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- #1

#### pankaz712

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How to prove that root n is irrational, if n is not a perfect square. Also, if n is a perfect square then how does it affect the proof.

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- #2

#### zketrouble

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If a root n is a perfect square such as 4, 9, 16, 25, etc., then it evaluating the square root quickly proves it is rational. The square root of the perfect square 25 is 5, which is clearly a rational number.

To prove a root is irrational, you must prove that it is inexpressible in terms of a fraction a/b, where a and b are whole numbers.

For the nth root of x to be rational:

nth root of x must equal (a^n)/(b^n), where a and b are integers and a/b is in lowest terms.

So the square root of 3 for example would be done like this:

2nd root of 3 must equal (a^2)/(b^2)

3 = (a^2)/(b^2)

3(b^2) = (a^2) So we know a^2 is divisible by three.

(b^2) = (a^2)/3 So we know b^2 is divisble by three.

Since this is neither of these are in lowest terms, the sqrt(3) is an irrational number.

- #3

#### VietDao29

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pankaz712 said:

How to prove that root n is irrational, if n is not a perfect square. Also, if n is a perfect square then how does it affect the proof.

Can you show us what you tried, so that we may know where you got stuck, and help you? It's against the forums' policies to provide help, unless you show us that you did put some efforts in solving the problem. So, just give the problem a try, and let's see how far you can get.

In case, you don't know where to start, then I'll give you a little push: try *Proof by Contradiction*.

- #4

#### pankaz712

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VietDao29 said:

Can you show us what you tried, so that we may know where you got stuck, and help you? It's against the forums' policies to provide help, unless you show us that you did put some efforts in solving the problem. So, just give the problem a try, and let's see how far you can get.

In case, you don't know where to start, then I'll give you a little push: try

Proof by Contradiction.

OK here is how i approached it:

n can either be a prime number or a composite

1) if n is prime:

a) assume root n is rational. therefore root n = p/q( where p, q are co-prime)

then n= p^2/q^2,

p^2= n *q^2

n divides p^2, therefore n divides p. Now for some integer k, p=nk

(nk)^2 = n*q^2

nk^2= q^2

n divides q^2, therefore n divides q. Now n is a common factor for p & q. But we know that p & q are co-prime. Hence our assumption is wrong, root n is irrational.

2) if n is a composite. Let's say it is product of two primes c1 & c2. The proof remains the same for any number of primes.

a) assume root n is rational. therefore root n =p/q( where p,q are co-prime)

c1c2=p^2/q^2

p^2= c1c2*q^2

c1 divides p^2 therefore divides p. Now for some k, p= c1k

(c1k)^2 = c1c2*q^2

c1k^2 = c2* q^2

Now c1 divides L.H.S. , therefore must divide R.H.S. It won't divide c2( its a prime). Therefore it divides q^2. Therefore c1 divides q.

Now c1 is a common factor for p&q. But we know that p&q are co-prime. Hence our assumption is wrong.

- #5

#### VietDao29

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pankaz712 said:

OK here is how i approached it:

n can either be a prime number or a composite

1) if n is prime:

a) assumeroot n is rational. therefore root n = p/q( where p, q are co-prime)

then n= p^2/q^2,

p^2= n *q^2

n divides p^2, therefore n divides p. Now for some integer k, p=nk

(nk)^2 = n*q^2

nk^2= q^2

n divides q^2, therefore n divides q. Now n is a common factor for p & q. But we know that p & q are co-prime. Hence our assumption is wrong, root n is irrational.(Video) Proof that square root of 2 is irrational | Algebra I | Khan Academy

Yup, this is good. :) There's only one error, the word '*root*' should read '*square root*' instead.

2) if n is a composite. Let's say it is product of two primes c1 & c2.

Well, no, this is **not** true. Not every composite number can be split into the product of 2 primes. It can be a product of 3, 4 or even more primes. Say: 30 = 2.3.5 (a product of 3 primes)

Well, you don't really need to split it into 2 separated cases like that. Assume that: [tex]\sqrt{n}[/tex] is rational, i.e:

[tex]\sqrt{n} = \frac{a}{b} \mbox{ , where } a, b \mbox{ are co-prime}[/tex]

[tex]\Rightarrow n = \frac{a ^ 2}{b ^ 2}[/tex]

[tex]\Rightarrow a ^ 2 \vdots b ^ 2[/tex]

Now, let's think a little bit. From here, what can you say about *a*, and *b*?

- #6

#### pankaz712

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VietDao29 said:

Well, no, this is

nottrue. Not every composite number can be split into the product of 2 primes. It can be a product of 3, 4 or even more primes. Say: 30 = 2.3.5 (a product of 3 primes)

yes i see that not all composite numbers can be split into the products of only 2 primes. But the proof remains the same even if when the number can be split into " n" number of primes.

