## Category Archives: IITJEE Foundation mathematics

### A cute illustrative problem on basics of inequalities: IITJEE Foundation Maths

Problem:

Graph the functions $f(x)=\frac{3}{x-1}$ and $\frac{2}{x+1}$ together to identify the values of x for which we can get $\frac{3}{x-1} < \frac{2}{x+1}$

Also, confirm your findings in the above graph algebraically/analytically.

Also, then find the points of intersection of these graphs using a TI graphing calculator and also algebraically/analytically.

Solution:

Just a suggestion, it helps to have TI graphing calculator, a bit pricey, but …it might help a lot…If you wish, you can grab one from the following:

http://www.amazon.in/TEXAS-GRAPHIC-CALCULATOR-NSPIRE-CX/dp/B00A49F98U/ref=sr_1_fkmr0_1?s=electronics&ie=UTF8&qid=1493101690&sr=1-1-fkmr0&keywords=TI+graphing+calculator+Non+CAS

Of course, there might be excellent freeware software packages that plot graphs, but I am strongly biased: I like the TI graphing calculator toy very much!! I go to the extent of claiming that it is better for a kid to have a nice graphing calculator like a TI than a smartphone! 🙂 🙂 🙂

Later on, I hope to show how a detailed study of graphs helps right from high school level. In fact, an immortal Russian mathematician, I. M. Gel’fand had written a book titled “Functions and Graphs” for high-school children. You can check if it is available in Amazon India or Flipkart or Infibeam, etc.

Oh, one more thing: I love this topic of inequalities because of several reasons. Just now, one of my students, Darpan Gajra asked me certain questions about inequalities when he tried to solve the above problem —- I think his questions were good, and I hope this little explanation that I gave him helps many of other students also.

Explanation:

Consider the given inequality: $\frac{3}{x-1} < \frac{2}{x+1}$.

You might be tempted to solve this very fast in the following way:

Just cross-multiply: So, we get $3(x+1)<2(x-1)$, which in turn means $3x+3<2x-2$, that is, $x<-5$. This is the answer by “fluke”!!!

In fact, the problem requires a detailed solution as follows:

Firstly, let us see what is wrong with the above fast solution:

Consider any inequality $a. Now, we know that only if $x>0$, then $ax. But, if $x<0$, then $ax>bx$. Now, in the present question, if you simply cross-multiply, you are not clearing assuming whether $(x-1)>0$ or $(x-1)<0$; $(x+1)>0$ or $(x+1)<0$.

Also, one simple yet, I call it a golden rule is: even though it is not mentioned explicitly in the question, when you solve the question, immediately write the big restrictions: $x \neq 1$, $x \neq -1$ as these values make the denominators of the two fractions zero. Develop this good habit.

So, now, coming to the solution:

$\frac{3}{x-1} < \frac{2}{x+1}$, where $x \neq \pm 1$

$\frac{3}{x-1} - \frac{2}{x+1}<0$

$\frac{3x+3-2x+2}{x^{2}-1}<0$

$\frac{x+5}{x^{2}-1}<0$

Case I: $(x^{2}-1)>0$: Multiplying both sides of the above fraction inequality by $(x^{2}-1)$ gives us:

$x+5<0$, that is, $x<-5$, but also $(x^{2}-1)>0$ is a restriction.  So, it means that $x<-5$ AND $(x-1)(x+1)>0$.

Subcase Ia: $(x+1)>0$ and $(x-1)>0$, which gives, $x>-1$ and $x>1$, which together imply that $x>1$, but we need $x<-5$ also as a restriction/assumption. So, in this subcase Ia, solution set is empty.

Subcase Ib: $x+1<0$ and $x-1<0$, which in turn imply, $x<-1$ and $x<1$, that is, $x<-1$, but we need $x<-5$ also. Hence, in subcase Ib: $x<-5$ and $x<-1$, we get $x<-5$ as the solution set.

Case II: $(x^{2}-1)<0$: Multiplying both sides of $\frac{x+5}{x^{2}-1}<0$ by $(x^{2}-1)$, we get:

$(x+5)>0$, hence, $x>-5$. But, we also need the restriction/assumption: $(x^{2}-1)<0$, which implies that $(x+1)(x-1)<0$.

Subcase IIa: $(x+1)>0$ and $(x-1)<0$, that is, $-1

So, in subcase IIa, we have $(x>-5) \bigcap (-1, that is, $(-1.

Subcase IIb: $(x+1)<0$, and $(x-1)>0$, that is, $x<-1$ and $x>1$. But, this itself is an empty set. Hence, in subcase IIb, the solution set is empty.

Now, the final solution is $\{ case I\}$ OR $\{ case II\}$, that is, $\{ case I\} \bigcup \{ case II\}$, that is, $\{ x<-5\}\bigcup {-1

Homework:

a) Graph the function $f(x)=\frac{x}{2}$ and $g(x)=1+\frac{4}{x}$ together to identify the values of x for which $\frac{x}{2}>1+\frac{4}{x}$

(b) Confirm your findings in (a) algebraically.

More later,

Nalin Pithwa.

