The left side of the equation is \large{a^2}. Then, we can write , where and are both between and . As we have shown before in this lesson, the prime factorization of \large{a^2} is a product of unique prime numbers with even powers. Notice that each of them is only divisible by \large{1} and itself. To determine whether a number is prime or not, we have to divide it by all numbers between 1 and itself . Rearrange the equation by moving the expression on the left to the right, and the one on the right to the left. After the middle number, the pairs are mere repetitions of the other pairs. (i) Decompose the number inside the square root into prime factors. Notice how we removed duplicates of prime numbers by expressing it as factors of primes with each unique prime having the appropriate power. This is what we got. (To verify further, try factors of other composite numbers.). Dividing a number by all numbers between 1 and itself is burdensome especially for large numbers. This is obviously a contradiction. So the next logical step is to consider the two possible cases/scenarios below. An essential part of this proof is to further assume that the greatest common divisor of a and b is 1. (ii) Inside the square root, for every two same numbers multiplied, one number can be taken out of the square root. it is not always possible to get the square root as an integer. The Bottom Line: In both cases, we have contradictions because a^2 implies that its unique prime factors must have even powers. However, we expect a contradiction such that we discard the assumption, and therefore claim that the original statement must be true, which in this case, the square root of a prime number is irrational. Answer: By prime factorisation, we know: 81 = 3 x 3 x 3 x 3. A prime number is a integer greater than that is divisible only by 1 and itself. This is also the similar in the second example. We can condense the prime factorization by rewriting it as a = {2^2} \cdot {3^3} \cdot 5 \cdot 7. From above, we can conjecture that every composite number has a factor less than or equal to its square root. Observe: The effect or consequence of squaring the unique prime factors of integer \large{a} (raising to the power of 2) is that all exponents become even numbers since 2 multiplied by any integer will always be an even number. The exponents of the prime factors of \large{b} are either even or odd. To prove this theorem, we will use the method of Proof by Contradiction. 1. The result includes lots of numbers after the decimal point. In a nutshell, this is the meaning of Equation #3 above. a = 2 2 ⋅ 3 3 ⋅ 5 ⋅ 7. 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. How you test. This time, we are going to prove a more general and interesting fact. Now, let’s prime factorize both integers a and b. PROOF: Let’s assume by contradiction that \large{\sqrt p } is rational where \large{p} is prime. Now, squaring both sides of the equation, we obtain \large{p =} \Large{{{{a^2}} \over {{b^2}}}}. Prime or Not: Determining Primes Through Square Root. Thew following steps will be useful to find square root of a number by prime factorization. That is, let p be a prime number then prove that \sqrt p is irrational. Answer: If \large\color{red}p does not occur in the prime factorization of \large{b^2}, then \large\color{red}p must stand on its own. Conclusion: Since we have reached contradictions in both cases, we shall reject the assumption that \large\color{red}{\sqrt p } is rational and therefore the original statement must be true that \large\color{red}{\sqrt p } is irrational. This is the result. 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}}}. To see it for yourself, below is the list of the first ten (10) prime numbers. We can condense the prime factorization by rewriting it as. Apply the Power of a Power Rule of Exponent. The reason is to demonstrate or illustrate by example the Fundamental Theorem of Arithmetic which is central to the proof of this theorem. Proof: We use proof by contradiction. Typically, what you do is you pick a number and test. Suppose a = 3,780. Case 1: Consider that \color{red}p occurs in the prime factorization of integer b^2, that means we have p\left( {p_n^{2{n_j}}} \right) = {p^{1 + 2{n_j}}} which is a prime number with an odd power. Now, the line of thought is to prove that \sqrt {\color{red}p} is rational. 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. To determine whether a number is prime or not, we have to divide it by all numbers between 1 and itself . For instance, we have the pairs (1,12), (2,6), and so on as factors of 12 as shown in the first table. Let us see some examples here: Square root of 81. But first, let’s define a prime number. Breaking it down as a product of prime numbers, we get a = 2 \cdot 2 \cdot 3 \cdot 3 \cdot 3 \cdot 5 \cdot 7. Answer: If \large\color{red}p occurs in the prime factorization of \large{b^2}, then we have \large\color{red}p times \large{{p^{\,2k}}} which is equal to \large{{p^{\,2k + 1}}}. Since a is a positive integer greater than 1 then you can express it as a product of unique prime numbers with even or odd powers. To do that, we will need to revisit Equation #3. For the numbers above, the square root was equal to an integer. Four is the square root of 16, and testing more perfect squares (the reader is encouraged to do so) will confirm the observation. It implies that the right-hand side of the equation with the expression \large{p\,{b^2}} must also be a product of unique prime factors with even exponents. Conjecture: Every composite number has a proper factor less than or equal to its square root. The prime number \large{\color{red}p} occurs in the unique prime factorization of \large{b^2}. Breaking it down as a product of prime numbers, we get. We are now ready to put the strategy of the proof together. If $p$ is prime, then $s=p-1$. Since we assume that \sqrt p is rational, it means there exists two positive integers a and b but b \ne 0 that we can express as a ratio like the one below. Observe: The exponents of the unique prime factors of integer \large{a} are either even or odd integers.