- #7

#### VietDao29

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pankaz712 said:

yes i see that not all composite numbers can be split into the products of only 2 primes. But the proof remains the same even if when the number can be split into " n" number of primes.

Well, but what if the power of the prime is not 1? What I mean is, something like this:

200 = 2^{3}5^{2}.

I doubt that the proof still remains the same.

- #8

#### pankaz712

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VietDao29 said:

Well, but what if the power of the prime is not 1? What I mean is, something like this:

200 = 2^{3}5^{2}.

I doubt that the proof still remains the same.

Yes...i believe u are right. I will surely check again and see. Please could u explain your method of proving it.

- #9

#### VietDao29

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VietDao29 said:

Well, you don't really need to split it into 2 separated cases like that. Assume that: [tex]\sqrt{n}[/tex] is rational, i.e:

[tex]\sqrt{n} = \frac{a}{b} \mbox{ , where } a, b \mbox{ are co-prime}[/tex]

[tex]\Rightarrow n = \frac{a ^ 2}{b ^ 2}[/tex]

[tex]\Rightarrow a ^ 2 \vdots b ^ 2[/tex]Now, let's think a little bit. From here, what can you say about

a, andb?

You can look at the post #5 above. Note that, what you are trying to show is that *n* **must** be a *perfect square*.

Now if [tex]a ^ 2 \vdots b ^ 2[/tex], what can you say about *a*, and *b*? In my proof, the symbol [tex]\vdots[/tex] stands for 'divisible by'.

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## FAQs

### Proving Root n is Irrational: Perfect Square Affects Proof? ›

To prove a root is irrational, you must **prove that it is inexpressible in terms of a fraction a/b, where a and b are whole numbers**. For the nth root of x to be rational: nth root of x must equal (a^n)/(b^n), where a and b are integers and a/b is in lowest terms. 3(b^2) = (a^2) So we know a^2 is divisible by three.

**Is the root n irrational if n is a perfect square? ›**

Answer: **False** because if n is a perfect square the answer would in whole number or a integer.

**Are all non perfect square roots irrational proof? ›**

**The square roots of numbers that are not a perfect square are members of the irrational numbers**. This means that they can't be written as the quotient of two integers. The decimal form of an irrational number will neither terminate nor repeat.

**How to use proof by contradiction to show that √ 3 is irrational? ›**

How do you Prove that Root 3 is Irrational? Root 3 is irrational is proved by the method of contradiction. **If root 3 is a rational number, then it should be represented as a ratio of two integers**. We can prove that we cannot represent root is as p/q and therefore it is an irrational number.

**How do you prove 3 2 √ 5 is irrational? ›**

Since (a - 3b)/2b is a rational number, then √5 is also a rational number. But, we know that √5 is irrational. Therefore, **our assumption was wrong that 3 + 2√5 is rational**. Hence, 3 + 2√5 is irrational.

**What is the probability that n is a perfect square? ›**

**If n tends to ∞ , the probability that a random number in the range [1,n] is a perfect square, tends to 0**. So , if we allow the random number to be any positive integer, the probability is 0.

**When can you say that a square root is rational or irrational? ›**

Sometimes, a square root — that is, a factor of a number that, when multiplied by itself, produces the number that you started with — is irrational number, **unless it's a perfect square that's a whole number**, such as 4, the square root of 16.

**Is every irrational number a perfect square? ›**

**No perfect square is an irrational number**, and no irrational number is a perfect square.

**How do you prove that a number is not a perfect square? ›**

All perfect squares end in 1, 4, 5, 6, 9 or 00 (i.e. Even number of zeros). Therefore, **a number that ends in 2, 3, 7 or 8 is not a perfect square**.

**Is N √ N is irrational? ›**

**If n is a natural number then √n may sometimes a natural number and sometime an irrational number**. E.g. If n = 2 then √n = √2 which is irrational and if n = 4, then √n = √4 = 2 which is a natural number.

### How do you prove that √ 2 is irrational by method of contradiction? ›

√2 = p/q, where 'p' and 'q' are integers, q ≠ 0 and p, q have no common factors (except 1). Thus, p and q have a common factor 2. This statement contradicts that 'p' and 'q' have no common factors (except 1). We can say that √2 is not a rational number.

**How can you prove that 2 3 √ 5 is an irrational number while it is given that √ 5 is an irrational number? ›**

2 + 3 5 = p q ⇒ 3 5 = p q - 2 ⇒ 3 5 = p - 2 q q ⇒ 5 = p - 2 q 3 q Since , p - 2 q 3 q is rational ⇒ 5 is rational But , it is given that 5 is an irrational number . Therefore , our assumption is wrong . Hence , 2 + 3 5 is an irrational number .