### Are kids no longer learning multiplication tables in school?

Today, I suddenly found some “echoes” or “resemblances” to my views. Kindly let me re-blog, or share the opinions about this from today’s EETimes, authored by Max Maxfield, Designline Editor. The URL is pasted below:

Thanks Max and EETimes !

From Nalin Pithwa.

### System of Equations — IITJEE Foundation Maths

Some problems are presented below as warm-up exercises:

Question 1:

Solve : $x+y = 15$ and $xy=36$

Question 2:

Solve: $x-y=12$ and $xy=85$

Question 3:

Solve: $x^{2}+y^{2}=74$ and $xy=35$

Question 4:

Solve : $x^{2}+y^{2}=185$ and $x+y=17$

Some other standard set of systems of equations and techniques for solving them are presented below:

Any pair of equations of the form

$x^{2} \pm pxy + y^{2}=a^{2}$….Equation I

$x \pm y = b$…Equation II

where p is any numerical quantity, can be reduced to one of the cases in the warm-up exercises; for, by squaring II and combining with I, an equation to find xy is obtained, the solution can then be completed by the aid of equation II.

Example 1:

Solve $x^{3}-y^{3}=999$…Equation 1

and $x-y=3$…Equation 2.

By division, $x^{2}+xy+y^{2}=333$…Equation 3.

From Equation 2, $x^{2} - 2xy + y^{2} = 9$;

by subtraction, $3xy=324$ and $xy=108$…Equation 4.

From Equation (2) and (4), $x=12, y=9$ or $x=9, y = -12$

Example 2:

Solve: $x^{4}+x^{2}y^{2}+y^{4}=2613$…Equation I

and $x^{2}+xy+y^{2}=67$…Equation 2

Dividing (1) by (2), $x^{2}-xy+y^{2}=39$…Equation 3.

From Equation (2) and (3), by addition, $x^{2}+y^{2}=33$; by subtraction $xy=14$, hence, we get the following: $x = \pm 7, x = \pm 2$, or $y = \pm 2, y = \pm 7$

Example 3:

Solve: $\frac{1}{x} - \frac{1}{y} = \frac{1}{3}$…Equation (1)

and $\frac{1}{x^{2}} + \frac{1}{y^{2}} = \frac{5}{9}$…Equation (2)

From (1), by squaring, $\frac{1}{x^{2}} - \frac{2}{xy} + \frac{1}{y^{2}} = \frac{1}{9}$,

and by subtraction, $\frac{2}{xy} = \frac{4}{9}$

adding to (2), $\frac{1}{x^{2}} + \frac{2}{xy} + \frac{1}{y^{2}} = 1$

and hence, $\frac{1}{x} + \frac{1}{y} = \pm 1$.

Combining with (1), $\frac{1}{x} = \frac{2}{3}$, or $-\frac{1}{3}$; $\frac{1}{y} = - \frac{1}{3}$, or $-\frac{2}{3}$; so, the solutions are: $x = \frac{3}{2}, -3$ and $y =3, -\frac{3}{2}$.

The following method of solution may always be used when the equations are of the same degree and homogeneous.

Example:

Solve $x^{2}+ xy + 2y^{2} = 74$ …Equation 1

and $2x^{2} + 2xy + y^{2} = 73$…Equation 2.

Put $y = mx$, and substitute in both equations. Thus,

$x^{2}(1+m+2m^{2}) =74$…Equation 3.

$x^{2}(2 + 2m + m^{2}) =73$…Equation 4.

By division, $\frac{1+m+2m^{2}}{2+2m+m^{2}} = \frac{74}{73}$

$73 + 73m + 146m^{2} = 148 + 148m + 74m^{2}$

Hence, $72m^{2} -75m - 75=0$; or $24m^{2} - 25m - 25 = 0$; hence, $(8m + 5)(3m - 5) = 0$; so $m = -\frac{5}{8}$ or $\frac{5}{3}$.

Case (i):

Take $m= -\frac{5}{8}$ and substitute in either (3) or (4).

From (3): $x^{2}(1- \frac{5}{8} + \frac{50}{64}) = 74$

$x^{2} = 64$, so $x = \pm 8$; so $y = mx = - \frac{5}{8}x = \mp 5$.

Case (ii):

Take $m = \frac{5}{3}$, then from (3) $x^{2}(1 + \frac{5}{3} + \frac{50}{9}) = 74$, so we get $x^{2} = 9$, so $x = \pm 3$; and hence, $y = mx = \frac{5}{3}x = \pm 5$.

When one of the equations is of first degree and the other of a higher degree, we may from the simple equation, find the value of one of the unknowns in terms of the other, and substitute in the second equation.

Example:

Solve $3x - 4y = 5$…Equation (1)

and $3x^{2} - xy - 3y^{2} = 21$…Equation (2)

From (1), we have $x = \frac{5+4y}{3}$

and substituting in (2), $3\frac{(5+4y)^{2}}{9} - \frac{y(5+4y)}{3} - 3y^{2}= 21$

$75 + 120y +48y^{2} - 15y - 12y^{2} - 27y^{2} = 189$

$9y^{2} + 105y -114 = 0$

$3y^{2} + 35y - 38 = 0$

$(y-1)(3y+31) = 0$

Hence, $y = 1, -\frac{38}{3}$, and $x = 3, -\frac{137}{9}$.

