Revision Notes
1.
0 is a Whole number but not a Natural number.
Thus if we need Whole Number solution to 3x + 4y = 24, then (x = 0 and y = 6) is an acceptable solution, but if we needed Natural Number solution to the same equation, then (x = 0 and y = 6) is not acceptable, since x = 0 is not a Natural number.
2.
0 is neither positive nor negative
Positive Integers are {1, 2, 3, ….} i.e. same as Natural Numbers
Non-negative Integers are {0, 1, 2, 3, …} i.e. same as Whole Numbers
3.
Irrational number forms are of the type:
… \(\sqrt{2}\), \(\sqrt{3}\) , \(\sqrt[3]{4}\) …. i.e. roots or n-th roots which cannot be found out to be Natural Numbers (i.e. \(\sqrt{4}\) is not irrational because it becomes 2)
… or many of the constants we use like π, e, ϕ (no further details needed)
All other real numbers that we use, Integers or Decimals, are Rational. And text-book definition of Rational Numbers are those that can be written in the form of p/q where p and q are Integers and q ≠ 0.
4.
Terminating v/s Non-terminating
Terminating decimals can be written as p/q by ignoring the decimal point and adding 0s to the denominator e.g. 0.1025 = 1025/10000, and hence all terminating decimals are Rational.
Non-terminating decimals can be of two types ….
…. non-terminating but recurring. These types can also be written as p/q form and are hence Rational Numbers.
…. non-terminating and also non-recurring. These numbers are Irrational.
5.
From points 3 and 4 …..
Any Rational number will be …
…. an Integer, or
…. a terminating decimal, or
…. a non-terminating but recurring decimal.
Thus, 1/7 or 1/13 or any of this form … which obviously is not Integer and also doesn’t seem to terminate …. HAS to then be recurring, even if it may recur after numerous digits.
And Irrational numbers i.e. \(\sqrt{2}\), \(\sqrt{3}\) , \(\sqrt[3]{4}\) …. will be non-terminating and non-recurring.
6.
Converting recurring decimals to p/q form
The following are just the final results. If not clear, please refer to the concept videos for details.
Purely recurring decimal:
\(0.abcabcabc … = 0.\overline{abc} = \frac{abc}{999}\); there are 3 digits that recur, hence 3 9s in the denominator
Recurring decimal in which a starting part of the decimal does not recur:
\(0.pqabcabcabc… = 0/pq\overline{abc} = \frac{pqabc-pq}{99900}\); there are 2 digits that do not recur, hence 2 0s in the denominator, after the required number of 9s. Notice the numerator carefully.
To do any calculation involving recurring decimals, convert them to the p/q form and then do the calculation.
7.
Even & Odd
Scope: Integers.
So 0 is even. And so are −2, −4 −6 ….. but usually when we talk of numbers, it is limited to Whole or Natural, so we usually don’t bother that these negatives are also Even and the common take-way is that Even means 0, 2, 4, 6, …..
Operations on Even & Odd
7.1 Adding/Subtracting Evens and Odds
Even ± Even ± Even ± ….. = EVEN
Odd ± Odd ± Odd ± ….. = EVEN OR ODD, depends on the number of terms in the LHS … if there are even number of terms then the result is EVEN and if there are odd number of terms then the result is ODD.
And if some of the terms on the LHS are Even and some are Odd, then the result depends on the number of odd terms on the LHS … if the number of Odd terms are even, then result is EVEN; if the number of Odd terms are odd, then the result is ODD.
This follows directly from the earlier two relations.
7.2 Multiplying Evens and Odds
…. ? × ? × … × Even × ? × …. = EVEN
If any one multiplicand is Even, the product is EVEN.
… Odd × Odd × Odd × …. × Odd = ODD
If all the multiplicands are Odd, then the product is ODD.
Notice how the above two take care of every possibility.
From the second relation it follows that ….
…. If a × b × c × …. × n = Odd, then each of a, b, c, …., n is Odd.
…. If a × b × c × …. × n = Even, then ATLEAST 1 out of a, b, c, …., n is Even, and the others could be either Even or Odd.
7.3 Exponential
Evenany natural number = EVEN
Oddany natural number = ODD
A rare case could be when the exponent could be 0, in which case,
(Even or Odd)0 = 1 i.e. Odd
8.
Prime & Composite
Scope: Prime and Composite categorisation is limited to just Natural numbers.
Natural numbers are actually categorised into 3 categories – prime, composite and 1, because 1 is neither prime nor composite.
Prime numbers have to have exactly 2 different factors; Composite numbers have more than 2 different factors. The number 1 is unique, it has only 1 factor, 1 itself.
The first composite number is 4.2 is the only even prime number. Rest all prime numbers are odd.
9.
Writing prime as a product of multiplicands
If a × b × c × …. × n = Prime, then exactly one of a, b, c, …., n is equal to the prime number and all the rest are 1. This can be used to solve questions like, if n × (n+2) × (n+4) is prime, where n is an integer, what value/s can n be?
10.
Co-prime numbers
Two numbers are said to be co-prime if the two numbers do not have any common factor other than 1.
Note, the numbers themselves need not be prime e.g. 8 & 9 (It should be obvious that if both the numbers themselves are prime, then ANY pair of prime numbers will also be co-prime numbers).Also, 1 and (any other natural number) are co-prime.
11
RARE & Difficult Stuff, not relevant to CAT, but is used in some difficult questions that float around in groups.
11.1
The number of ways of writing a natural number, n, as a sum of two co-prime numbers = ϕ(n), where ϕ(n) is called Totient Function and is explained below.
\( \phi\left(n\right)=n \times \left(1-\frac{1}{p_1}\right) \times \left(1-\frac{1}{p_2}\right) \times \left(1-\frac{1}{p_3}\right) \times \ldots \)
, where p1, p2, p3 … are all the prime factors of n.
This appears scary, but can be calculated orally after a few trials.
E.g. to find ϕ(240), we first find the prime factors of 240.
240 = 16 × 15 = 24 × 3 × 5.
We are only interested in the prime factors, 2, 3 & 5 and not the exponents, so one can find prime factors even without complete factorisation. Next,
\(\phi\left(240\right)=n \times \left(1-\frac{1}{2}\right) \times \left(1-\frac{1}{3}\right) \times \left(1-\frac{1}{5}\right)\)
This above step can also be eliminated once one knows what (1 – 1/p) will be and one can directly write
\(\phi\left(240\right)=n \times \left(\frac{1}{2}\right) \times \left(\frac{2}{3}\right) \times \left(\frac{4}{5}\right)\)
This calculation will always result in a natural number and can be orally done sequentially 240 × ½ = 120; 120 × 2/3 = 80; and 80 × 4/5 = 64; thus, ϕ(240) = 64.
ϕ(n) has one more use …
11.2
The number of natural numbers that are co-prime to n and less than n = ϕ(n)
11.3
The number of ways of writing a natural number, n, as a product of two co-prime numbers = 2p−1, where p is the number of prime factors of n.
12. Must Know Questions from the Exercise
Solve all and bookmark any question that you may have found to teach you something new, but nothing stands out as a question that is unique and needs special mention.
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