样运行的。但是，我们不是计算101/10, 而是要计算 101/2 , 101/4等。在我们开始之前，我们要解释一下，为什么我们必须用10，来做这种工作，而不是用其他的数。当然，我们意识到，对数表，除了数学的取根问题外，还有着巨大的用途，由于，对于任何的底，都有：{？}
logb(ac)=logba+logbc. (22.3)

We are all familiar with the fact that one can use this fact in apractical way to multiply numbers if we have a table of logarithms. The onlyquestion is, with what base b shall we compute? It makes no difference what base is used; we can usethe same principle all the time, and if we are using logarithms to any particularbase, we can find logarithms to any other base merely by a change in scale, amultiplying factor. If we multiply Eq. (22.3)by 61 , it is just as true, and if we had a table of logs with a base b, and somebody else multiplied all of our table by 61 , there would be no essential difference. Suppose that we know thelogarithms of all the numbers to the base b . In other words, we can solve the equation ba=cfor any c because we have a table. The problem is to find the logarithm of thesame number c to some other base, let us say the base x . We would like to solve xa′=c . It is easy to do, because we can always write x=bt, which defines t , knowing x and b . As a matter of fact, t=logbx . Then if we put that in and solve for a′ , we see that (bt)a′=ba′t=c. In other words, ta′ is the logarithm of c in base b . Thus a′=a/t . Thus logs to base x are just 1/t , which is a constant, times the logs to the base, b . Therefore any log table is equivalent to any other log table if wemultiply by a constant, and the constant is 1/logbx . This permits us to choose a particular base, and for convenience wetake the base 10 . (The question may arise as to whether there is any natural base, anybase in which things are somehow simpler, and we shall try to find an answer tothat later. At the moment we shall just use the base 10 .)
下面事实，我们都熟悉，即如果我们有算法表的话，那么，就可以用一种实践的方式，把若干数，乘起来{？}。唯一的问题就是，我们用什么样的底b，来做计算呢？用什么做底，并无不同；我们可以一直使用同样的原理，如果我们对数中，使用某个具体的底，那么，对于任何其他的底，我们都可以通过改变尺度、即一个可乘的因子，来找到算法。如果我们用61，乘以方程（22.3），它还一样是真的，如果我们有个对数表，以b为底，而其它的人，用61，乘以整个表，那么，将没有任何本质上的不同。假设，对于所有的数，我们已经知道了它们以b为底的对数。换句话说，对于任何c，我们都可以解方程ba=c，因为，我们已经有表了。问题是，对于同一个数c，要找出，以其他的数、比如说x，为底的对数。我们希望解xa′=c 。这很容易做，因为，如果知道了x和b的话，我们总是可以写出x=bt ，它定义了t。事实上，t=logbx。因此，如果我们把它带入，求解a′，我们看到(bt)a′=ba′t=c。换句话说，ta′，就是c的以b为底的对数，这样a′=a/t。这样，以x为底的对数，就是常数1/t，乘以以b为底的对数。因此，任何对数表，都等价于另外一个对数表乘以一个常数，此常数就是1/logbx。这就允许我们，可以选一个具体的底，为了方便，我们选10（可能会产生一个问题，即是否还有其他自然的底，在这种底中，事情在某种程度上，会更简单些，对此，稍后我们将尝试去寻找答案。现在，我们将只以10为底。）