The examples we have presented above will be sufficient as a general explanation of the methods to be employed; but, in some cases, special artifices are necessary.

Example:

Solve: $x^{2} + 4xy + 3x = 40 - 6y -4y^{2}$…Equation (1)

and $2xy - x^{2} = 3$…Equation (2)

From (1) and (2), we have $x^{2} + 4xy + 4y^{2} + 3x + 6y = 40$;

that is, $(x+2y)^{2} + 3(x+2y) - 40 = 0$;

or, $((x+2y)+8)((x+2y)-5) = 0$, hence, $x+2y = -8, 5$

Case:

Combining $x+2y =5$ with (2), we obtain $2x^{2} - 5x + 3=0$, so $x = 1, \frac{3}{2}$; and by substituting in $x + 2y = 5$, we get $y = 2, -\frac{7}{4}$.

Case:

Combining  $x+2y = -8$ with (2), we obtain $2x^{2} + 8x + 3 = 0$; hence, $x = \frac{-4 \pm \sqrt{10}}{2}$, and $y = \frac{-12 \mp \sqrt{10}}{4}$

Example:

Solve: $x^{2}y^{2} - 6x = 34 - 3y$…Equation (1)

and $3xy + y = 2((9+x)$…Equation (2)

From (1), we have $x^{2}y^{2} -6x + 3y = 34$;

From (2), we have $9xy - 6x + 3y = 54$;

By subtraction, $x^{2}y^{2} - 9xy + 20 = 0$; that is, $(xy - 5)(xy -4) = 0$; so $xy = 5, 4$.

Case:

Substituting $xy = 5$ in (2), gives $y -2x = 3$. From these equations, we obtain $x =1$ or $-\frac{5}{2}$; $y = 5$, or $-2$.

Case:

Substituting $xy =4$ in (2), gives $y - 2x = 6$. From these equations, we obtain $x = \frac{-3 \pm \sqrt{17}}{2}$, and $y = 3 \pm \sqrt{17}$.

Hope you enjoyed it,

Nalin Pithwa.

### Some problems for IITJEE Foundation mathematics

Problem 1.

A cistern can be filled by two pipes in 33$\frac{1}{3}$; if the larger pipe takes 15 minutes less than the smaller to fill the cistern, find in what time it will be filled by each pipe singly.

Problem 2.

By rowing half the distance and walking the other half, a man can travel 24 km. on a river in 5 hours with the stream, and in 7 hours against the stream. If there were no current, the journey would take 5$\frac{2}{3}$ hours; find the rate of his walking, and rowing and the rate of the stream.

Problem 3.

Factorize: $2a^{2}x^{2}-2(3b-4c)(b-c)y^{2}+abxy$

Problem 4:

Find the fourth root of $81x^{4}-216x^{3}y+216x^{2}y^{2}-96x^{3}y+16y^{4}$

Problem 5:

Find the sixth root of

$(x^{3}-\frac{1}{x^{3}})^{2}-6(x-\frac{1}{x})(x^{3}-\frac{1}{x^{3}})+9(x-\frac{1}{x})$

More interesting stuff on IITJEE foundation maths later,

Nalin Pithwa

### A little note : IITJEE foundation maths

Example:

Solve $2(x^{2}-6)=3(x-4)$

Solution:

We have $2x^{2}-12=3x-12$

that is, $2x^{2}=3x$ call this as Equation (1)

Transposing, $2x^{2}-3x=0$

$x(2x-3)=0$

Hence, $x=0$, or $2x-3=0$

Thus, the roots are 0 and $\frac{3}{2}$

Note:

In equation (1), above, we might have divided both sides by x and obtained the simple equation $2x=3$, hence, $x=\frac{3}{2}$, which is one of the solutions of the given equation. But, the student must be particularly careful to notice that whenever an x, or a factor containing x, is removed by division from every term of an equation, it must not be neglected, since the equation is satisfied by $x=0$, which is therefore, one of the roots.

More on IITJEE foundation maths later,

Nalin Pithwa

### Cyclic Fractions for IITJEE foundation maths

Consider the expression

$\frac{1}{(a-b)(a-c)}+\frac{1}{(b-c)(b-a)}+\frac{1}{(c-a)(c-b)}$

Here, in finding the LCM of the denominators, it must be observed that there are not six different compound factors to be considered; for, three of them differ from the other three only in sign.

