The Fundamental Theorem of Arithmetic

Fundamental theorem of arithmetic
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Mathematical Statement (or Proposition)

A mathematical statement is a sentence which is either true or false . It can not be ambiguous.

Conjecture

It is a conclusion or proposition based on incomplete information, for which no proof has been found.

Axiom

A statement or proposition which is regarded as being established, accepted, or self-evidently true. It need not to be proved.

Theorem

It is a statement that has been proved on the basis of previously established statements, such as other theorems, and generally accepted statements, such as axioms. A theorem is a logical consequence of the axioms.

Proof

The process which can establish the truth of a mathematical statement based purely on logical arguments is called a mathematical proof.

❕ The Fundamental Theorem of Arithmetic was first recorded as Proposition 14 of Book IX in Euclid’s Elements, before it came to be known as such.

Carl Friedrich

It’s first correct proof was given by Carl Friedrich Gauss (1777-1855) in his Disquisitiones Arithmeticae (1801).

Carl Friedrich Gauss is often referred to as the Prince of Mathematicians’ and is considered one of the three greatest mathematicians of all time, along with Archimedes and Newton. He has made fundamental contributions to both mathematics and science.

The Fundamental Theorem of Arithmetic

“Every composite number can be expressed ( factorised ) as a product of primes, and this factorisation is unique, apart from the order in which the prime factors occur.”

➧ Given any composite number, there is one and only one way to write it as a product of primes, as long as we are not particular about the order in which the primes occur. e.g., we regard 2 × 3 × 5 × 7 as the same as 3 × 5 × 7 × 2, or any other possible order in which these primes are written.

➧ In general, given a composite number x, we factorise it as x = p_1p_2 ... p_n, where p_1p_2 ... p_n are primes and written in ascending order, i.e. p_1\leq p_2 \leq . . .\leq p_n.

➢ When we combine the same primes, we get powers of primes. e.g.,

32760 = 2 × 2 × 2 × 3 × 3 × 5 × 7 × 13 = 2^3 \times 3^2 \times 5 \times 7 \times 13

➢ Once we have decided that the order will be ascending, then the way the number is factorised, is unique.

Applications of the Fundamental Theorem of Arithmetic

➧ The Fundamental Theorem of Arithmetic has many applications, both within mathematics and in other fields

➢ One of its application is in Prime factorisation method for finding HCF & LCM of two or more integers.

➢ It is applied to prove the irrationality of many irrational numbers.

➢ It is applied to explore when exactly the decimal expansion of rational number is terminating and when it is non-terminating repeating.

Finding HCF and LCM of Integers

➢ For any two positive integers p and q,

HCF(p,q) \times LCM(p,q) = p \times q

➢ For any three positive integers p, q and r,

HCF(p,q,r)= \frac{p\times q\times r\times LCM(p,q,r)}{LCM(p,q)\times LCM(q,r)\times LCM(r,p)}
LCM(p,q,r)= \frac{p\times q\times r\times HCF(p,q,r)}{HCF(p,q)\times HCF(q,r)\times HCF(r,p)}

(Q) Find the HCF and LCM of 6, 72 and 120, using the prime factorisation method.

(A) 6 = 2 \times 3,
72 = 2^3 \times 3^2,
120 = 2^3 \times 3 \times 5

HCF_{(6,\,72\, and\, 120)} = 2 \times 3 (common prime factors with their lowest power)
= 6

LCM_{(6,\,72\,and\,120)} = 2^3 \times 3^2 \times 5 ( all prime factors with their highest power)
= 360

Proving Irrationality of an Irrational Number

To prove that a given number is irrational, we need the following theorem, whose proof is based on the Fundamental Theorem of Arithmetic.

Theorem : Let p be a prime number. If p divides a2, then p divides a, where a is a positive integer.

Proof :

Let, a = p_1p_2...p_n , where p_1p_2...p_n are primes not necessarily distinct.

\implies a^2= {p_1}^2{p_2}^2 ...{p_n}^2

\implies p_1p_2...p_n are prime factors of a^2

p divides a^2(given)

p is one of the prime factor of a^2 (from fundamental theorem of arithmetic)

But Prime factorisation of any number is unique

p is one of p_1p_2...p_n

a = p_1p_2...p_n

\implies p\,divides\, a.

Theorem : √2 is irrational.

Proof:

Let us assume that √2 is rational, such that for integers a and b (b≠0), √2 = a/b , where a and b are co-primes, i.e. ratio is in standard form.

\implies b\sqrt{2} = a

Squaring both sides, we get

\implies (b\sqrt{2})^2 = (a)^2

\implies 2b^2 = a^2

\implies 2\, divides\, a^2

\implies 2\, divides\, a

So, we can write a = 2c for some integer c,

2b^2 = 4c^2 \left(\because 2b^2 = a^2\right)

\implies b^2 = 2c^2

\implies 2\, divides\, b^2

\implies 2\, divides\, b

So, 2 is a common factor of a and b.

But this contradicts the fact that a & b are co-primes.

Hence, this contradicts our assumption that √2 is rational.

So, we conclude that √2 is irrational.

➤ The technique we used above is known as Proof by Contradiction.

