Approximating irrational numbers
The next problem comes with what happens with the irrational powers. Suppose that we want to define, for instance, $10^{\sqrt{2}}$. In principle, the answer is simple enough. If we approximate the square root of $2$ to a certain number of decimal places, then the power is rational, and we can take the approximate root, using the above method, and get an approximation to $10^{\sqrt{2}}$. Then we may run it up a few more decimal places (it is again rational), take the appropriate root, this time a much higher root because there is 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 the work is quite difficult. How can we cope with this problem? In the computations of square roots, cube roots, and other small roots, there is an arithmetical process available by which we can get one decimal place after another. But the amount of labor needed to calculate irrational powers and the logarithms that go with them (the inverse problem) is so great that there is no simple arithmetical process we can use. Therefore tables have been built up which permit us to calculate these powers, 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 saving time; if we must raise some number to an irrational power, we can look it up rather than having to compute it. Of course, such a computation is just a technical 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 $x=10^{\sqrt{2}}$, but we also have the problem of solving $10^x = 2$, or $x = \log_{10} 2$. This is not a problem where we have to define a new kind of number for the result, it is merely a computational problem. The answer is simply an irrational number, an unending decimal, not a new kind of a number. Let us now discuss the problem of calculating solutions of such equations. The general idea is really very simple. If we could calculate $10^1$, and $10^{4/10}$, and $10^{1/100}$, and $10^{4/1000}$ and so on, and multiply them all together, we would get $10^{1.414\dots}$ or $10^{\sqrt{2}}$, and that is the general idea on which things work. But instead of calculating $10^{1/10}$ and so on, we shall calculate $10^{1/2}$, $10^{1/4}$, and so on. Before we start, we should explain why we make so much work with $10$, instead of some other number. Of course, we realize that logarithm tables are of great practical utility, quite aside from the mathematical problem of taking roots, since with any base at all, \begin{equation} \label{Eq:I:22:3} \log_b(ac)=\log_ba+\log_bc. \end{equation} We are all familiar with the fact that one can use this fact in a practical way to multiply numbers if we have a table of logarithms. The only question is, with what base $b$ shall we compute? It makes no difference what base is used; we can use the same principle all the time, and if we are using logarithms to any particular base, we can find logarithms to any other base merely by a change in scale, a multiplying 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 the logarithms of all the numbers to the base $b$. In other words, we can solve the equation $b^a = c$ for any $c$ because we have a table. The problem is to find the logarithm of the same number $c$ to some other base, let us say the base $x$. We would like to solve $x^{a'} = c$. It is easy to do, because we can always write $x = b^t$, which defines $t$, knowing $x$ and $b$. As a matter of fact, $t = \log_b x$. Then if we put that in and solve for $a'$, we see that $(b^t)^{a'} = b^{a't} = c$. 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 we multiply by a constant, and the constant is $1/\log_b x$. This permits us to choose a particular base, and for convenience we take the base $10$. (The question may arise as to whether there is any natural base, any base in which things are somehow simpler, and we shall try to find an answer to that later. At the moment we shall just use the base $10$.) 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. The powers of $10$ are given in the first column, and the result, $10^s$, is given in the third column. Thus $10^1 = 10$. 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 any number.1 Using 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 to take $10^{0.5}$, so we now know at least one logarithm, if we happen to need the 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 more information. So we take the square root again, and find $10^{1/4}$, 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 $10^{0.75}$, we can get it, because that is $10^{(0.5+0.25)}$; it is therefore the product of the second and third numbers. If we can get enough numbers in column $s$ to be able to make up almost any number, then by multiplying the proper things in column 3, we can get $10$ to any power; that is the plan. So we evaluate ten successive square roots of $10$, and that is the main work which is involved in the calculations. Why don’t we keep on going for more and more accuracy? Because we begin to notice something. When we raise $10$ to a very small power, we get $1$ plus a small amount. The reason for this is clear, because we are going to have to take the $1000$th power of $10^{1/1000}$ to get back to $10$, so we had better not start with too big a number; it has to be close to $1$. What we notice is that the small numbers that are added to $1$ begin to look as though we are merely dividing by $2$ each time; we see $1815$ becomes $903$, then $450$, $225$; so it is clear that, to an excellent approximation, if we take another root, we shall get $1.00112$ something, and rather than actually take all the square roots, we guess at the ultimate limit. When