B.2 - Mathematical Modelling Involving Inequalities

In the previous section, we saw the mathematical model for the motion of a simple pendulum. In this section, we will see another example.

Example 3:
• The situation is:
A farm house uses atleast 800 kg of special food daily. The special food is a mixture of corn and soyabean with the following compositions:
$\begin{array}{cc}{}    &{\textbf{Material}}    &{\textbf{Protein present per Kg}}    &{\textbf{Fibre present per Kg}}    &{\textbf{Cost per Kg}}\\
{}    &{\textbf{Corn}}    &{0.09}    &{0.02}    &{\text{Rs 10}}\\
{}    &{\textbf{Soyabean}}    &{0.60}    &{0.06}    &{\text{Rs 20}}\\
\end{array}$
The dietary requirements of the special food stipulate atleast 30% protein and at most 5% fibre. How can this dietary requirement be satisfied at the least possible cost?
• This situation can be converted into a mathematical problem. For that, we engage in the process of mathematical modeling. The final result can be obtained in 4 steps:
 

Step 1:
First we study the situation. We will write it in order:

(i) “Atleast 800 kg” means:
    ♦ The total quantity should not be less than 800 kg.
    ♦ The total quantity can be greater than 800 kg.
(ii) The available nutrients and costs can be listed:
• One kg corn will
    ♦ give 0.09 kg protein.
    ♦ give 0.02 kg fibre.
    ♦ cost Rs 10
• One kg soya bean will give
    ♦ give 0.60 kg protein.
    ♦ give 0.06 kg fibre.
    ♦ cost Rs 20
(iii) Our aim is:
• The 800 kg (or more) food must
    ♦ contain atleast 30% protein.
    ♦ contain at most 5% fibre.
    ♦ cost the least possible amount.

• We can write:
Step 1 involves studying the situation and identifying the parameters. Essential parameters should be carefully identified.

Step 2:
This can be written in order:
(i) Let the food be made up of x kg corn and y kg of soya bean.
• Then from (i) of step 1, we can write: x + y ≥ 800

(ii) Available protein and fibre:
• From the x kg corn, we will get some protein and fibre.
• From the y kg soya bean also, we will get some protein and fibre.
• Then from (ii) of step 1, we can write:
    ♦ Total protein than can be obtained from x kg corn and y kg soya bean is:
0.09x + 0.60y   
    ♦ Total fibre than can be obtained from x kg corn and y kg soya bean is:
0.02x + 0.06y

(iii) Checking whether the dietary requirements are fulfilled:
• We are taking (x+y) kg food. This will contain (0.09x + 0.60y) kg of protein.
Then from (iii) of step 1, we can write:
$\begin{array}{ll}{}    &{\frac{0.09x + 0.60y}{x+y} \times 100}    & {~\ge~}    &{30}    &{} \\
{\Rightarrow}    &{\frac{0.09x + 0.60y}{x+y}}    & {~\ge~}    &{0.30}    &{} \\
{\Rightarrow}    &{0.09x + 0.60y}    & {~\ge~}    &{0.30 x + 0.30y}    &{} \\
{\Rightarrow}    &{0.30y}    & {~\ge~}    &{0.21 x}    &{} \\
{\Rightarrow}    &{0}    & {~\ge~}    &{0.21 x – 0.30 y}    &{} \\
{\Rightarrow}    &{0.21 x – 0.30 y}    & {~\le~}    &{0}    &{} \\
\end{array}$

• We are taking (x+y) kg food. This will contain (0.02x + 0.06y) kg of fibre.
Then from (iii) of step 1, we can write:
$\begin{array}{ll}{}    &{\frac{0.02x + 0.06y}{x+y} \times 100}    & {~\le~}    &{5}    &{} \\
{\Rightarrow}    &{\frac{0.02x + 0.06y}{x+y}}    & {~\le~}    &{0.05}    &{} \\
{\Rightarrow}    &{0.02x + 0.06y}    & {~\le~}    &{0.05 x + 0.05y}    &{} \\
{\Rightarrow}    &{0.01y}    & {~\le~}    &{0.03 x}    &{} \\
{\Rightarrow}    &{0}    & {~\le~}    &{0.03 x – 0.01 y}    &{} \\
{\Rightarrow}    &{0.03 x – 0.01 y}    & {~\ge~}    &{0}    &{} \\
\end{array}$

(iv) Checking cost:
We are taking (x+y) kg food. That food will cost (10x + 20y)
Then from (iii) of step 1, we can write: (10x + 20y) should be as small as possible.