Irrationality of an expression with Rational and Irrational factors

➢ Sum and Difference of a rational number and a irrational number is irrational.

➣ The product and quotient of a non-zero rational number and irrational number is irrational.

Show that: 2−√3 is irrational.

Proof : Let us assume that 2-√3 is rational, such that for co-primes a and b\, (b\neq 0), 2-\sqrt{3} = \frac{a}{b}.

\implie 2 - \frac{a}{b} = \sqrt{3}

\implies \sqrt{3} = 2 - \frac{a}{b} =\frac{2b - a}{b}

a and b are integers

\implies \frac{2b-a}{b} is rational

\implies \sqrt{3} is rational

But this contradicts the fact that \sqrt{3} is irrational.

So, we conclude that 2-\sqrt{3} is irrational.

Rational Numbers & Their Decimal Expansions

Decimal expansion of a Rational number can be either

terminating or

non-terminating reccurring (repeating).

❔ Can we predict by looking at denominator when the given rational is terminating and when it is non-terminating reoccurring?

Rational numbers whose decimal expansions are terminating

↬ Fraction whose denominator is power of 10 will terminate in its decimal expansion.

e.g., 375/100 = 3.75, 23/1000 = 0.023 etc.

∵ Prime factorisation of 10 = 2×5

⇒ Prime factorisation of 10^k = 2^k \times 5^k , where k is any positive integer.

⇒ Denominator of a fraction which is a power of 10 can be expressed as 2^k\times 5^k.

If we simplify these fraction into a standard rational number, where p & q are co-primes then some powers of two’s or five’s or both in the numerator & the denominator will cancel each other out, then Prime factorisation of q will have the form 2^n\times 5^m, where m and n are some non-negative integers (can be 0).

e.g, 0.375 = \frac{375}{1000} = \frac{3\times 5^3}{10^3}

= \frac {3\times 5^3}{2^3\times 5^3}= \frac{3}{2^3}

here, q = 2^3\times 5^0

↬ Fraction whose denominator is power of 2 only will be terminating in its decimal expansion.

e.g., 27/2 = 13.5, 11/4 = 2.75, 7/8 = 0.875 etc.

↬ Fraction whose denominator is power of 5 only will be terminating in its decimal expansion.

e.g., 1/5 = 0.2, 200/125 = 1.6, etc.

Theorem: Let x = p/q be a rational number, such that the prime factorisation of q is of the form 2n×5m, where n, m are non-negative integers. Then x has a decimal expansion which terminates.

Example: Among following rational numbers, select which will terminate.

(i)\, \frac{27}{320}, (ii)\, \frac{81}{90}\, (iii)\, \frac{45}{210}\, (iv)\, \frac{127}{750}

Solution:

(i) \frac{27}{320} = \frac{3^3}{2^6\times 5} ,

Here, q = 2^6\times 5^1\, (= 2^n\times 5^m)

\implies \frac{27}{320} will terminate and,

= 0.084375

(ii) \frac{81}{90} = \frac{3^4}{2\times 3^2\times 5} = \frac{3^2}{2\times 5}

Here, q = 2^1×5^1 (= 2^n×5^m, n=m)

\implies \frac{ 81}{90} will terminate and,

= 0.9

(iii) \frac{45}{210} = \frac{3^2 \times 5}{2\times 3\times 5\times 7} = \frac{3}{2\times 7}

Here, q = 2^1× 7\, (\neq 2^n\times 5^m)

\implies \frac{45}{210} will not terminate.

(iv) \frac{127}{750} = \frac {127}{2×3×5^3}
Here, q = 2\times 3\times 5^3 (≠ 2^n\times 5^m)

\implies \frac{127}{750} will not terminate.

Theorem : Let x be a rational number whose decimal expansion terminates. Then x can be expressed in the form p/q ,where p and q are co-prime, and the prime factorisation of q is of the form 2n×5m, where n, m are non-negative integers.

e.g, 375/1000 (= 0.375)

dividing by 125 (HCF), we get

= \frac{3}{8} = \frac{3}{2^3}\,(= \frac{p}{q}\, and\, q = 2^3\times 5^0)

We can convert the rational number in form of p/q to equivalent rational number of form a/b , where b is power of 10 .We can then easily write its decimal expansion which is terminating.

e.g,
\frac{3}{8} = \frac{3}{2^3} (form of \frac{p}{q})

= \frac{3\times 5^3}{2^3\times 5^3} (multiplying both the numerator and denominator with 5^3)

= \frac{375}{1000} = 0.375

Rational numbers whose decimal expansions are non-terminating recurring (repeating)

Rational numbers in the form p/q where q is not of the form 2n×5m, will not have terminating decimal expansion.

e.g.,\frac{1}{7} = 0.\overline{142857}

Theorem: Let x = p/q be a rational number, such that the prime factorisation of q is not of the form 2^n\times 5^m, where n, m are non-negative integers. Then x has a decimal expansion which is non-terminating recurring (repeating).

e.g., \frac{45}{210} = \frac{3^2 \times 5}{2\times 3\times 5\times 7} = \frac{3}{2\times 7}

Here, q = 2^1\times 7\, (\neq 2^n\times 5^m)

\implies \frac{45}{210} will not terminate and,

=0.169333... = 0.169\overline3

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