we take a small fraction $\Delta/1024$ as $\Delta$ approaches zero, what will the answer be? Of course it will be some number close to $1+0.0022511\,\Delta$. Not exactly $1+0.0022511\,\Delta$, however—we can get a better value by the following trick: we subtract the $1$, and then divide by the power $s$. This ought to correct all the excesses to the same value. We see that they are very closely equal. At the top of the table they are not equal, but as they come down, they get closer and closer to a constant value. What is the value? Again we look to see how the series is going, how it has changed with $s$. It changed by $211$, by $104$, by $53$, by $26$. These changes are obviously half of each other, very closely, as we go down. Therefore, if we kept going, the changes would be $13$, $7$, $3$, $2$ and $1$, more or less, or a total of $26$. Thus we have only $26$ more to go, and so we find that the true number is $2.3025$. (Actually, we shall later see that the exact number should be $2.3026$, but to keep it realistic, we shall not alter anything in the arithmetic.) From this table we can now calculate any power of $10$, by compounding the power out of $1024$ths. 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, and the numerical values are shown in Table 22–1 (columns 2 and 3). 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 the answer is going to be bigger than $1/4$, and less than $1/2$. Let us take the factor $10^{1/4}$ out; we divide $2$ by $1.778\dots$, and get $1.124\dots$, and so on, and now we know that we have taken away $0.250000$ from the logarithm. The number $1.124\dots$, is now the number whose logarithm we need. When we are finished we shall add back the $1/4$, or $256/1024$. Now we look in the table for the next number just below $1.124\dots$, 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, as follows:  \begin{equation*} 2 = (1.77828)(1.074607)(1.036633)(1.0090350)(1.000573). \end{equation*} 
 \begin{gather*} 2 = (1.77828)(1.074607)(1.036633)\;\times\\ (1.0090350)(1.000573). \end{gather*}     There was one factor $(1.000573)$ left over, naturally, which is beyond the range of our table. To get the logarithm of this factor, we use our result that $10^{\Delta/1024} \approx 1+ 2.3025 \Delta/1024$. We find $\Delta= 0.254$. 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 $\log_{10} 2 = 0.30103$, which happens to be right to $5$ figures! This is how logarithms were originally computed by Mr. Briggs of Halifax, in 1620. He said, “I computed successively $54$ square roots of $10$.” We know he really computed only the first $27$, because the rest of them can be obtained by this trick with $\Delta$. His work involved calculating the square root of $10$ twenty-seven times, which is not much more than the ten times we did; however, it was more work because he calculated to sixteen decimal places, and then reduced his answer to fourteen when he published it, so that there were no rounding errors. He made tables of logarithms to fourteen decimal places by this method, which is quite tedious. But all logarithm tables for three hundred years were borrowed from Mr. Briggs’ tables by reducing the number of decimal places. Only in modern times, with the WPA and computing machines, have new tables been independently computed. There are much more efficient methods of computing logarithms today, using certain series expansions.  In the above process, we discovered something rather interesting, and that is that for very small powers $\epsilon$ we can calculate $10^\epsilon$ easily; we have discovered that $10^\epsilon = 1+ 2.3025\epsilon$, by sheer numerical analysis. Of course this also means that $10^{n/2.3025} = 1+ n$ if $n$ is very small. Now logarithms to any other base are merely multiples of logarithms to the base $10$. The base $10$ was used only because we have $10$ fingers, and the arithmetic of it is easy, but if we ask for a mathematically natural base, one that has nothing to do with the number of fingers on human beings, we might try to change our scale of logarithms in some convenient and natural manner, and the method which people have chosen is to redefine the logarithms by multiplying all the logarithms to the base $10$ by $2.3025\dots$ This then corresponds to using some other base, and this is called the natural base, or base $e$. Note that $\log_e (1 + n) \approx n$, or $e^n \approx 1+ n$ as $n\to0$. It is easy enough to find out what $e$ is: $e = 10^{1/2.3025\dots}$ or $10^{0.434310\dots}$, an irrational power. Our table of the successive square roots of $10$ can be used to compute, not just logarithms, but also $10$ to any power, so let us use it to calculate this natural base $e$. For convenience we transform $0.434310\dots$ into $444.73/1024$. Now, $444.73$ is $256 + 128 + 32 + 16 + 8 + 4 + 0.73$. Therefore $e$, since it is an exponent of a sum, will be a product of the numbers  \begin{equation*} (1.77828)\!(1.33352)\!(1.074607)\!(1.036633)\!(1.018152)\! (1.009035)\!(1.001643) = 2.7184. \end{equation*} 
 \begin{align*} (1.&77828)\!(1.33352)\!(1.074607)\!(1.036633)\;\times\\ &(1.018152)\!(1.009035)\!(1.001643)= 2.7184. \end{align*}  (The only problem is the last one, which is $0.73$, and which is not in the table, but we know that if $\Delta$ is small enough, the answer is $1 + 0.0022486\,\Delta$.) When we multiply all these together, we get $2.7184$ (it should be $2.7183$, but it is good enough). The use of such tables, then, is the way in which irrational powers and the logarithms of irrational numbers are all calculated. That takes care of the irrationals.