Answer: By prime factorisation, we know: 81 = 3 x 3 x 3 x 3. Why did we spend some time prime factorizing the integer above? We can write it symbolically in math as \gcd \left( {a,b} \right) = 1. 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}}}. You have probably noticed that there is some sort of “middle number” in both tables denoted by the red-colored text. Now, let’s prime factorize both integers a and b. Prime or Not: Determining Primes Through Square Root. We can condense the prime factorization by rewriting it as. This is an important observation that we will take advantage of later. Notice: The same observation from integer \large{a} can be drawn for integer \large{b}. This equation is asking to be squared on both sides, and see what we can make sense of it after doing so. A number that is not prime is composite. We can always find the square root of perfect numbers using the prime factorisation method. There is no general formula to find a primitive root. Proof: We use proof by contradiction. From above, we can conjecture that every composite number has a factor less than or equal to its square root. To determine whether a number is prime or not, we have to divide it by all numbers between 1 and itself . Observe: The exponents of the unique prime factors of integer \large{a} are either even or odd integers. The next logical step is to generalize the factorization of any integer greater than 1 using the fundamental theorem of arithmetic. This is also the similar in the second example. However, by raising a to the power of 2, a^2 must have prime factorizations wherein each unique prime number will have an even exponent. Since {a^2}={\color{red}p}\,{b^2}, it implies that right hand side of the equation which is {\color{red}p}\,{b^2} is comprised of unique prime factors with even powers. The powers of the unique prime factors are even or odd numbers. Please click OK or SCROLL DOWN to use this site with cookies. This contradicts our assumption that . Then, we can write , where and are both between and . To see it for yourself, below is the list of the first ten (10) prime numbers. Four is the square root of 16, and testing more perfect squares (the reader is encouraged to do so) will confirm the observation. Now, squaring both sides of the equation, we obtain \large{p =} \Large{{{{a^2}} \over {{b^2}}}}. The exponents of the prime factors of \large{b} are either even or odd. Sometimes, you may get a real number when finding the square root. But upon squaring it, the exponents all become even numbers. In this post, we discuss a shorter way of determining if a number is prime and explain why the method works. A prime number is a integer greater than that is divisible only by 1 and itself. Case 2: Consider that \color{red}p does not occur in the prime factorization of integer b^2, this means the prime number \color{red}p is unique and does not have the same copy in the prime factorization of b^2 which means \color{red}p has an odd power of one since \color{red}p^1. Thus, after squaring the integer a, we can clearly see that the exponents of all unique prime factors become all even numbers since any integer multiplied by 2 is always an even number. If $p$ is prime, then $s=p-1$. First, let $s=\phi(p)$ where $\phi()$ is the Euler's totient function. Conjecture: Every composite number has a proper factor less than or equal to its square root. Suppose a = 3,780. 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. I want to move around the equation so that it is much easier to understand what it is trying to say. We use cookies to give you the best experience on our website. This is the result. We are now ready to put the strategy of the proof together. In this discussion, the word “numbers” refer to positive integers. The left side of the equation is \large{a^2}. 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. After the middle number, the pairs are mere repetitions of the other pairs. Notice that each of them is only divisible by \large{1} and itself. We will use this property to square the integer \color{blue}\large{a}. The Fundamental Theorem of Arithmetic states that for every integer n greater than one, n > 1, we can express it as a prime number or product of prime numbers. Notice how we removed duplicates of prime numbers by expressing it as factors of primes with each unique prime having the appropriate power. Once you find one primitive root, you find all the others. For a, we have \large{a = 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}}}. A number that is not prime is composite. We will assume the negation (or opposite) of the original statement to be true. Here is its unique prime factorizations: \large{b = 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}}}. In our previous lesson, we proved by contradiction that the square root of 2 is irrational. In effect, we just need to include the first three pairs to be sure that we have all the factors. Suppose is composite. How you test. 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. Before proceeding with primes, let us examine the behavior of the factors (or divisors) of composite numbers. 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. This gives us {a^2}={\color{red}p}\,{b^2}. So the next logical step is to consider the two possible cases/scenarios below. 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. A prime number is a positive integer greater than 1 that has exactly two positive integer divisors: namely, 1 and itself. For example, use the square root calculator below to find the square root of 5. THEOREM: If \large{p} is a prime number, then \large{\sqrt p } is irrational. The prime number \large\color{red}p is not included (excluded) in the unique prime factorization of \large{b^2}. For instance, we have the pairs (1,12), (2,6), and so on as factors of 12 as shown in the first table. 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. This is obviously a contradiction. After squaring the integer \color{blue}\large{a}, the exponents of the unique prime factors of \color{blue}\large{a} are now ALL even numbers. We only need to test the divisibility up to 4 and we have already all the factors. The next step is to square the integer \color{blue}\large{a}, thus we have \color{blue}\large{a^2}. Now, the line of thought is to prove that \sqrt {\color{red}p} is rational. 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. In a nutshell, this is the meaning of Equation #3 above. An essential part of this proof is to further assume that the greatest common divisor of a and b is 1. Thus, \large\color{red}p has an odd power which is 1. We can condense the prime factorization by rewriting it as a = {2^2} \cdot {3^3} \cdot 5 \cdot 7. (To verify further, try factors of other composite numbers.). We will also use the proof by contradiction to prove this theorem. 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. This is unquestionably a contradiction as well. This time, we are going to prove a more general and interesting fact. 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.