Thus,

$(a-c) = -(c-a)$

$(b-a) = -(a-b)$

$(c-b) = -(b-c)$

Hence, replacing the second factor in each denominator by its equivalent, we may write the expression in the form

$-\frac{1}{(a-b)(c-b)}-\frac{1}{(b-c)(a-b)}-\frac{1}{(c-a)(b-c)}$ call this expression 1

Now, the LCM is $(b-c)(c-a)(a-b)$

and the expression is $\frac{-(b-c)-(c-a)-(a-b)}{(b-c)(c-a)(a-b)}=0$.,

Some Remarks:

There is a peculiarity in the arrangement of this example, which is desirable to notice. In the expression 1, the letters occur in what is known as cyclic order; that is, b follows a, a follows c, c follows b. Thus, if a, b, c are arranged round the circumference of a circle, if we may start from any letter and move round in the direction of  the arrows, the other letters follow in cyclic  order; namely, abc, bca, cab.

The observance of this principle is especially important in a large class of examples in which the differences of three letters are involved. Thus, we are observing cyclic order when we write $b-c$, $c-a$, $a-b$, whereas we are violating order by the use of arrangements such as $b-c$, $a-c$, $a-b$, etc. It will always be found that the work is rendered shorter and easier by following cyclic order from the beginning, and adhering to it throughout the question.

Homework:

(1) Find the value of $\frac{a}{(a-b)(a-c)} + \frac{b}{(b-c)(b-a)} + \frac{c}{(c-a)(c-b)}$

2) Find the value of $\frac{b}{(a-b)(a-c)} + \frac{c}{(b-c)(b-a)} + \frac{a}{(c-a)(c-b)}$

3) Find the value of $\frac{z}{(x-y)(x-z)} + \frac{x}{(y-z)(y-x)} + \frac{y}{(z-x)(z-y)}$

4) Find the value of $\frac{y+z}{(x-y)(x-z)} + \frac{z+x}{(y-z)(y-x)} + \frac{x+y}{(z-x)(z-y)}$

5) Find the value of $\frac{b-c}{(a-b)(a-c)} + \frac{c-a}{(b-c)(b-a)} + \frac{a-b}{(c-a)(c-b)}$

More later,

Nalin Pithwa

### Some properties of continuous functions

Theorem:

If $g,f : [a,b] \rightarrow \Re$ are continuous functions and c is a constant, then

a) $f+g$ is a continuous functions.

b) $f-g$ is a continuous functions.

c) cf is a continuous function.

d) fg is a continuous function.

Proof:

We shall only prove statement d. Choose and fix any $\varepsilon >0$. Since is continuous at $x_{0}$, we have that for the positive number $\frac{\varepsilon}{(2|g(x_{0})|+1}$ there exists a $\delta_{1}>0$ such that

$|f(x)-f(x_{0}|< \frac{\varepsilon}{2(|g(x_{0}|+1)}$ whenever $|x-x_{0}|<\delta_{1}$

Since $||f(x)|-|f(x_{0})|| \leq |f(x)-f(x_{0})|$, we conclude that

$|f(x)|<|f(x_{0})|+\frac{\varepsilon}{2(|g(x_{0})|+1)}$, whenever $|x-s_{0}|<\delta_{1}$. Let $|f(x_{0})|+\varepsilon 2(|g(x_{0})|+1)=M$. Also, since g is continuous at $x_{0}$, for the positive number $\frac{\varepsilon}{2M}$, there is a $\delta_{2}>0$ such that $|g(x)-g(x_{0})|< \frac{\varepsilon}{2M}$ whenever $|x-x_{0}|<\delta_{2}$. Put $\delta=min(\delta_{1},\delta_{2})$. Then, whenever $|x-x_{0}|<\delta$, we have

$|f(x)g(x)-f(x_{0})g(x_{0})|$ equals

$|f(x)g(x)-f(x)g(x_{0})+f(x)g(x_{0})-f(x_{0})g(x_{0})|$

$\leq |f(x)||g(x)-g(x_{0})|+|g(x_{0})(f(x)-f(x_{0}))|$

which equals

$|f(x)||g(x)-g(x_{0})|+|g(x_{0})||f(x)-f(x_{0})|$

$< M.\frac{\varepsilon}{2M}+|g(x_{0})|.\frac{\varepsilon}{2(|g(x_{0}|+1)}$

which equals $\frac{\varepsilon}{2}+\frac{\varepsilon}{2}=\varepsilon$

Observe that we have not claimed that the quotient of two continuous functions is continuous. The problem is obvious: $\frac{f(x)}{g(x)}$ cannot have any meaning at x for which $g(x)=0$. So, the question would be, if $g(x) \neq 0$ for every $x \in [a,b]$, is the function $h:[a,b] \rightarrow \Re$, defined by $h(x)=\frac{f(x)}{g(x)}$, continuous? The answer is yes. For a proof, we need a preliminary result.

Lemma.

if $g:[a,b] \rightarrow \Re$ is continuous and $g(x_{0}) \neq 0$, then there is an $m > 0$ and $\delta >0$ such that if $x_{0}-\delta, then $|g(x)|>m$.

Proof.

Let $|g(x_{0})|=2m$. Now, $m>0$. By continuity of g, there is a $\delta>0$ such that

$|g(x)-g(x_{0})| for $x_{0}-\delta

But, $|g(x)-g(x_{0})| \geq ||g(x)|-|g(x_{0})||$ and hence, $-m<|g(x)|-|g(x_{0})|<m$, giving us

$m=|g(x_{0})|-m<|g(x)|$ for $x_{0}-\delta. Hence, the proof.

