Fig. 4
Yet I can not leave this subject without showing how we can make a very child understand some of the geometrical theorems that have acquired a bad reputation in the world of candidates for degrees, including even such as the pons asinorum of Pythagoras; the demonstration, that is, that if we construct the triangles B and C on the sides of a right-angled triangle, their sum will be equal to the square A constructed on the hypotenuse. The usual demonstration of this theorem is not very complicated, but there is something tiresome, artificial, and hard in it. The demonstration I propose is almost intuitive, and the reasoning of it is both simple and rigorous.
Suppose we take two equal squares, and, making equal lengths on the four sides of one of them, join the points so obtained as indicated in the first of the two figures (Figs. 5 and 6) so as to form four right-angled triangles, and then place four other squares in the corners of the original square. These right-angled triangles are of such sort that the sum of their sides is equal to the side of the square. This can be demonstrated, but it strikes the eyes without that. We see, too, that the interior figure is a square, and that it is constructed on the hypotenuse of the triangles in question.
| Fig. 5. | Fig. 6. |
It is easy to see in the other figure, which is formed after the same measures as its alternate, that the triangles 1, 2, 3, 4 can be arranged so as to occupy the positions 1', 2', 3', 4' in such way as to leave in the main square two smaller squares constructed on the sides of one of the right-angled triangles. It follows that the square A is equivalent to the sum of the squares B and C. The theorem thus becomes a kind of intuition, a thing evidently indisputable.
It is a curious fact that the origin of this demonstration is lost in the obscurity of the past; it probably goes back to thirty or forty centuries, at least, before the Christian era, and apparently to India. Bhascara, in his Bija Ganita, after tracing a figure, a simple combination of these two, says, "There you see it." I remark that such a demonstration, even if dressed with geometrical terms, assuming a character that conforms to existing ways of teaching, would be vastly superior, even in secondary schools, to the demonstrations of Legendre and others, which are much harder. The return to what was done very long ago in this case constitutes a great advance upon what we are doing now.
Fig. 7.
Having given our little one an initiation into the mysteries of arithmetic and geometry, we introduce him to algebra, a branch which passes in the majority of families as the hardest, most complicated, and most abstruse that can be imagined. I do not pretend that algebraic theories enter easily into the child's delicate brain; rather the contrary; but I declare that some ideas in algebra can be made comprehensible to children without fatigue. We can, for instance, make them understand, in the way of amusement and without great difficulty, the formula that gives the sum of the first numbers. We take a sheet of paper ruled in squares and shade the first square of the first line, then the first two squares of the second line, the first three of the third, etc. (Fig. 7). The whole number of squares shaded in this manner represents visibly the sum of the first whole numbers up to any one we may choose—to 7 in the figure. If we give this paper to the child and ask him to return it, he will very easily perceive that the figures formed by the white and the black squares are alike. The number sought for will therefore be equal to half the sum of the squares—that is, in the present example
1 + 2 + 3 + 4 + 5 + 6 + 7 = (7 X 8) : 2 = 28,