*Mathematics is one of the most important cultural components of every modern society. Its influence on other cultural elements has been so fundamental and wide-spread as to warrant the statement that her “most modern” ways of life would hardly have been possibly without mathematics. Appeal to such obvious examples as electronics radio, television, computing machines and space travel, to substantiate this statement is unnecessary : the elementary art of calculating is evidence enough. Imagine trying to get through three day without using numbers in some fashion or other*!**– R.L. Wilder**

Welcome to the blog **Math1089 – Mathematics for All!**.

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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 *a*^{2} + 2*ab* + *b*^{2}.

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 *a*^{2} and that of *GFID* is *b*^{2}. 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} **=

*a*

^{2}+

*ab*+

*ab*+

*b*

^{2}

**=**, as expected.

*a*^{2}+ 2*ab*+*b*^{2}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 *a*^{2} − 2*ab* + *b*^{2}.

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

*a*^{2} = (*a *− *b*)^{2} + *b*^{2} + (*a *− *b*)*b* + (*a *− *b*)*b*, so that (*a *− *b*)^{2} = *a*^{2} − *b*^{2} − 2(*a *− *b*)*b.*

After simplification, **( a − b)^{2} = a^{2} − 2ab + b^{2}**, 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 *a*^{2} − *b*^{2}. We will show that, this is equal to (*a* + *b*)(*a* – *b*).

Observe that, area of *ABCD* − area of *YZCU* = area of *AWUD* − area of *WBZY*. This means that *a*^{2} − *b*^{2} = *a*(*a *− *b*) + (*a *− *b*)*b*.

After simplification, ** a^{2} − b^{2} = (a + b)(a − b)**, as wanted.

Next, consider the product, dealing with three variables, given by (*a *+ *b *+ *c*)^{2} = *a*^{2} + *b*^{2} + *c*^{2} + 2*ab *+ 2*ac *+ 2*bc*. 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 *a*^{2} + *b*^{2} + *c*^{2} + 2*ab *+ 2*ac *+ 2*bc*.

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} = *a*^{2} + *b*^{2} + *c*^{2} + *ab *+ *ac *+ *bc *+ *ab *+ *ac *+ *bc*

This means that **( a + b + c)^{2} = a^{2} + b^{2} + c^{2} + 2ab + 2ac + 2b**

*, as required.*

**c**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 *x*^{2} + *x(a + b)* + *b*^{2}.

Note that, the area of the bigger rectangle is equal to the sum of the area of a square and three rectangles, which equals *x*^{2} + *ax* + *bx* + *ab.*

It follows that, **( x + a)(x + b) = x^{2} + (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, b* ≤ *x* without any loss of generality. The figure left with is a rectangle with area (*x* − *a*)(*x* − *b*) (**blue** one). We now consider the geometrical representation of the product (*x* − *a*)(*x* − *b*). We will show that, this is equal to *x*^{2} − *x*(*a + b*) + *b*^{2}.

Observe that, area of *PQRS* = sum of areas of the three rectangles (**red**, **green** and **blue**). This means that *x*^{2} = *ax *+ *b*(*x *− *a*) + (*x *− *a*)(*x *− *b*).

After simplification, **( x − a)(x − b) = x^{2} – 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**.

Your suggestions are eagerly and respectfully welcome! See you soon with a new mathematics blog that you and I call **“****Math1089 – Mathematics for All!**“.