• We can write:
Step 2 involves drawing the necessary diagrams and writing the relevant mathematical equations/inequalities. In short, we write a mathematical problem in this step. Any mathematical problem will have a definite solution. So in this step, it is necessary to recheck all the works done thus far.

Step 3:
In this step, we put the equations and inequalities into actual use. This can be written in order:

(i) From (i) of step 2, we have:
x + y ≥ 800
• The red line in fig.B.2 below represents
x + y = 800

Fig.B.2

• So any (x,y) in the upper half plane II of the red line will satisfy the inequality. See section 6.3.

(ii) From (iii) of step 2, we have:
0.21x - 0.30y ≤ 0
• The magenta line in fig.B.2 above represents
0.21x - 0.30y = 0
• So any (x,y) in the upper half plane II of the magenta line will satisfy the inequality.

(iii) Again from (iii) of step 2, we have:
0.03x - 0.01y ≥ 0
• The green line in fig.B.2 above represents
0.03x - 0.01y = 0
• So any (x,y) in the lower half plane I of the green line will satisfy the inequality.

(iv) So we have three regions:
    ♦ upper half plane II of the red line.
    ♦ upper half plane II of the magenta line.
    ♦ lower half plane I of the green line.
• The three regions will over lap in the yellow region shown in fig.B.2.
• So any (x,y) in the yellow region will satisfy all the three inequalities.

(v) From (iv) of step 2, we have:
(10x + 20y) must be as small as possible.
• For that, x and y must be as small as possible. At the same time, x and y must satisfy all three inequalities.
• This point is marked by a cyan dot in fig.B.2 above.
• If we draw a vertical line through the cyan dot, then that line will meet the x-axis at 470.59
• If we draw a horizontal line through the cyan dot, then that line will meet the y-axis at 329.41
• So the required point is (470.6,329.4)
• We can write:
Step 3 involves the actual application of the result obtained in step 2. Calculators or digital computers can be used for lengthy problems.

Step 4:
This can be written in order:
(i) First we do the validation.
• We are going to buy 470.6 kg corn and 329.4 kg soya bean.
• So total quantity = (470.6 + 329.4) = 800.
Thus “atleast 800” is satisfied.
• In this way all the dietary requirements can be checked. This will complete the validation process.
(ii) Next we do the interpretation.
• We state that:
If we buy 470.6 kg corn and 329.4 kg soya bean, then all the dietary requirements will be satisfied and at the same time, the cost will be the least possible value. The least possible cost is (470.6 × 10 + 329.4 × 20) = Rs 11294.00
• We can write:
In step 4, we do validation and interpretation.

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In the next section, we will see one more example.

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Chapter 14.5 - Miscellaneous Exercise

In the previous section, we saw validating statements. In this section, we will see some miscellaneous examples.

Solved example 14.14
Check whether “Or” used in the following compound statement is exclusive or inclusive. Write the component statements of the compound statement and use them to check whether the compound statement is true or not. Justify your answer.
t: you are wet when it rains or you are in a river.
Solution:
1. We have:
t: you are wet when it rains or you are in a river.
2. The component statements are:
p: you are wet when it rains.
q: you are wet when you are in a river.
   ♦ p and q are connected by “or”.
3. Let us analyze the component statements:
• During rain, you are wet.
• When in a river, you are wet.
• When in a river, if it rains, you are wet.
• So the “or” used in this case is inclusive “or”.
4. Both the component statements are true. So t is true.

Solved example 14.15
Write the negation of the following statements
(i) p: For every real number x, x² > x.
(ii) q: There exists a rational number x such that x² = 2.
(iii) r: All birds have wings.
(iv) s: All students study mathematics at the elementary level.
Solution:
Part (i)
1. Consider the statement:
p: For every real number x, x² > x.
2. We want ~p.
• p tells us that: For all real numbers, the given property is valid.
• ~p must tell us that "For all" is false.
3. This ~p can be achieved by saying that there is at least one real number for which the property is not valid.
• So we get:
~p: There exists one real number x for which x² ≤ x.

Part (ii)
1. Consider the statement:
q: There exists a rational number x such that x² = 2.
2. We want ~q.
• q tells us that: There is at least one rational number for which the given property is valid.
• ~q must tell us that "at least one" is false.
3. This ~q can be achieved by saying that for all rational numbers, the property is not valid.
• So we get:
~p: For all rational numbers x, x² ≠2.

Part (iii)
1. Consider the statement:
r: All birds have wings.
2. We want ~r.
• r tells us that: For all birds, the given property is valid.
• ~r must tell us that "For all" is false.
3. This ~r can be achieved by saying that there is at least one bird for which the property is not valid.
• So we get:
~r: There exists one bird which do not have wings.

