Prove that Root 3 is Irrational Number
Is root 3 an irrational number? Numbers that can be represented as the ratio of two integers are known as rational numbers, whereas numbers that cannot be represented in the form of a ratio or otherwise, those numbers that could be written as a decimal with nonterminating and nonrepeating digits after the decimal point are known as irrational numbers. The square root of 3 is irrational. It cannot be simplified further in its radical form and hence it is considered as a surd. Now let us take a look at the detailed discussion and prove that root 3 is irrational.
Prove that Root 3 is Irrational Number
The square root of a number is the number that when multiplied by itself gives the original number as the product. A rational number is defined as a number that can be expressed in the form of a division of two integers, i.e. p/q, where q is not equal to 0.
√3 = 1.7320508075688772... and it keeps extending. Since it does not terminate or repeat after the decimal point, √3 is an irrational number. We can learn to prove that root 3 is irrational by following different methods.
Prove That Root 3 is Irrational by Contradiction Method
There are many ways in which we can prove the root of 3 is irrational by contradiction. Let us get one such proof.
Given: Number 3
To Prove: Root 3 is irrational
Proof:
Let us assume the contrary that root 3 is rational. Then √3 = p/q, where p, q are the integers i.e., p, q ∈ Z and coprimes, i.e., GCD (p,q) = 1.
√3 = p/q
⇒ p = √3 q
By squaring both sides, we get,
p^{2 }= 3q^{2 }
p^{2 }/ 3 = q^{2 } (1)
(1) shows that 3 is a factor of p. (Since we know that by theorem, if a is a prime number and if a divides p^{2}, then a divides p, where a is a positive integer)
Here 3 is the prime number that divides p^{2}, then 3 divides p and thus 3 is a factor of p.
Since 3 is a factor of p, we can write p = 3c (where c is a constant). Substituting p = 3c in (1), we get,
(3c)^{2 }/ 3 = q^{2}
9c^{2}/3 = q^{2 }
3c^{2 } = q^{2 }
c^{2 } = q^{2 }/3  (2)
Hence 3 is a factor of q (from 2)
Equation 1 shows 3 is a factor of p and Equation 2 shows that 3 is a factor of q. This is the contradiction to our assumption that p and q are coprimes. So, √3 is not a rational number. Therefore, the root of 3 is irrational.
Prove That Root 3 is Irrational by Long Division Method
The irrational numbers are nonterminating decimals and this can be proved in the case of root 3 as well. Divide 3 using the long division algorithm.
 Write 3 as 3 00 00 00. Consider the number in pairs from the right. So 3 stands alone.
 Now divide 3 with a number such that the number × number gives 3 or a number lesser than that. We determine 1 × 1 = 1
 Subtract this product from 3 and get the remainder as 2. Obtain 1 as the quotient. Bring down the first pair of zeros. 200 becomes the new dividend.
 Double the quotient obtained. It is 2 and now let us have our new divisor, 2n which when multiplied by n should get the product less than or equal to 200.
 We determine 27 × 7 = 189. Subtract this from 200 and get the remainder as 11. Bring down the next pair of zeros. 1100 becomes the new dividend.
 Double the quotient obtained. It is 34 and now let us have our new divisor, 34n. Find a digit 'n' such that 34n × n gives the product less than or equal to 1100.
 We determine 343 × 3 = 1029. Subtract this from 1100 and get the remainder as 71. Bring down the next pair of zeros. 7100 becomes the new dividend.
 Repeat the same division process until we get the quotient approximated to 3 decimal places. Thus, we have evaluated √3 = 1.732
 This is a neverending process. We get nonterminating and nonrepeating digits after the decimal point. √3 = 1.7320508075688772 and it goes on.
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Solved Examples

Example 1: Ana wants to prove that √48 is an irrational number. Can you use the fact that the square root of 3 is irrational to prove it?
Solution:Let us do the prime factorization of 48.
48 = 2 × 2 × 2 × 2 × 3
Adding square root on both sides, we get
√48 = √(2^{4} × 3) = √(2 × 2 × 2 × 2 × 3)
= √(16 × 3)
= 4 √3
=4 × 1.7320508075688772... = 6.928203230275508...
Since √3 cannot be simplified any further and the numbers after the decimal point are nonterminating, 48 = 4 √3 is irrational. Thus proved.

Example 2: Peter said to his friend Frank that the square root of 3 is an irrational number. He then asked him to find √3 using a number line. Can you help him?
Solution:Look at the image below showing root 3 on the number line and read the explanation given below it to understand the process.
Let us construct root 2 on the number line. Draw the number line and number it as 1, 0, 1, and 2 at equal intervals of 1 unit each. Now consider OA = 1 unit on the number line. Construct a perpendicular line AB = 1 unit. Join OB. Applying the Pythagorean Theorem, we get (√(OA^{2} + AB^{2}) = √(1^{2 }+ 1^{2}) = √2
Thus OB = √2
We can conclude that the line segment OB = √2. Construct BC perpendicular to OB, and join OC. Applying the Pythagorean theorem, we get OB^{2} + BC^{2 }= OC^{2}
(√2)^{2 } + (√1)^{2 }= OC^{2}
OC^{2} = 2 +1 = 3 ⇒ OC = √3
We can conclude that OC is the line that represents root 3. Now, with OC as the radius, and O as the center, construct an arc CD, such that the arc meets the number line at D.
OC = √3 = OD (same radius as OC)
Thus √3 is represented on the number line as OD.
FAQs on Is Root 3 an 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.
Is 2 times the square root of 3 Irrational?
Yes, 2√3 is irrational. 2 × √3 = 2 × 1.7320508075688772 = 3.464101615137754..... and the product is a nonterminating decimal. This shows 2√3 is irrational. The other way to prove this is by using a postulate which says that if we multiply any rational number with an irrational number, the product is always an irrational number. That is why 2√3 is irrational.
How to Prove Root 3 is Irrational by Contradiction?
Let us assume, to the contrary, that root 3 is rational. That is, we can find co primes p and q where q ≠ 0, such that √3 = p/q.
Rewriting, we get √3q = p. Squaring on both sides, we get the equation 3q^{2} = p^{2}. Hence p^{2 }is divisible by 3, and so p is also divisible by 3. Hence we can write p = 3c for some integer c. Substituting p = 3c in our equation, we get 3q^{2} = 9c^{2 }⇒ q^{2} =3c^{2} . This shows q is also divisible by 3. Thus p and q have at least 3 as a common factor. But this contradicts our assumption that p and q are coprimes. Therefore, our assumption was wrong, and root 3 is an irrational number.
Is 3 Times the Square Root of 3 Irrational?
Yes, 3√3 is irrational. 3 × √3 = 3 × 1.7320508075688772... = 5.196152422706631.......... and the product is a nonterminating decimal. This shows 3√3 is irrational. The other way to prove this is by using a postulate which says that if we multiply any rational number with an irrational number, the product is always an irrational number. That is why 3√3 is irrational.
How to Prove that 1 by Root 3 is irrational?
1 by root 3 can be rationalized as, \(\dfrac{1}{√3} \times \dfrac{√3}{√3} = \dfrac{√3}{3}\). Now, we know that the product of a rational and an irrational number is always irrational. Here, 1/3 is a rational number and √3 is an irrational number. Therefore, it is proved that 1 by Root 3 is irrational.
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