# Reasoning - Inequality

Combination of two elementary problems are involved in the problems based on inequality and coded inequality.

In this type of problems, coding scheme is told entirely in the question itself. To decode inequalities in a given problem wouldn’t mean anymore headache than a couple of extra seconds.

Essentially, it is a problem of inequalities and it is this aspect that should be mastered. Hence we first learn the basics of inequalities.

We know the result of multiplication between 5 and 3 and the number 15 are equal. Since they are equal, it is equality but in case 5 × 5 ≠ 15, the product of 5 and 5 is not equal to the number 15, it is an inequality.

Greater than − It is denoted by >. For example, 5 × 5 > 15

Less than − It is denoted by <. For example, 5 × 2 < 15

Greater than or equal to − It is denoted by ≥. When we don’t know the exact condition of inequality between two numbers, we use this symbol. For instance, consider two numbers x and q. We know that x is not less than q. In this case x can either be equal to q or greater than q .So we use ≥ sign.

Less than or equal to − It is denoted by ≤. When one number is either less than another number or equal to that number then this symbol is used. For example, consider two numbers X and B where X is not greater than B. In this case X is less than or equal to B. So it can be represented as X ≤ B.

Two golden rules for combining inequalities are as follows −

A common term can combine two inequalities.

Example 1

Inequality − A > B, C > D

Here four terms are used but there is no common term. So these two inequalities cannot be combined.

Example 2

Inequality − A ≤ B, X ≥ Y

So here also common term is missing. So they cannot be combined.

If the common term is higher than one and less than the other, both the inequalities can be combined.

Example 1

Inequality − P > X, X > C.

Here, common term is X. X is greater than C but less than P. So the combination will be like this – P > X > C or C < X < P.

Example 2

Inequality − X < P, X ≥ C

Here X is less than P, and higher than or equal to term C. Since X is common, the combination is possible. That is - P > X ≥ C or C ≤ X < P.

Deriving a conclusion from a combined inequality −

Another rule, the third golden rule, is used to derive a conclusion from a combined inequality is as follows −

Add two inequalities and derive a conclusion by letting the middle term vanish. The conclusion inequality has ≥ sign if and only if both the signs in the combined inequality were ≥ and vice-versa.

Hence the conclusion will normally have a > sign strictly, unless the ≥ sign appears twice in the combined inequality.

Example 1 − Derive a conclusion from following combined inequalities.

i. x > y > z

ii. x < y < z

Solution

i. x > z

ii. x < z

### Strategy to Solve Problems on Inequality and Coded Inequality

Steps involved in solving the problems are as follows −

Step 1 − Neatly and quickly decode the symbol that refers arithmetical operation.

Example − Given that P α Q. Means P > Q. Therefore replace α by >. You should take one code at a time and replace it by its original mathematical symbol before going to next code and you should do it quickly.

Step 2 − Take one conclusion at a time and decide which statements are relevant for evaluating the conclusion.

Now, this needs some thinking. What do you mean by a relevant statement? Here we mean the statement which is not useless to derive a conclusion. If there is a conclusion say x > y then a statement like a > b is useless because it doesn’t contain either x or y. Therefore any analysis can’t tell us anything about this conclusion. The relevant statements are those that can be combined to prove or disprove that conclusion. So this statement is not relevant for x > y.

To decide which statement is relevant for a conclusion, take two terms of a given conclusion and see if each of them separately appears with a single common term. These statements will be relevant statements.

Example − Suppose after performing step 1, we have following statement;

M > N, L = M, O > N, L ≤ K

Conclusion

a) M < K, b) L > N

Step 3 − Use three golden rules to combine relevant statements and derive a conclusion from it. Golden rules are;

Rule 1 − There must be a common term.

Rule 2 − The common term must be less than or equal to one term and greater than or equal to another.

Rule 3 − The conclusion is that inequality is obtained by letting the common-term disappear and it has a ≤ or a ≥ sign if and only if both inequalities in second step had a ≤ sign or a ≥ sign. In all other cases, there will be a < or a > sign in the conclusion.

For conclusion a (M < K) the relevant statements are

M = L and L ≤ K.

By combining we get M = L < K

So, M ≤ K (according to step 3)

Now M ≤ K doesn’t imply that M < K because M ≤ K allows for M to be less than or equal to K which is not true in case of M < K.

For conclusion b, relevant statements are

M > N and L = M

After combining we get, L = M > N L > N

Hence conclusion is verified, well and good. So L > N. If not, perform the following checks.

Check 1 − Check if the conclusion directly follows from only single given statement.

Sometimes statement may be in form of A ≥ B and one conclusion may be in the form of B ≤ A. Obviously both are completely identical but sometimes we are prone to ignore such minor tricks of the examiner.

Example − Consider the following: (Let α means >, β means ≥, γ means =, δ means <, η means ≤)

Let, given statement: E γ F, C δ D, F δ g, D β F

Conclusion1. G η F.

Here conclusion is G η F or G ≤ F and it is identical to F β G or F ≥ G. Hence it is directly following from one single statement.

Check 2 − The conclusion you reach after third step may be identical to the given conclusion although it may not look so in first glance.

Check 3 − If after third step you get a conclusion that has a ≥ sign and two given conclusions have a > sign and an = sign between same terms, the choice either 1 or 2 is correct.

For Example − Suppose you reach A ≥ B after performing the third step. Now suppose the given conclusions are - I) A > B and II) A = B. Then the choice “either I or II follows” is correct.

Similarly if you conclude that M ≤ N and given conclusions are I) M < N and II) M = N then again same answer follows.

Check 4 − If two given conclusions have the signs given below between the same terms

a) ≤ and > signs, or

b) < and > signs, or

c) > and ≤ signs, or

d) ≥ and < signs

and if neither of the conclusion has been accepted in any of the steps above; the choice either of the two follows is correct.

Suppose, in a given question the conclusions are

a) A ≥ B b) A < B

Now suppose that neither of them have been proved to be true by virtue of any preceding steps. Since they have the same pair (A and B) and the signs are ≥ and <; the choice either follows is correct.

Note − Check 4 merely tells that one number can only have three positions vis-avis another number. It can be either less than or equal to or greater than the other.

This is true universally for any two numbers. That is, [A ≤ B or A > B] is a universally correct statement, because A can be either (less than or equal to) or (greater than) B.

Thus for any two numbers A and B, the following are always correct −

I. A ≤ B or A < B

II. A < B or A > B

III. A > B or A ≤ B

IV. A ≥ B or A < B

These four pairs are called complementary pairs. In such cases, one out of two statements will always be true. We choose “either follows” as answer. But remember, we choose this as our answer only if neither of the two statements have been otherwise proved on any previous step.

reasoning_inequality.htm
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