Part (iv)
1. Consider the statement:
s: All students study mathematics at the elementary level.
2. We want ~s.
• s tells us that: For all students, the given property is valid.
• ~s must tell us that "For all" is false.
3. This ~r can be achieved by saying that there is at least one student for which the property is not valid.
• So we get:
~r: There exists one student who does not study mathematics at the elementary level.

Solved example 14.16
Using the words “necessary and sufficient”, rewrite the statement “The integer n is odd if and only if n² is odd”. Also check whether the statement is true.
Solution:
Part (i): Rewriting the statement
Using the words “necessary and sufficient”, we can write:
“The integer n is odd” is necessary and sufficient condition for “n² to be odd” and vice versa. 

Part (ii): Checking the validity.
• The given compound statement is in the form “p if and only if q”.
The component statements are:
p: The integer n is odd.
q: n² is odd.
• In such cases, we know that:
    ♦ Whenever p is true, q is also true.
    ♦ Whenever q is true, p is also true.
• So we need to show two items:
Case (i) If p is true, then q is true.
Case (ii) If q is true, then p is true.

Case (i): If p is true, then q is true.
    ♦ First, we assume that p is true.
    ♦ Based on this assumption, we check q.
1. Assuming p to be true:
• Let us assume that, integer n is indeed odd.
2. Since it is an odd integer, we can write:
    ♦ n = 2m + 1
        ✰ Where m is some integer.
3. Now we calculate n². We get:
$\begin{array}{ll}
{}&{n^2} & {~=~}& {(2m+1)^2} &{} \\
{}&{} & {~=~}& {4m^2 + 4m + 1} &{} \\
{}&{} & {~=~}& {2(2m^2 + 2m) + 1} &{} \\
\end{array}$

4. Consider the result in (3):
• m is an integer.
⇒ m² will be an integer.
⇒ (2m² + 2m) will be even.
⇒ 2(2m² + 2m) will be even.
⇒ 2(2m² + 2m) + 1 will be odd.

5. Based on (3) and (4), we can write:
n² is odd.
• So the statement q is true.
6. We obtained the truth value of q as T.
• We obtained this by assuming that the truth value of p is T.
• So the truth value of “if p then q” is T.

Case (ii): If q is true, then p is true.
Here we will use the contrapositive method. That is., if ~p then ~q.
    ♦ First, we assume that ~p is true.
    ♦ Based on this assumption, we check ~q.
The statements are:
~p: The integer n is even.
~q: n² is even.
1. Assuming ~p to be true:
• Let us assume that, n is an even integer and n² is indeed even.
2. Since it is an even integer, we can write:
    ♦ n = 2m
        ✰ Where m is some integer.
3. Now we calculate n². We get:
$\begin{array}{ll}
{}&{n^2} & {~=~}& {(2m)^2} &{} \\
{}&{} & {~=~}& {4m^2} &{} \\
{}&{} & {~=~}& {2(2m^2)} &{} \\
  \end{array}$

4. Consider the result in (3):
• m is an integer.
⇒ m² will be an integer.
⇒ 2m² will be even.
⇒ 2(2m²) will also be even.

5. Based on (3) and (4), we can write:
n² is even.
• So the statement ~q is true.
6. We obtained the truth value of ~q as T.
• We obtained this by assuming that the truth value of ~p is T.
• So the truth value of “if ~p then ~q” is T.
• So the truth value of “if q then p” is T.

◼ From case (i), we obtained:
• The truth value of “if p then q” is T.
◼ From case (ii), we obtained:
• The truth value of “if q then p” is T.
◼ So we can write: The given "p if and only if q" is true.
• That is.,
The statement “The integer n is odd if and only if n² is odd” is true.

Solved example 14.17
For the given statement, identify the necessary and sufficient conditions
t: If you drive over 80 km per hour, then you will get a fine. 
Solution:
1. Consider the statement:
t: If you drive over 80 km per hour, then you will get a fine.
2. The component statements are:
p: You drive over 80 km per hour.
q: You will get a fine.
3. This is a statement with "if p then q".
• So we can write:
   ♦ p is a sufficient condition for q.
   ♦ q is a necessary condition for p.
(see example 2 at the beginning of section 14.3)
4. Thus we get:
• The sufficient condition is:
Driving over 80 km per hour.
(This is sufficient to get a fine)
• The necessary condition is:
Getting a fine.
(This is necessary if the speed is above 80 km per hour)

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The link below gives some more solved examples:

Miscellaneous Exercise

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In the next chapter, we will see statistics.

