## Geometric Proofs of Few Algebraic Identities – Part 1

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Algebra is full of identities which can be proved in many ways. One idea is to prove them using geometry. As the name suggests, we will prove several algebraic identities using geometric concepts.

We all know that, the area of a square is the square of the length of its side. Consider a square ABCD of side a + b, as shown in the figure. Area of this square is (a + b)2. We will show that, this is equal to a2 + 2ab + b2.

However, the square ABCD can be divided in four quadrilaterals as shown in the right side figure. Here, AEFG and FHCI are squares of sides a and b respectively. Area of the square AEFG is a2 and that of GFID is b2. Areas of the rectangles EBHF and GFID are each ab. We have

area of ABCD = area of AEFG + area of EBHF + area of FHCI + area of GFID

This means that, (a + b)2 = a2 + ab + ab + b2 = a2 + 2ab + b2, as expected.

Now we give a geometric representation of the square of the difference of two numbers a and b, where we can assume b a without any loss of generality. Now the problem is to find the area of the square of sides a − b, as shown in the figure. To find this, we need to consider the square AXYV. Since the length of the sides is a b, area of this square is (a b)2. We will show that, this is equal to a2 − 2ab + b2.

Observe that the area of the square of side ABCD is equal to the sum of the areas of the square of sides AXYV and YZCW, respectively, plus the area of two equal rectangles XBZY and VYWD. This means that

a2 = (a b)2 + b2 + (a b)b + (a b)b, so that (a b)2 = a2b2 − 2(a b)b.

After simplification, (a b)2 = a2 − 2ab + b2, as required.

Now we give a geometric representation of the difference of the squares of two numbers a and b, where we can assume b a without any loss of generality. The problem now is to find the area of the shaded portion of the following figure, which is same as the difference of the areas ABCD and YZCU. This is same as a2b2. We will show that, this is equal to (a + b)(ab).

Observe that, area of ABCD − area of YZCU = area of AWUD − area of WBZY. This means that a2b2 = a(a b) + (a b)b.

After simplification, a2b2 = (a + b)(a b), as wanted.

Next, consider the product, dealing with three variables, given by (a + b + c)2 = a2 + b2 + c2 + 2ab + 2ac + 2bc. To establish the equality, consider a square of sides a + b + c. Since the length of the sides is a + b + c, the area of the square is (a + b + c)2. We will show that, this is equal to a2 + b2 + c2 + 2ab + 2ac + 2bc.

Clearly, area of the square with side length a + b + c and the sum of the areas of the nine rectangles in which the square is partitioned, are same. It follows that

(a + b + c)2 = a2 + b2 + c2 + ab + ac + bc + ab + ac + bc

This means that (a + b + c)2 = a2 + b2 + c2 + 2ab + 2ac + 2bc, as required.

We now consider the geometrical representation of the product (x + a)(x + b). To begin with, consider a rectangle of sides x + a and x + b. Area of this rectangle is (x + a)(x + b) as shown below. We will show that, this is equal to x2 + x(a + b) + b2.

Note that, the area of the bigger rectangle is equal to the sum of the area of a square and three rectangles, which equals x2 + ax + bx + ab.

It follows that, (x + a)(x + b) = x2 + (a + b)x + ab, as wanted.

To start with, consider a square of side x and cut a rectangle of width a (red one) from it. From the remaining portion, cut another rectangle of width b (green one), where we can assume a, bx without any loss of generality. The figure left with is a rectangle with area (xa)(xb) (blue one). We now consider the geometrical representation of the product (xa)(xb). We will show that, this is equal to x2x(a + b) + b2.

Observe that, area of PQRS = sum of areas of the three rectangles (red, green and blue). This means that x2 = ax + b(x a) + (x a)(x b).

After simplification, (x a)(x b) = x2 – x(a + b) + ab, as required.

Consider a rectangle of length b + c and width a. Area of this rectangle is (b + c) a. Next, the length b + c is divided into two parts of lengths b and c, so that we obtain two rectangles (blue and green). Geometrically we will show that, this is equal to ba + ca.

It is evident from the figure that,

area of the bigger rectangle = area of blue rectangle + area of green rectangle.

This means that (b + c) a = ba + ca, as required.

That’s all in this blog. Don’t get disappointed, as the name suggests, there will be a second part of this article, will come under the heading Geometric Proofs of Few Algebraic Identities – Part 2.

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