The lemma says that if a continuous function does not vanish at a point, then there is an interval containing it in which it does not vanish at any point.

Theorem.

If $f,g :[a,b] \rightarrow \Re$ are continuous and $g(x) \neq 0$ for all x, then $h:[a,b] \rightarrow \Re$ defined by $h(x)=\frac{f(x)}{g(x)}$ is continuous.

The proof of the above theorem using the lemma above is left as an exercise.

Examples.

a) $f:\Re \rightarrow \Re$ defined by $f(x)=a_{0}$ for all $x \in \Re$, where $a_{0}$ is continuous.

b) $f:\Re \rightarrow \Re$ defined by $f(x)=x$ is continuous.

c) $g:\Re \rightarrow \Re$ defined by $g(x)=x^{2}$ is a continuous function because $g(x)=f(x)f(x)$, where $f(x)=x$. Since f is continuous by (b), g must be continuous.

d) $h:\Re \rightarrow \Re$ by $h(x)=x^{n}$, n being a positive integer, is continuous by repeated application of the above reasoning.

e) $p: \Re \rightarrow \Re$ defined by $p(x)=a_{0}+a_{1}x+\ldots +a_{n}x^{n}$, where $a_{0}, a_{1}, \ldots , a_{n}$ is also continuous. This is because of the fact that if

$f_{1}, f_{2}, f_{3} \ldots, f_{n}:\Re \rightarrow \Re$ are defined by $f_{1}(x)=x$, $f_{2}=x^{2}$, …, $f_{n}=x^{n}$, then $a_{1}f_{1}, a_{2}f_{2}, \ldots a_{n}f_{n}$ are also continuous functions. Hence,

$a_{0}+a_{1}f_{1}+ \ldots +a_{n}f_{n}=p$ is also a continuous function as the sum of continuous functions is a continuous function. Thus, we have shown that a polynomial is a continuous function.

f) Let p and q be polynomials. Let $\alpha_{1}, \alpha_{2}, \ldots, \alpha_{n} \in \Re$ be such that $q(\alpha_{1})=q(\alpha_{2})=\ldots=q(\alpha_{n})$  and $q(\alpha) \neq 0$ for $\alpha \neq \alpha_{1}, \alpha \neq \alpha_{2}, \ldots, \alpha neq \alpha_{n}$.

Now, let $D = \Re - \{ \alpha_{1}, \alpha_{2}, \ldots , \alpha_{n}\}$.

Then, $h:D \rightarrow \Re$ defined by $h(x)=\frac{p(x)}{q(x)}$ is a continuous function. What we have said is that a rational function which is defined everywhere except on the finite set of zeroes of the denominator is continuous.

g) $f:\Re \rightarrow \Re$ defined by $f(x)=\sin{x}$ is continuous everywhere. Indeed, $f(x)-f(x_{0})=\sin{x}-\sin{x_{0}}=2\sin{\frac{x-x_{0}}{2}}\cos{\frac{x+x_{0}}{2}}$. Therefore,

$|f(x)-f(x_{0})|=2|\sin{\frac{(x-x_{0})}{2}}| |\cos{\frac{(x+x_{0})}{2}}|\leq |x-x_{0}|$ (because $|\sin{x}| \leq |x|$, where x is measured in radians)

h) $f:\Re \rightarrow \Re$ defined by $f(x)=\cos{x}$ is continuous since

$|f(x)-f(x_{0})|=|\cos{x}-\cos{x_{0}}|=2|\sin{\frac{(x_{0}-x)}{2}}\sin{\frac{x+x_{0}}{2}}| \leq |x-x_{0}|$

i) $f:\Re - \{ (2n+1)\frac{\pi}{2}: n \in \mathbf{Z}\} \rightarrow \Re$ defined by $f(x)=\tan{x}$ is continuous. We had to omit numbers like $\ldots, \frac{-3\pi}{2}, \frac{-\pi}{2}, \frac{\pi}{2}, \frac{3\pi}{2}, \ldots$ from the domain of f as $\tan{x}$ cannot be defined for these values of x.

j) $f:\Re_{+} \rightarrow \Re$ defined by $f(x)=x^{1/n}$ is a continuous function. Indeed,

$f(x)-f(a)=x^{1/n}-a^{1/m}$ which equals

$\frac{(x-a)}{x^{\frac{n-1}{n}}+x^{\frac{n-2}{n}}\frac{1}{a^{n}}+\ldots +a^{\frac{n-1}{n}} }$

Choose $|x-a|<|a/2|$ to start with, so that $|a/2|<|x|<(3/2)|a|$. Thus,

$|x^{\frac{n-1}{n}}+x^{\frac{n-2}{n}}a^{1/n}+\ldots+a^{\frac{n-1}{n}}|>|a|^{\frac{n-1}{n}} \times ((1/2)^{\frac{n-1}{n}}+(1/2)^{\frac{n-2}{n}}+\ldots+1)$

Given an $\varepsilon >0$, let

$\delta=min\{\frac{|a|}{2}, \varepsilon \times |a|^{\frac{n-1}{n}} \times \left( (1/2)^{\frac{n-1}{n}}+\ldots+1 \right)\}$.

