We are now ready to put the strategy of the proof together. Now, squaring both sides of the equation, we obtain \large{p =} \Large{{{{a^2}} \over {{b^2}}}}. That is, let p be a prime number then prove that \sqrt p is irrational. Apply the Power of a Power Rule of Exponent. In order to have uniformity in our application of the fundamental theorem of arithmetic, we have to agree that we write the prime factors of an integer in ascending order. (To verify further, try factors of other composite numbers.). 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. 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. We will assume the negation (or opposite) of the original statement to be true. 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. In our previous lesson, we proved by contradiction that the square root of 2 is irrational. This equation is asking to be squared on both sides, and see what we can make sense of it after doing so. This contradicts our assumption that . Similarly, we can do the same for integer b. 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. I hope you agree that the equation above is exactly the same as the one below. 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. No need to test divisibility by 5 up to 16. Now, let’s prime factorize both integers a and b. Let’s have an example to amplify what I meant above. To determine whether a number is prime or not, we have to divide it by all numbers between 1 and itself . 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. Pairing the numbers to get the perfect squares we get; 81 = 9 x 9 = 9 2… I want to move around the equation so that it is much easier to understand what it is trying to say. Suppose a = 3,780. A number that is not prime is composite. You have probably noticed that there is some sort of “middle number” in both tables denoted by the red-colored text. Before proceeding with primes, let us examine the behavior of the factors (or divisors) of composite numbers. The reason is to demonstrate or illustrate by example the Fundamental Theorem of Arithmetic which is central to the proof of this theorem. To see it for yourself, below is the list of the first ten (10) prime numbers. A number that is not prime is composite. So the next logical step is to consider the two possible cases/scenarios below. THEOREM: If \large{p} is a prime number, then \large{\sqrt p } is irrational. First, let $s=\phi(p)$ where $\phi()$ is the Euler's totient function. The next step is to square the integer \color{blue}\large{a}, thus we have \color{blue}\large{a^2}. For instance, we have the pairs (1,12), (2,6), and so on as factors of 12 as shown in the first table. The next logical step is to generalize the factorization of any integer greater than 1 using the fundamental theorem of arithmetic. A prime number is a integer greater than that is divisible only by 1 and itself. To test that $a$ is a primitive root of $p$ you need to do the following. An essential part of this proof is to further assume that the greatest common divisor of a and b is 1. Why did we spend some time prime factorizing the integer above? To do that, we will need to revisit Equation #3. We can condense the prime factorization by rewriting it as. The exponent tells how many times the prime number appears in the prime factorization. This is obviously a contradiction. Notice that each of them is only divisible by \large{1} and itself. ◼︎, a = 2 \cdot 2 \cdot 3 \cdot 3 \cdot 3 \cdot 5 \cdot 7, 100 = 2 \cdot 2 \cdot 5 \cdot 5 = {2^2} \cdot {5^2}, 126 = 2 \cdot 3 \cdot 3 \cdot 7 = 2 \cdot {3^2} \cdot 7, 5,070 = 2 \cdot 3 \cdot 5 \cdot 13 \cdot 13 = 2 \cdot 3 \cdot 5 \cdot {13^2}, 12,375 = 3 \cdot 3 \cdot 5 \cdot 5 \cdot 5 \cdot 11 = {3^2} \cdot {5^3} \cdot 11, \large{p =} \Large{{{{a^2}} \over {{b^2}}}}, \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}}}, \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}}, \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}}}, \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}}}, \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}}, \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}}}, p\left( {p_n^{2{n_j}}} \right) = {p^{1 + 2{n_j}}}. School math, multimedia, and technology tutorials. This means, we will have to examine if in {\color{red}p}\,{b^2} that \color{red}p occurs or does not occur in the prime factorizations of b^2. Notice that in the prime factorization of integer \color{blue}\large{a}, the prime numbers can either have an odd or even exponent. Thew following steps will be useful to find square root of a number by prime factorization. In this discussion, the word “numbers” refer to positive integers. We multiply both sides by b^2 to get rid of the denominator, thus we get p\,{b^2} = {a^2}. Four is the square root of 16, and testing more perfect squares (the reader is encouraged to do so) will confirm the observation. This gives us {a^2}={\color{red}p}\,{b^2}. Multiply both sides of the equation by b^2 to get rid of the denominator. 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. 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. How to Create Math Expressions in Google Forms, 5 Free Online Whiteboard Tools for Classroom Use, 50 Mathematics Quotes by Mathematicians, Philosophers, and Enthusiasts, 8 Amazing Mechanical Calculators Before Modern Computers, More than 20,000 mathematics contest problems and solutions, Romantic Mathematics: Cheesy, Corny, and Geeky Love Quotes, 29 Tagalog Math Terms I Bet You Don't Know, Prime or Not: Determining Primes Through Square Root, Solving Rational Inequalities and the Sign Analysis Test, On the Job Training Part 2: Framework for Teaching with Technology, On the Job Training: Using GeoGebra in Teaching Math, Compass and Straightedge Construction Using GeoGebra. Please click OK or SCROLL DOWN to use this site with cookies. 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. Answer: By prime factorisation, we know: 81 = 3 x 3 x 3 x 3. The left side of the equation is \large{a^2}. This time, we are going to prove a more general and interesting fact. To prove this theorem, we will use the method of Proof by Contradiction. The Bottom Line: In both cases, we have contradictions because a^2 implies that its unique prime factors must have even powers. This is the most important observation that we can take from this step. 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. Notice: The same observation from integer \large{a} can be drawn for integer \large{b}. How you test. For example, use the square root calculator below to find the square root of 5. We can always find the square root of perfect numbers using the prime factorisation method. Typically, what you do is you pick a number and test. 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. Otherwise, check your browser settings to turn cookies off or discontinue using the site. Conjecture: Every composite number has a proper factor less than or equal to its square root. Observe: The exponents of the unique prime factors of integer \large{a} are either even or odd integers. In a nutshell, this is the meaning of Equation #3 above. The prime number \large\color{red}p is not included (excluded) in the unique prime factorization of \large{b^2}. a = 3,780 a = 3,780. The result includes lots of numbers after the decimal point. The theorem further asserts that each integer has a unique prime factorization thus it has a distinct combination or mix of prime factors. a = 2 2 ⋅ 3 3 ⋅ 5 ⋅ 7. 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. We only need to test the divisibility up to 4 and we have already all the factors. Then we square the prime factorization of integer \large{a}. 1. 2. Just a side-note, the number 2 is the only EVEN prime number. Thus, \large\color{red}p has an odd power which is 1. By the same token, we will factorize integer. Dividing a number by all numbers between 1 and itself is burdensome especially for large numbers. Proof: We use proof by contradiction. This is what we got. For the numbers above, the square root was equal to an integer. For example, to say that 257 is prime, we must be sure that it is not divisible by any number between 1 and 257. 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. This is an important observation that we will take advantage of later. The powers of the unique prime factors are even or odd numbers. \left( {\sqrt p } \right)^2 ={ \Large{{\left( {{a \over b}} \right)^2}}}. Recall that in Equations #1, #2, and #3, it is clear that we must show how to factorize integers a and b in general form. Let us see some examples here: Square root of 81. A prime number is a integer greater than that is divisible only by 1 and itself. 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 also the similar in the second example. PROOF: Let’s assume by contradiction that \large{\sqrt p } is rational where \large{p} is prime. We will also use the proof by contradiction to prove this theorem. That is, let \color{red}p be a prime number and \sqrt {\color{red}p} is a rational number. Notice that the 4th, 5th, and 6th pairs are just “reverses” of the first three pairs. 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