Prove that √5 is irrational

3:55, p wouldn't even need to be prime to continue with the proof. Sal chose p to be one single factor, but really p could be a series of factors, where at least one of the factors is not a duplicate. Ie, in the list of factors of a^2 he has there, p could be f1 * f1 * f3. This new p isn't prime since it has three factors, and in fact even 2 of the factors are duplicated (f1 and f1), but as long as one factor is unique (f3) then we can say a must be a multiple of p and the proof continues on. Doesn't this show that the square root of a number is irrational if that number is made of anything other than pairs of duplicated factors? No, I believe the proof would break down in this case, and here's why. You wrote "in the list of factors of a^2 he has there, p could be f1 * f1 * f3." Obviously, this would mean that p is a factor of a^2. But in the case of your p, which isn't a prime number anymore, we wouldn't automatically know whether p is a factor of a as well. In your example, we'd need to know that the factors of a^2 contained f1 * f1 * f1 * f1 * f3 * f3 i.e. we'd need to know that p^2 is a factor of a^2 to know that p is a factor of a Otherwise it could happen for example that the factors of a^2 are p * f3 = (f1 * f1 * f3) * f3, making a = f1 * f3 a is still rational, but p is not among the factors of a, ergo a is not a multiple of p, and the proof doesn't work from this point on EDIT the proof works as long as you set your new p as only the unpaired factor (in this examp...

Transcript Question 7 (OR 2nd question) Show that 7 − √5 is irrational, give that √5 is irrational. We have to prove 7 − √5 is irrational Let us assume the opposite, i.e., 7 − √5 is rational Hence, 7 − √5 can be written in the form 𝑎/𝑏 where a and b (b≠ 0) are co-prime (no common factor other than 1) Hence, 7 − √5 = 𝑎/𝑏 7 – 𝑎/𝑏 = √5 √5 = 7 – 𝑎/𝑏 √5 = (7𝑏 − 𝑎)/𝑏 Here, (7𝑏 − 𝑎)/𝑏 is a rational number But √5 is irrational Since, Rational ≠ Irrational This is a contradiction ∴ Our assumption is incorrect Hence, 7 − √5 is irrational Hence proved. Show More

Prove that √5 is irrational and hence prove that (2 - √5) is also irrational. Rational numbers are integers that are expressed in the form of p / q where p and q are both co-prime numbers and q is non zero. Answer: Hence proved that√5 and(2 - √5) areirrational numbers Let's find if(2 - √5)is Explanation: Let us assume that√5is a ⇒√5= p / q On squaring both sides we get, ⇒ 5q 2 = p 2 ⇒ p 2is a primenumber that divides q. Therefore, p is a primenumber that divides q Let p = 5x where x is a By substituting the value of p in 5q 2 = p 2, we get ⇒ 5q 2 = (5x) 2 ⇒ 5q 2 = 25 x 2 ⇒q 2 = 5 x 2 ⇒q 2is a primenumber that divides x. Therefore, qis a Since p and q both are primenumbers with 5 as a common multiple which means that p and q are not This leads to a contradiction that root 5is a rational number in the form of p / q with p and q both co-prime numbers andq≠ 0 Now, to prove that(2 - √5) is an irrational number, we will again use the contradiction method Let us assume that(2 - √5)is a rational number with p and q asco-prime integers and q≠ 0 ⇒(2 - √5) = p / q ⇒√5= 2 - p/ q ⇒√5= (2q - p) / q ⇒(2q - p) / q is a rational number However,√5 is an irrational number This leads to a contradiction that (2 - √5)is a rational number Thus,(2 - √5) is an irrational number by contradiction method

Let us assume, to the contrary, that 2√5 − 3 is a rational number ∴ 2√5 − 3 = p/q, where pand qare integers and q ≠ 0 ⇒√5 = p+3q/2q... (1) Since pand qare integers ∴ p +3a/2qis a rational number ∴ √5 is a rational number which is a contradiction as √5 is an irrational number Hence our assumption is wrong and hence 2√5 − 3 is an irrational number.

Prove that Root 5 is Irrational Number Is root 5 an irrational number? A number that can be represented in p/q form where q is not equal to 0 is known as a rational number whereas numbers that cannot be represented in p/q form are known as irrational numbers. An irrational number can also be denoted as a number that does not terminate and keeps extending after the decimal point. Now that we know about rational and irrational numbers let us take a look at the detailed discussion and prove that root 5 is irrational. 1. Prove that Root 5 is Irrational Number 2. Prove That Root 5 is Irrational by Contradiction Method 3. Prove That Root 5 is Irrational by Long Division Method 4. Solved Examples 5. FAQs on Is Root 5 an Irrational? Prove that Root 5 is Irrational Number Problem statement: Prove that Root 5 is Irrational Number Given: The number 5 Proof: On calculating the value of √5, we get the value √5 = 2.23606797749979...As discussed above a decimal number that does not terminate after the decimal point is also an irrational number. The value obtained for the root of 5 does not terminate and keeps extending further after the decimal point. This satisfies the condition of √5 being an irrational number. Hence, √5 is an irrational number. The square root of 5 is commonly also called "root 5". The root of a number "n" is represented as √n. Thus, we define the root of a number as the number that on multiplication to itself gives the original number. For example, √5 on multiplicati...

Please help me prove that $(√5 - 1)/2$ is irrational. I know how to prove √5 is irrational: Assume that √5 is rational meaning √5 = $p/q$ $p,q$ $are$ $Z$ $and$ $q≠0$ $p^2/q^2 = 5$ $q^2 = p^2/5$ Therefore, 5 must be a factor of p. $Let$ $p = 5c$ $q^2 = (5c)^2/5$ $q^2 = 25c^2/5$ $c^2 = q^2/5$ Therefore, 5 must also be a factor of q. CONTRADICTION! Thanks. Try This : (If you know how to prove $\sqrt $ is irrational.

Transcript Ex 1.2 , 1 Prove that 5 is irrational. We have to prove 5 is irrational Let us assume the opposite, i.e., 5 is rational Hence, 5 can be written in the form / where a and b (b 0) are co-prime (no common factor other than 1) Hence, 5 = / 5b = a Squaring both sides ( 5b)2 = a2 5b2 = a2 ^2/5 = b2 Hence, 5 divides a2 So, 5 shall divide a also Hence, we can say /5 = c where c is some integer So, a = 5c Now we know that 5b2 = a2 Putting a = 5c 5b2 = (5c)2 5b2 = 25c2 5b2 = 25c2 b2 = 1/5 25c2 b2 = 5c2 ^2/5 = c2 Hence 5 divides b2 So, 5 divides b also By (1) and (2) 5 divides both a & b Hence 5 is a factor of a and b So, a & b have a factor 5 Therefore, a & b are not co-prime.Hence, our assumption is wrong By contradiction, 5 is irrational Show More