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Chapter 14.4 - Validating Statements

In the previous section, we saw Implications. In this section, we will see validating statements.

• Truth value of a statement tells us whether a statement is true or false.
• If a statement is true, then its truth value is true.
    ♦ We write it as T.
• If a statement is false, then its truth value is false.
    ♦ We write it as F.
• Validating statements is the process of finding the truth value of a statement.

---

• Validating simple statements is easy. But for compound statements, we have to follow certain rules.
• The rules depend on the connectives/implications present in the compound statement.

Rule 1: Statements with “and”.
• If the connective in a compound statement is “and”, then the truth value of each of the component statements must be T. Then only we get T for the compound state.
• So when the component statements are p and q, we can use the following two steps:
Step 1: Show that p is true.
Step 2: Show that q is true. 

Rule 2: Statements with “or”.
• If the connective in a compound statement is “or”, then the truth value of at least one component statement must be T. Then only we get T for the compound state.
• So when the component statements are p and q, we can use the following three cases:
Case 1: p is false but q is true.
Case 2: p is true but q is false.
Case 3: Both p and q are true.
• If we get any of the above three cases, then the truth value of the compound statement is T.  

Rule 3: Statements with “if-then”.
• When the compound statement is in the form “if p then q”, we know that:
Whenever p is true, q is also true.
• So we can think of a method to validate such statements:
    ♦ First, we assume that p is true.
    ♦ Based on this assumption, we check q.
    ♦ If q is true, then the compound statement can be given the truth value T
• This method is known as direct method.

• When the compound statement is in the form “if p then q”, we know that:
if ~q then ~p is applicable.
• So we can think of another method to validate such statements:
    ♦ First, we assume that ~q is true.
    ♦ Based on this assumption, we check ~p.
    ♦ If ~p is true, then the compound statement can be given the truth value T
• This method is known as contrapositive method.
• So for validating “if p then q” statements, we have two methods:
(i) Direct method  (ii) Contrapositive method.

Rule 4: Statements with “if and only if”.
• When the compound statement is in the form “p if and only if q”, we know that:
    ♦ Whenever p is true, q is also true.
    ♦ Whenever q is true, p is also true.
• So we can think of a method to validate such statements:
In this method, we need to show two items:
(i) If p is true, then q is true.
(ii) If q is true, then p is true.

Rule 5
This rule is used to prove that, a statement is true. In this rule, we make use of contradiction. The steps are:
(i) We assume that p is false.
This is same as assuming ~p to be true.
(ii) Then we arrive at some result which contradicts the assumption.
(iii) Due to the contradiction, we conclude that, p is true.

Rule 6
This rule is used to prove that, a statement is false. In this rule, we simply present an example where the statement is not valid. Such an example is called counter example. When the counter example is presented, the statement is immediately written off as false.
◼ We may be able to present numerous examples where a statement is valid. Whatever be the number of such examples we present, we cannot conclude that the statement is true.
◼ We may be able to present numerous counter examples where a statement is not valid.  Just one counter example is sufficient to conclude that the statement is false.

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Let us see some solved examples:

Solved example 14.10
Check whether the following statement is true or not.
If x, y ∈ Z are such that x and y are odd, then xy is odd.
Solution:
1. Consider the statement:
r: If x, y ∈ Z are such that x and y are odd, then xy is odd.
2. The component statements are:
p: x, y ∈ Z are such that x and y are odd.
q: xy is odd.
3. This is a statement with "if-then". So we will apply Rule 3. We will use the direct method.
    ♦ First, we assume that p is true.
    ♦ Based on this assumption, we check q.
4. Assuming p to be true:
• Let us assume that, both x and y are indeed elements of the set Z and they are indeed odd.
• We know that, Z is the set of integers. So we can write:
We assume that both x and y are indeed odd integers.
5. Since they are odd integers, we can write:
    ♦ x = 2m + 1
    ♦ y = 2n + 1
        ✰ Where m and n are some integers.
6. Now we calculate the product xy. We get:
$\begin{array}{ll}
{}&{xy}
& {~=~}& {(2m+1)(2n+1)} &{} \\

{}&{}
& {~=~}& {4mn + 2m + 2n + 1} &{} \\

{}&{}
& {~=~}& {2(2mn + m + n) + 1} &{} \\

\end{array}$

7. Consider the result in (6):
• m and n are integers
⇒ (2mn + m + n) can be odd or even.
⇒ 2(2mn + m + n) will be even.
⇒ 2(2mn + m + n) + 1 will be odd.