Then, for $|x-a|<\delta$, we have

$|f(x)-f(a)|=\frac{|x-a|}{|x^{\frac{n-1}{n}}+x^{\frac{n-2}{n}} \times a^{1/n}+\ldots+a^{\frac{n-1}{n}}|}< \varepsilon$.

It can be shown that f defined by $f(x)=x^{r}$ is also a continuous function for every real $r \in \Re$.

k) Consider the function $f:\Re \rightarrow \Re$ defined by $f(x)=a^{x}$. Is f a continuous function? This is left as an exercise. (Hint: It will suffice to prove continuity at $x=0$. This would follow from $\lim_{m \rightarrow \infty}a^{1/m}$).

k) Suppose $f:\Re \rightarrow \Re$ is defined by $f(x)=1/x$, if $x \neq 0$ and $f(0)=0$. We can see that f is not continuous at 0 as $f(x)$ changes abruptly when x goes over from negative to positive values.

More later,

Nalin Pithwa

### Composition of functions

Suppose A, B and C are sets and $f:A \rightarrow B$ and $g:B \rightarrow C$ are functions. We define a function

$h: A \rightarrow C$ by

$h(a)=g(f(a))$ for every $a \in A$.

It is easily seen that h is a well-defined function, as $f(a) \in B$ for every $a \in A$ and $g(b) \in C$ for every $b \in B$. The function h is called the composition of the functions g and f and is denoted by $g \circ f$.

Examples.

a) Suppose that in a forest, carnivorous animals sustain themselves by feeding only on herbivorous animals and the nutrition level of a herbivorous animal depends on the vegetation around the animal. So the nutrition level of a carnivorous animal ultimately depends on the vegetation around the population of herbivorous animals it feeds on. Thus, if V is the density of vegetation around the herbivorous animals n is the level of nutrition of the herbivorous animal (for simplicity measured by its weight, though there are often more parameters depicting the level of nutrition of an animal), $n: V \rightarrow \Re$ is a function. Similarly, if c is the level of nutrition of a carnivorous animal, then c is a function depending on the level of nutrition of herbivorous animals it feeds on:

$c: \Re \rightarrow \Re$

Thus, $c \circ n: V \rightarrow \Re$

is the level of nutrition of the carnivorous animal ultimately depending on the density of vegetation.

b) Take for example the force experienced by a moving charged particle in a magnetic field which is varying in time. We know that the force on the charged particle depends on the strength of the magnetic field in which it moves. Again, as the magnetic field strength varies with time, the force experienced is ultimately a function of time and position.

c) Suppose there is a lamp  in the room. The intensity of illumination at point in the room depends on the illuminating power of the lamp. The illuminating power of the lamp again depends on the voltage of the electricity supply which makes the lamp glow. So, ultimately, the intensity of illumination depends on the voltage of the power supply.

Exercise.

Give five more examples of composition of functions.

More later,

Nalin Pithwa

### Basic Set Theory: Functions

Among the kinds of relations we have been talking about, there are some which are very useful in mathematics. A relation $f \subset A \times B$ is called a function from A to B if

i) domain of $f=A$

ii) $(a,b) \in f$ and $(a,b^{'}) \in f \Longrightarrow b=b^{'}$.

Thus a function from A to B is a relation whose domain is the whole of A and which is not one-to-many. The set B is called the codomain or range of the function f.

If $f \subset A \times B$ is a function from A to B, then we write $f: A \rightarrow B$ and say that f is a function from A to B. We also say that f maps A into B and often a function is called a map or mapping. If

$(a,b) \in f$, we write

$b=f(a)$.

We call b the value of the function f at a or image of a under f. Observe that there can be no ambiguity in writing f(a) because it is impossible that $(a,b) \in f$ and $(a,b^{'}) \in f$ and $b \neq b^{'}$. So by definition of a function $f: A \rightarrow B$, we have for every $a \in A$ a unique $f(a) \in B$. Thus, a function would be completely determined if we knew $f(a) \in B$ for every $a \in A$. That is why sometimes a function is defined as a “rule” which associates to every element $a \in A$ a unique element $f(a)$ of B. We are obliged to put the word rule within quotation marks for the simple reason that the rule is often elusive. In fact, the aim in many situations is to “discover” the rule.

