\left( {\sqrt p } \right)^2 ={ \Large{{\left( {{a \over b}} \right)^2}}}. An essential part of this proof is to further assume that the greatest common divisor of a and b is 1. Answer: By prime factorisation, we know: 81 = 3 x 3 x 3 x 3. Since \large{\sqrt p } is a rational number, we can express it as a ratio/fraction of two positive integers \large{\sqrt p = }\Large{{a \over b}} where a and b belong to the set of positive integers, b is not equal to zero, and the Greatest Common Divisor (GCD) of a and b is 1. Therefore, we have proved that the square root of a prime number is irrational. Suppose a = 3,780. The reason is to demonstrate or illustrate by example the Fundamental Theorem of Arithmetic which is central to the proof of this theorem. The next step is to square the integer \color{blue}\large{a}, thus we have \color{blue}\large{a^2}. This contradicts our assumption of our main equation {a^2}={\color{red}p}\,{b^2} where {\color{red}p}\,{b^2} must contain only prime numbers with even powers. Conjecture: Every composite number has a proper factor less than or equal to its square root. PROOF: Let’s assume by contradiction that \large{\sqrt p } is rational where \large{p} is prime. Observe: The exponents of the unique prime factors of integer \large{a} are either even or odd integers. THEOREM: If \large{p} is a prime number, then \large{\sqrt p } is irrational. I want to move around the equation so that it is much easier to understand what it is trying to say. In a nutshell, this exponent rule will allow us to distribute the outermost exponent 2 to the exponents of the unique prime numbers inside the parenthesis. That is, \large{{b^2} = {\left( {q_1^{{m_1}}\,q_2^{{m_2}}\,q_3^{{m_3}}\,q_4^{{m_4}}\,q_5^{{m_5}} \cdot \cdot \cdot q_m^{{m_k}}} \right)^2}} which can be simplified as \large{{b^2} = q_1^{2{m_1}}\,q_2^{2{m_2}}\,q_3^{2{m_3}}\,q_4^{2{m_4}}\,q_5^{2{m_5}} \cdot \cdot \cdot q_m^{2{m_k}}}. In other words, the prime factorization of an integer is so unique because each prime factor always appears in the same amount or quantity thereby the arrangement doesn’t matter. Let us see some examples here: Square root of 81. Notice how we removed duplicates of prime numbers by expressing it as factors of primes with each unique prime having the appropriate power. Now, if we square a, we get \large{{a^2} = {\left( {p_1^{{n_1}}\,p_2^{{n_2}}\,p_3^{{n_3}}\,p_4^{{n_4}}\,p_5^{{n_5}} \cdot \cdot \cdot p_n^{{n_j}}} \right)^2}} then simplify by multiplying the outermost exponent which is \color{red}2 to each and every exponent of the unique prime factor to obtain \large{{a^2} = p_1^{2{n_1}}\,p_2^{2{n_2}}\,p_3^{2{n_3}}\,p_4^{2{n_4}}\,p_5^{2{n_5}} \cdot \cdot \cdot p_n^{2{n_j}}}. a = 2 ⋅ 2 ⋅ 3 ⋅ 3 ⋅ 3 ⋅ 5 ⋅ 7. a = 2 \cdot 2 \cdot 3 \cdot 3 \cdot 3 \cdot 5 \cdot 7 a = 2 ⋅ 2 ⋅ 3 ⋅ 3 ⋅ 3 ⋅ 5 ⋅ 7. Again, this contradicts the supposition of our main equation {a^2}={\color{red}p}\,{b^2} where {\color{red}p}\,{b^2} must contain only prime numbers with even powers. The Bottom Line: In both cases, we have contradictions because a^2 implies that its unique prime factors must have even powers. This equation is asking to be squared on both sides, and see what we can make sense of it after doing so. Notice that in the prime factorization of integer \color{blue}\large{a}, the prime numbers can either have an odd or even exponent. That is, let \color{red}p be a prime number and \sqrt {\color{red}p} is a rational number. a = 2 2 ⋅ 3 3 ⋅ 5 ⋅ 7. Notice that each of them is only divisible by \large{1} and itself. Thus, the resulting \large{p} has an odd power which is 2k+1. It is the case since the product of 2 and any integer will always be an even number.