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32 votes
32 votes

A partial order $P$ is defined on the set of natural numbers as follows. Here $\frac{x}{y}$ denotes integer division.

  1. $(0, 0) \in P.$
  2. $(a, b) \in P$ if and only if $(a \% 10) \leq (b \% 10$) and $(\frac{a}{10},\frac{b}{10})\in  P.$

Consider the following ordered pairs:

  1. $(101, 22)$
  2. $(22, 101)$
  3. $(145, 265)$
  4. $(0, 153)$

Which of these ordered pairs of natural numbers are contained in $P$?

  1. (i) and (iii)
  2. (ii) and (iv)
  3. (i) and (iv)
  4. (iii) and (iv)
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7 Answers

Best answer
42 votes
42 votes

Ans. D
For ordered pair $(a, b),$ to be in $P,$ each digit in a starting from unit place must not be larger than the corresponding digit in $b.$

This condition is satisfied by options

  • (iii) $(145, 265) \to 5 ≤ 5, 4 < 6$ and $1 < 2$
  • (iv) $(0, 153) \to 0 < 3$ and no need to examine further
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13 votes
13 votes

it is just like recursion for more understandbility we hve following ideas as.

We will have to check each and every condition recursive untill condition become completed.

Now check for (145,265)      a%10<=b%10 now remaining(14,26) again check a%10<=b%10 yes again check .

Similraly u can apply for option four .

11 votes
11 votes

(i)      (101, 22)

                 |

    -----------------

      |                |

   (1,2)         (10,2)

                      |

            --------------

              |              |

           (0,2)         (1,0)  ---> fails here bcz (a !<= b)

likewise you can check other options.

(iii) & (iv) are correct, hence option D    

1 votes
1 votes

Since P is given to be a partial order relation, options (i) & (ii) are ruled out as they violate  antisymmetry which says that if (a,b)∊R & (b,a)∊R => a=b but 22 ≠101.

Now 145%10≤265%10 and the evaluating the second condition gives (0,0) which is given to be in P. In similar vein (iv) can also be checked to be in P. 

Hence option iii & iv is correct.

Answer:

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