Examples

a) Suppose we are observing the position of a particle moving in a straight line. We know that at every instant of time the particle has a unique position on the line. If we agree that every point of the line corresponds to a real number with some convenient point on it decided to represent the real number zero, then the position of the particle at the time t is represented by the real number $x(t)$. We observe that our with our ordinary notion of time and position, we do not expect a particle to occupy two different positions at the same time (nothing prevents, though,the particle from occupying the same position at two different times. Indeed, if the particle is at rest for a certain length of time, then it would have the same position for the entire length of time.) Suppose every time is represented by a real number, zero being a certain time deemed as the present time, then all future times are represented by positive real numbers and past times by negative real numbers. Here, a real number is used as an intuitive object, say, as points on a line, postponing a mathematical discussion of real numbers later. Let $\Re$ be the set of real numbers, then the position of the particle moving in a straight line is, in fact, a function

$x: \Re \rightarrow \Re$

which associates to every time t, the position $x(t)$.

b) Look at a railway time table. Below a particular train there are times recorded in a column. It records the times of arrival and departure of a train at certain stations. We may look upon this as a function whose domain consists of disjoint intervals of time and range consists of certain cities where the train stops. Thus, for every time in the mentioned interval we have a city where the train is at that time. The table does not say anything about the train’s position at a time after its departure and before its arrival at the next. This does not disqualify it to be a function if we take the domain to be the intervals of time depicting the times of arrival and departure at certain stations.

c) Consider the distance of a particle, falling freely under gravity, measured from the point from where the particle started its fall. if we record the time since it began to fall, we know that the distance $x(t)$ at a time is given by the formula

$x(t)=(1/2)gt^{2}$

(assuming, of course that the gravity does not vary and there is no air resistance).

Here we have in fact a “rule” which tells us where the particle would be at any time. This certainly represents a function. However, we are often not so lucky to have a neat formula like this for other functions.

d) The incoming news on a particular day that tells us the temperature of the four metros recorded at 5.30 am on that day. Here, we have a definite temperature for every such metro at 5.30am of that day. So it is in fact a function whose domain consists of the set of four metros and range the real numbers quantifying temperature.

e) Look at your school time table. What does it record on a particular day? it records the subjects to be taught at different times of the day. A particular student has a particular subject being taught during a particular period,as a student is not expected to be in two different classes at the same time. So, for each student, the time table is a function whose domain is the set of periods and the range the set of subjects.

f) When we want to mail a letter enclosed in an envelope, we usually go to the post master with it to tell us the denomination of the stamp to be affixed. The post master weighs the letter and tells us the postage according to the weight. Here, we have a definite postage for a definite weight, we don’t have different rates for the same weight (for same kind of mail like registered or ordinary). The rate chart, with the post master, for a particular kind of mail, truly, is a function whose domain is the set of positive real numbers representing the weights of the mail and the range again consists of positive real numbers representing the postage. We write the chart as a function f such that

$p-f(a)$

meaning p is the postage to mail a letter of weight w.

g) We know that the solubility of a salt varies with temperature. That is to say that a given salt has a definite solubility at a definite temperature. So the relation which associates to every temperature the definite solubility of a salt is in fact a function.

h) Consider the population of a community of biological species at different times. If $N(t)$ is the population of the community at time t, then surely

$N: \Re \rightarrow \Re$

represents a function.

i) If we measure the voltage of an AC power supply over a period of time, we observe that it is different at different times. At every instant of time, it has a voltage Sometimes, it is positive, sometimes it is negative and sometimes zero. In fact, for ordinary domestic supply in India, the voltage varies between 330 volts and -330 volts. But this change from 330 volts to -330 volts occurs within an interval of $1/100$th of a second. (The ordinary DC voltmeters of your lab won’t be able to record this quick change in voltage). In fact, if $V(t)$ is the volrage at time t, we have

$V(t)$=E_{0}\sin{(\omega t + \phi)}

where $\omega=2\pi/50$ and $E_{0}=330$ and $\phi$ is a constant known as the phase.

j) Let S be a sphere resting on a horizontal plane H. Let n be the north pole and s be the southpole, where the sphere is touching the plane. Now let us join any other point $p \in S$ to n and extend it to meet the plane at some point q. Now to every point $p \in S - \{ n \}$, we have a unique point q on H. This defines a function

$f: S - \{ n \} \rightarrow H$. This function is called the stereographic projection of the sphere on the plane.

k) The trivial map or function $id: A \rightarrow A$ which sends every element of the set A to itself is called the identity map.

Look around, see and think…are there any functions that come to your notice? Let me know…

More later,

Nalin Pithwa

### Basic Set Theory: Relations

Among the various kinds of relations, we are familiar with, father, mother, son, brother, sister, husband, wife, less than, equal to, parallel to, etc., are a few examples. We would like to fix our ideas as to what exactly is meant by a relation. Let us see how we use the terms in common parlance. We say Dasaratha is the father of Rama. Similarly, Babar is the father of Humayun and Janaka is the father of Sita. The above statements like Dasaratha is the father or Rama are but examples fatherhood. But, what do we exactly mean by fatherhood? We may attempt to define fatherhood as the collection of pairs like Dasaratha and Rama, Babar and Humayun, Janaka and Sita, etc. But in our usage “Dasaratha is the father of Rama” we cannot interchange the positions of Dasaratha and Rama. Indeed the statement “Rama is the father of Dasaratha” does not convey the same meaning as the statement “Dasaratha is the father of Rama”. Again, in the statement “Dasaratha is the father of Rama”, if we drop  the word Rama, then too it won’t mean anything. For the statement about fatherhood, what is important is a pair in a definite order: (Dasartha, Rama). Thus, a certain collection of ordered pairs stands for a definite relation, as in this case the ordered pairs

(Dasaratha, Rama), (Babar, Humayun), (Janaka, Sita)

are examples of relation of fatherhood. So fatherhood would stand for a certain collection of ordered pairs of human beings. Similarly, the relation of brotherhood would stand for a certain other collection of ordered pairs. So would sisterhood, etc. In each case, they are different collections of ordered pairs. Now we proceed to define the term relation formally. For that we need the notion of a cartesian product.