8. Based on (6) and (7), we can write:
The product xy is odd.
• So the statement q is true.
9. We obtained the truth value of q as T.
• We obtained this by assuming that the truth value of p is T.
• So the truth value of “if p then q” is T.
• We can write:
The truth value of the given compound statement r is T.

Alternatively, using the contrapositive method:
1. We have: if ~q then ~p
• The contrapositive statements are:
~q: xy is even.
~p: One among x, y is even.
2. For applying the contrapositive method,
    ♦ First, we assume that ~q is true.
    ♦ Based on this assumption, we check ~p.
3. Assuming ~q to be true:
~q: xy is even.
4. When xy is even, x or y has to be even.
(supposing x = 2n, we get xy = 2ny, which is even)
• xy will never become even if both x and y are odd.
• So ~p is true.
5. We obtained the truth value of ~p as T.
• We obtained this by assuming that the truth value of ~q is T.
• So the truth value of “if ~q then ~p” is T.
• We can write:
The truth value of the given compound statement r is T.

Solved example 14.11
Check whether the following statement is true or false by proving its contrapositive.
If x, y ∈ Z such that xy is odd, then both x and y are odd.
Solution:
1. Consider the statement:
r: If x, y ∈ Z such that xy is odd, then both x and y are odd.
2. The component statements are:
p: x, y ∈ Z such that xy is odd.
q: both x and y are odd.
3. This is a statement with "if-then". So we will apply the contrapositive method for validation.
• We have: if ~q then ~p
• The contrapositive statements are:
~q: One among x, y is even.
~p: xy is even.
2. For applying the contrapositive method,
    ♦ First, we assume that ~q is true.
    ♦ Based on this assumption, we check ~p.
3. Assuming ~q to be true:
~q: One among x, y is even.
4. When one among x, y is even, xy will be even.
(supposing x = 2n, we get xy = 2ny, which is even)
• xy will never become even if both x and y are odd.
• So ~p is true.
5. We obtained the truth value of ~p as T.
• We obtained this by assuming that the truth value of ~q is T.
• So the truth value of “if ~q then ~p” is T.
• We can write:
The truth value of the given compound statement r is T.

Solved example 14.12
Prove that the following statement is true.
p: √7 is irrational
Solution:
1. Assume that √7 is rational.
2. If √7 is rational, we can write: $\sqrt{7} = \frac{a}{b}$
Where the ratio $\frac{a}{b}$ is in the simplest form.
• That is, a and b do not have any common factors.
3. Squaring both sides, we get: $7 = \frac{a^2}{b^2}$
4. Based on this we can write: $b^2 = \frac{a^2}{7}$
• b is an integer. So b² will be an integer.   
5. Let us analyze the result in (4):
$\text{An integer}(b^2)~=~\frac{\text{An integer}(a) \times \text{An integer}(a)}{7}$
• So $\frac{a}{7}$ must be an integer. Let it be c.
6. We can write: $\frac{a}{7}= c$
• Thus we get a = 7c
• Squaring both sides, we get: a² = 49c².
7. But from (3) we have a² = 7b².
8. Equating the results in (6) and (7), we get:
49c² = 7b².
• From this we get: 7c² = b².
9. Based on this we can write: $c^2 = \frac{b^2}{7}$
• c is an integer. So c² will be an integer.
10. Let us analyze the result in (9):
$\text{An integer}(c^2)~=~\frac{\text{An integer}(b) \times \text{An integer}(b)}{7}$
• So $\frac{b}{7}$ must be an integer. Let it be d.   
11. Let us compare two results:
   ♦ In (5), we saw that, a is divided by 7 to give an integer.
   ♦ In (10), we saw that, b is divided by 7 to give an integer.
• That means, a and b has a common factor 7. This contradicts our assumption in (2) that, a and b have no common factors.
12. So it is clear that the assumption that √7 is rational is wrong.
• We can write:
p: √7 is irrational
is true.

Solved example 14.13
By giving a counter example, prove that the following statement is false.
If n is an odd integer, then n is prime.
Solution:
1. Consider the statement:
r: If n is an odd integer, then n is prime.
2. The component statements are:
p: n is an odd integer.
q: n is prime.
3. This is a statement with "if-then".
• We have to prove "if p then q" is false.
4. For statement p, we take n = 9 as an example. This is acceptable because 9 is an odd number.
5. Now we consider statement q.
• We know that 9 is not a prime number. So when n = 9, the statement q becomes false.
6. So we presented a counter example which proved "if p then q" to be false.
• That is., r: If n is an odd integer, then n is prime.
is false.

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The link to some solved examples is given below:

Exercise 14.5

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In the next section, we will see some miscellaneous examples.

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