**Is √ 3 an irrational number assertion? ›**

The square roots of 3 is not a perfect square. So, **it is an irrational number.**

**How do you prove that 3+ 2 √ 3 is irrational? ›**

Let (3+2√3) be a rational number which can be expressed in the form of p/q where p and q have no other common factor than 1. (p-3q)/2q is rational but √3 is irrational. Thus our contradiction is wrong. Hence, (3+2√3) is an irrational number.

**How do you prove √ 5 is irrational? ›**

**Assuming √5 as a rational number, i.e., can be written in the form a/b where a and b are integers with no common factors other than 1 and b is not equal to zero**. It means that 5 divides a^{2}. This has arisen due to the incorrect assumption as √5 is a rational number. Therefore, √5 is irrational.

**What is n if n is not a perfect square number answer? ›**

If n is not a perfect square number, then n is an **irrational number**.

**What is the rule for perfect squares? ›**

Perfect squares are numbers or expressions that are the product of a number or expression multiplied to itself. 7 times 7 is 49, so 49 is a perfect square. **x squared times x squared equals x to the fourth**, so x to the fourth is a perfect square.

**What does it mean when n is perfect square? ›**

**A number expressed as a product of an integer by itself** is called a perfect square. Since the same number is multiplied twice, the perfect square is also written as the second exponent of an integer. Thus, the squares of all integers are known as perfect squares.

**Does a square root have to be a perfect square to be rational? ›**

Real numbers have two categories: rational and irrational. **If a square root is not a perfect square, then it is considered an irrational number**.

**How do you prove the square root of 7 is an irrational number? ›**

Also, the square root of 7 will be an irrational number **if it gives a value after the decimal point that does not terminate and does not repeat**. The value of root 7 is 2.64575131106...it is clear that it is non-terminating and nonrepeating, hence √7 an irrational number.

### Could a number ever be both rational and irrational? ›

**A number cannot be both rational and irrational**. It has to be one or the other. All rational numbers can be written as a fraction with an integer as the numerator and a non-zero integer as the denominator. A number can either be written this way, making it a rational number, or it can't, making it an irrational number.

**Can a rational number be a perfect square? ›**

102.01 is a rational number, and since there is another rational number 10.1, such that (10.1)^{2} = 102.01, **102.01 is a perfect square**. Since 102.01 is a rational number and the square root of 102.01 is a rational number (10.1), 102.01 is a perfect square.

**What is an example of an irrational square root? ›**

For example, **√5** is an irrational number. We can prove that the square root of any prime number is irrational. So √2, √3, √5, √7, √11, √13, √17, √19 … are all irrational numbers.

**Are all perfect squares positive proof? ›**

A perfect square is an integer which is the square of another integer n , that is, n2 . (**Since a negative times a negative is positive, a perfect square is always positive**.

**How can you use factors to show if a number is a perfect square? ›**

**When an expression has the general form a²+2ab+b², then we can factor it as (a+b)²**. For example, x²+10x+25 can be factored as (x+5)².

**Which of the following Cannot be a perfect square explain why? ›**

The unit digit of a perfect square can be only 0, 1, 4, 5, 6 or 9. Since, the unit digit of a perfect square cannot be 8. Therefore, **198 is not a perfect square.**

**How do you show that the root n is an irrational number? ›**

√n=p/q (q is not equal to 0 where p nd q are coprime integers). this shows p divides q. which is a contradiction. hence √n is irrational **if n is not a perfect square**.

**Is √ n irrational if n is prime number? ›**

and are also not divisible by any other number than itself. Clearly, as n is prime. So, **√n is irrational**.

**Are all root numbers irrational? ›**

In fact, **all square roots of natural numbers, other than of perfect squares, are irrational**. Like all real numbers, irrational numbers can be expressed in positional notation, notably as a decimal number. In the case of irrational numbers, the decimal expansion does not terminate, nor end with a repeating sequence.

**How do you prove a contradiction proof? ›**

**The steps taken for a proof by contradiction (also called indirect proof) are:**

- Assume the opposite of your conclusion. ...
- Use the assumption to derive new consequences until one is the opposite of your premise. ...
- Conclude that the assumption must be false and that its opposite (your original conclusion) must be true.

### How do you prove a number is irrational direct proof? ›

(A real number x is irrational if x≠r for each rational number r.) A real number is a certain sequence of rationals. To prove it irrational, **show that it is not equal to any p/q considered as a constant sequence**. To show inequality of two reals, show that entries at some point in the sequences are far enough apart.

**How do you prove 3 ✓ 5 irrational? ›**

- Let us assume that 3 + √ 5 is a rational number.
- So, it contradicts our assumption. Our assumption of 3 + √ 5 is a rational number is incorrect.
- 3 + √ 5 is an irrational number.
- Hence, it is proved that 3 + √ 5 is irrational.