Cartesian Product.

For sets A and B, the set of all ordered pairs formed by the first element from A and the second element from B, is called the Cartesian product of A and B. It is denoted by $A \times B$. We have

$A \times B= \{ (a,b): a \in A \hspace{0.1in} \textup{and} \hspace{0.1in} b \in B\}$

Relation.

By a relation or correspondence between the elements of a set A and those of a set B, we mean a subset f of the Cartesian product $A \times B$. Thus, f is a relation between set A and set B if

$f \subset A \times B$

If $(a,b) \in f$, then we say that a is related to b through f. We write this as afb. Thus, if f is the set of all human beings and M is the set of all male human beings, then fatherhood is a subset of $M \times H$, whereas brotherhood is another subset of $M \times H$. Motherhood is a subset of $W \times H$, where

$W = \{ w: w is a woman\}$. So if $f \subset M \times H$ is the fatherhood relation, $(a,b) \in f$ means a is the father of b. Similarly, if $g \in M \times H$ is the brotherhood relation, then $(a,b) \in G$ means a is a brother of b. (Observe that a is a brother of b does not imply b is a brother of a since b could be a sister of a. )

Definition:

For sets A and B if $f \subset A \times B$, then $\{ x: (x,u) \in f \hspace{0.1in} \textup{for some} \hspace{0.1in} u \in B\}$ is called the domain of the relation f, and $\{ y: (x,y) \in f \hspace{0.1in textup{for some} \hspace{0.1in} x \in A}\}$ is called the range of the relation f.

It is quite obvious that the domain of f, above, is a subset of A and the range. a subset of B.

In the above example of fatherhood, its domain cannot contain a woman. Nor did it contain all men. Indeed many men are not fathers.

Like father, mother, brother, sister we also have the relation less than, equal to, greater than, similar to, congruent to, perpendicular to, etc.

For the set $A=\{ 1,2,3,4\}$, let $f \subset A \times A$ be defined by

$f=\{ (1,2), (1,3), (1,4), (2,3), (2,4), (3,4)\}$.

We see that the relation f here is the relation less than in the set A. Here the domain of f is $\{ 1,2,3\}$ whereas the range is $\{ 2,3,4\}$.

In case, $f \subset A \times A$, we call f a relation in A. For A, the set of all lines parallel to is a subset of

$A \times A$. Similarly, for the set T of all triangles, similarity and congruence are subsets of $T \times T$. We enumerate a few relations.

1) A relation f in A{ that is, $f \subset A \times A$ is called a symmetric relation if $afb \Longleftrightarrow bfa$.

2) $f \subset A \times A$ is called a reflexive relation if afa for all $a \in A$.

3) $f \subset A \times A$ is called an antisymmetric relation if $afb \hspace{0.1in} \textup{and} \hspace{0.1in} bfa \Longrightarrow a=b$

4) A relation $f \subset A \times A$ is called a transitive relation if $afb, bfc \Longrightarrow afc$.

Examples.

a) Similarity of triangles is reflexive, symmetric and transitive.

b) Fatherhood is neither reflexive, nor symmetric, nor transitive.

c) “Ancestor of” is transitive but is neither reflexive nor symmetric.

d) The relationship “a is the spouse of b” is symmetric but neither reflexive nor symmetric.

e) “Less than” is transitive but not reflexive.

f) “Less than or equal to” is reflexive, antisymmetric and transitive.

g) If X is a class of sets, let $R \subset X \times X$ be defined by ARB if $A \subset B$. We see that R is a reflexive, antisymmetric and transitive relation.

Digression.

The way we have defined relation should strictly be called binary relation as it consists of ordered pairs. But, we have other kinds of relations too. Like collinearity of points, concurrence of lines, etc. They are not binary relations. But we don’t go into them here.

Equivalence relation.

A relation $\sim$ on A is called an equivalence relaiton if it is reflexive, symmetric and transitive, that is,

i) $a \sim a$ for all $a \in A$.

ii) $a \sim b \Longleftrightarrow b \sim a$

iii) $a \sim b \hspace{0.1in} \textup{and} \hspace{0.1in} b \sim c \Longrightarrow a \sim c$.

Examples.

a) Equality is certainly an equivalence relation.

b) Similarity between triangles is an equivalence relation.

c) Congruence of triangles is an equivalence relation.

d) Parallelism between lines is an equivalence relation if we agree that a line is parallel to itself.

e) Define a relation $\equiv (\mod 5)$ on Z by $m \equiv n (\mod 5)$ if $m - n$ is divisible by 5. Note that this is an equivalence relation.

More later,

Nalin Pithwa