Table 22–1 表22-1 10的连续平方根

Now let us see how to calculate logarithms.We begin by computing successive square roots of 10 , by cut and try. The results are shown in Table 22–1. Thepowers of 10 are given in the first column, and the result, 10s , is given in the third column. Thus 101=10 . The one-half power of 10 we can easily work out, because that is the square root of 10 , and there is a known, simple process for taking square roots of anynumber.1Using this process, we find the first square root to be 3.16228 . What good is that? It already tells us something, it tells us how totake 100.5 , so we now know at least one logarithm, if we happen to needthe logarithm of 3.16228 , we know the answer is close to 0.50000 . But we must do a little bit better than that; we clearly need moreinformation. So we take the square root again, and find 101/4 , which is 1.77828 . Now we have the logarithm of more numbers than we had before, 1.250 is the logarithm of 17.78 and, incidentally, if it happens that somebody asks for 100.75, we can get it, because that is 10(0.5+0.25) ; it is therefore the product of the second and third numbers. If wecan get enough numbers in column s to be able to make up almost any number, then by multiplying theproper things in column 3, we can get 10 to any power; that is the plan. So we evaluate ten successive squareroots of 10 , and that is the main work which is involved in the calculations.
现在,让我们看，如何计算对数。我们用实验的方式，从计算10的连续的平方根开始。结果见表22-1。10的次方数，在第一列，结果10s在第三列。这样101=10。10的二分之一次方，我们很容易得出，因为，它就是10的平方根。有一个已知的、简单的过程，可以用来取任何数的平方根。（脚注1）利用这个过程，我们发现，第一个平方根是3.16228。这有什么好处呢？它已经告诉了我们一些事情，它告诉我们，如何得到100.5 ，所以，现在我们至少知道了一个对数，如果我们碰巧需要 3.16228的对数，那么，我们就知道，答案接近于0.50000。但是，我们应该做的比这更好一些；很明显，我们需要更多信息。于是，我们再取平方根，发现101/4，是 1.77828。现在，我们比以前，有了更多数的对数，顺便说一下，如果碰巧有人问，100.75是多少的话，我们的可以得到它，因为，它就是10(0.5+0.25)；因此，它就是第二个和第三个数的积。如果在列s中，我们可以得到足够的数{？}，那么，我们可以造出任何数，因此，通过从第三列中，取出合适的数，让其相乘，我们就可以得到10的任何次方；这就是计划。所以，我们就估算了10的十个连续的平方根，这就是本计算中的主要工作。

Let us now actually calculate a logarithm,because the process we shall use is where logarithm tables actually come from.The procedure is shown in Table 22–2, andthe numerical values are shown in Table 22–1(columns 2 and 3).
现在，让我们实际计算一个对数，因为我们要用的过程，就是计算对数表的过程。程序见表22-2，数值见表22-1（列2和列3）。
Table 22–2 表22-2 对数log102的计算

Suppose we want the logarithm of 2 . That is, we want to know to what power we must raise 10 to get 2 . Can we raise 10 to the 1/2 power? No; that is too big. In other words, we can see that theanswer is going to be bigger than 1/4 , and less than 1/2 . Let us take the factor 101/4 out; we divide 2 by 1.778… , and get 1.124… , and so on, and now we know that we have taken away 0.250000 from the logarithm. The number 1.124… , is now the number whose logarithm we need. When we are finished weshall add back the 1/4 , or 256/1024 . Now we look in the table for the next number just below 1.124… , and that is 1.074607 . We therefore divide by 1.074607 and get 1.046598 . From that we discover that 2 can be made up of a product of numbers that are in Table 22–1, asfollows:
假设我们想得到2的对数。也就是说，我们想知道，10的几次方等于2。我们可以求10的1/2次方吗？不行；这个太大。换句话说，我们可以看到，答案将会比1/4大，而比1/2小。让我们把因子101/4取出，我们让2，除,1.778…，得到1.124…，等等，现在，我们知道，我们已经从对数中，把0.250000拿掉了。我们现在要求的，就是1.124…的对数。当我们完成之后，我们会把1/4、或256/1024，再加回来。现在，我们在表中查看，刚刚比1.124…小的数，它就是1.074607。因此，我们除以1.074607，得到1.046598。如此，我们发现，2可以是表22-1中的若干数之积，即：
2=(1.77828)(1.074607)(1.036633)(1.0090350)(1.000573).
There was one factor (1.000573) left over, naturally, which is beyond the range of our table. To getthe logarithm of this factor, we use our result that 10Δ/1024≈1+2.3025Δ/1024. We find Δ=0.254 . Therefore our answer is 10 to the following power: (256+32+16+4+0.254)/1024 . Adding those together, we get 308.254/1024 . Dividing, we get 0.30103 , so we know that the log102=0.30103 , which happens to be right to 5 figures!
有一个因子(1.000573)，剩下了，它不在表的范围之内。要得到这个因子的对数，我们用我们的结果，即10Δ/1024≈1+2.3025Δ/1024。我们发现Δ=0.254。因此，我们的答案就是，求10的(256+32+16+4+0.254)/1024次方。把这些加起来，我们得到308.254/1024。再除，得到0.30103，于是，我们知道log102=0.30103，刚好是5位！