**Is 2 √ 3 √ 5 irrational or rational? ›**

Expert-Verified Answer

we can get integers a&b in the form a/b. but this contradicts that 2√3+√5 is irrational. this contradiction has arisen due to the incorrect assumption that 2√3+√5 is rational . hence **it is irrational**.

**Is .3333 a irrational number? ›**

Answer and Explanation: The decimal 0.3333 is a **rational number**. It can be written as the fraction 3333/10,000.

**Is √ 2 √ 3 an irrational number assertion? ›**

Answer: Option (c) is correct. **The assertion that 2 √3 is an irrational number is true**, but the reason given that the sum of two irrational numbers is always an irrational number is false. It is possible for the sum of two irrational numbers to be a rational number.

**Is .33333 rational or irrational? ›**

The number 0. 33333… is a **rational number** because it can be re-written as 13 . Some numbers can't be rewritten as a fraction with integers, and so they are not rational numbers.

**How can you prove that 4 minus root 3 is an irrational number given that root 3 is an irrational number? ›**

**4 - 3 = p q ⇒ - 3 = p q - 4 ⇒ - 3 = p - 4 q q ⇒ 3 = 4 q - p q Since , 4 q - p q is rational ⇒ 3 is rational But , it is given that 3 is an irrational number .**

**Is 4 √ 5 an irrational number prove that? ›**

Here, 4q - p/q is rational but this equals to an irrational number i.e., √5 But this is impossible because a rational number can't be equal to an irrational number So this contradiction arrises because of our wrong assumption, Hence, 4 - √5 is irrational.

**Is ✓ 5 ✓ 3 an irrational number? ›**

Summary: Hence proved that 5 - √3 is an irrational number **using contradiction**.

**How do you prove that root 3 root 2 is an irrational number given that root 6 is an irrational number? ›**

Expert-Verified Answer

On squaring both sides , we get 2 + 3 + 2√6 = (a2/b2) So,5 + 2√6 = (a2/b2) a rational no. So, **2√6 = (a2/b2) – 5** Since, 2√6 is an irrational no. and (a2/b2) – 5 is a rational no.

### Is √ 3 √ 2 is rational or irrational? ›

∴ √3 + 2 is an **irrational number**. The correct option is 1 i.e. Irrational number. The Sum of an irrational number and a rational number is always an irrational number.

**How do you prove 4 2 root 3 is irrational number? ›**

4+ 2 √ 3 is a rational no., so it can be written in the form of p/q , where p and q are integers and q not equal to zero. **1/2 (p -4q/ q) is a rational no., as p and q are integers** . ... so, 4 +2 √ 3 is irrational.

**What if a perfect square is of n digits? ›**

If a perfect square is of n digits, then **its square root will have n/2 digit if n is even**. n is even. Therefore, the square root will have n/2 digits.

**Can an irrational number be a perfect square? ›**

**No perfect square is an irrational number**, and no irrational number is a perfect square. The square root of no perfect square is an irrational number, the square of no irrational number is a perfect square.

**Can we say that if a perfect square is of n digits then its square root will have n 1 2 digits if n is even or if n is odd? ›**

**The square root of a perfect square of n digits will have (n + 1)/2 digits, if n is odd**. Answer: True.

**Is it true that if n is a perfect square then 2n is also a perfect square? ›**

Answer: **2n cannot be a perfect square**.

**Is the nth root of n irrational? ›**

Does the n-th root of n have to be an irrational number, where n is a natural number greater than 1? **Yes**. Suppose n > 1 and ^{n}√n is rational = a/b, with gcd(a,b)=1. Since gcd(a,b) = 1, also gcd(a^{n},b^{n}) = 1, so a^{n} divides n, meaning a^{n} ≤ n.

**Is a number a natural number if it is irrational? ›**

Irrational numbers and rational numbers are two distinct classifications — **a rational number (and integers, whole numbers, or natural numbers) can't be irrational**. Rational numbers and irrational numbers together make up the real numbers.

**What is an irrational n? ›**

What are Irrational Numbers? An irrational number is **a real number that cannot be expressed as a ratio of integers**; for example, √2 is an irrational number. We cannot express any irrational number in the form of a ratio, such as p/q, where p and q are integers, q≠0.

**How do you prove that if M and n are perfect squares? ›**

Proof: Let m and n be integers and suppose that m and n are perfect squares. By the definition of “perfect square”, we know that **m = k2 and n = j2, for some integers k and j.** **So then mn is k2j2, which is equal to (kj)2.** **Since k and j are integers, so is kj**.

### What is the relationship between a perfect square and rational irrational numbers? ›

**Rational Numbers consist of numbers that are perfect squares** such as 4, 9, 16, 25, etc. Irrational Numbers consist of surds such as 2, 3, 5, 7 and so on. Both the numerator and denominator of rational numbers are whole numbers, in which the denominator of rational numbers is not equivalent to zero.