Prove that Root 2 is Irrational Number
Is root 2 an irrational number? Numbers that can be represented in the form of a/b, where a and b are integers and b does not equal to 0 are known as rational numbers whereas numbers that cannot be represented in the form of a fraction or ratio of two integers and whose decimal expansion do not terminate or repeat are known as irrational numbers. Now let us take a look at the detailed discussion and prove that root 2 is irrational.
Prove that Root 2 is Irrational Number
The square root of a number is the number that gets multiplied to itself to give the original number. The square root of 2 is represented as √2. The actual value of √2 is undetermined. The decimal expansion of √2 is infinite because it is nonterminating and nonrepeating. Any number that has a nonterminating and nonrepeating decimal expansion is always an irrational number. So, √2 is an irrational number.
Problem statement: Prove that Root 2 is Irrational Number
Given: The number 2
There are two methods to prove that √2 is an irrational number, and those methods are:
 By contradiction method
 By longdivision method
Let's learn about both methods in detail.
Prove That Root 2 is Irrational by Contradiction Method
In this method, we start with an assumption that is contrary to what we are actually required to prove. Then, using a series of logical deductions from this assumption, we reach an inconsistency – a mathematical or logical error – which enables us to conclude that our original assumption was incorrect.
In this case, we start by supposing that √2 is a rational number. Thus, there will exist integers p and q (where q is nonzero) such that p/q = √2. We also make the assumption that p and q have no common factors. As even if they have common factors we would cancel them to write it in the simplest form. So, let us assume that p and q are coprime numbers here having no common factor other than 1.
Now, squaring both sides, we have p^{2}/q^{2} = 2, which can be rewritten as,
p^{2} = 2{q^{2}}.........(1)
We note that the righthand side of the equation is multiplied by 2, which means that the lefthand side is a multiple of 2. So, we can say that p^{2} is a multiple of 2. This further means that p itself must be a multiple of 2, as when a prime number is a factor of a number, let's say, m^{2}, it is also a factor of m. Thus, we can assume that,
p=2m, m∈Z [Set of Integers]
⇒(2m)^{2}= 2q^{2} [From (1)]
⇒4m^{2}=2q^{2}
⇒q^{2}=2m^{2}
Now, the righthand side is a multiple of 2 again, which means that the lefthand side is a multiple of 2, which further means that q is a multiple of 2, i.e., q = 2n, where n ∈ Z. We have thus shown that both p and q are multiples of 2. But is that possible? This can only mean one thing: our original assumption of assuming √2 as p/q (where p and q are coprime integers) is wrong:
√2 ≠ p/q
Thus, √2 does not have a rational representation – √2 is irrational.
Prove That Root 2 is Irrational by Long Division Method
The value of the square root of 2 by long division method consists of the following steps:
 Step 1: Write 2 as dividend in the division format. Add a point and then attach 6 to 8 zeros after the point. It is to get the appropriate number of decimal places in the decimal expansion of √2.
 Step 2: Find the largest number whose square is less than or equal to the number 2. Take this number as the divisor and the quotient (1 in this case). Subtract and write the remainder.
 Step 3: In the quotient, put a decimal point after 1. Bring down two zeroes to the right of the remainder. So, the new dividend is 100.
 Step 4: Double the divisor and enter it as a new divisor with a blank on its right. Think about the greatest possible digit that can fill the blank which will also become the next digit in the quotient, such that when we multiply, the product should be either equal or less than the dividend. Subtract the numbers and write the remainder obtained.
 Step 5: Then, we have to repeat the same process to get the desired number of decimal places. If required, we can add more zeros on the right of the original dividend, i.e, 2, in pairs.
When you will continue this process, you will get the decimal expansion of √2 as 1.4142135623730950488016887...... which is clearly nonterminating and nonrepeating. Therefore, √2 is irrational.
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Solved Examples

Example 1: Sara wants to prove that √32 is an irrational number. Can the fact that the square root of 2 is irrational be used to prove it?
Solution: First, let us write √32 in terms of √2, which is, √32 = √(2×2×2×2×2), which is equal to 2×2×√2, or 4√2.
Now, we know that √2 is an irrational number, hence 4√2 is also an irrational number. Therefore, √32 is an irrational number. 
Example 2: Thomas said to his friend Elle that the square root of 2 is an irrational number. He then asked her to find √2 using a number line. Can you help her?
Solution: To locate √2 on a number line, we will use a unit square. Let's name the vertices of the square as shown. We know that √2 is the length of the diagonal of a unit square, or may use Pythagoras theorem to find the diagonal of a square with each side equals to 1 unit.Keep the vertex O at 0. We already discovered that OB= √2 units. So place the unit square from 0 to 1 unit on the number line. Using a compass with center O and radius OB, draw an arc intersecting the number line at point P.
Therefore, point P represents √2 on the number line.
FAQs on Is Root 2 an Irrational
How do you Prove that Root 2 is Irrational?
There are two methods to prove the irrationality of root 2 and those methods are:
 Contradiction method
 Long division or decimal expansion method
In the contradiction method, we first assume that √2 is a rational number and hence can be written in form of m/n, where m and n are coprime numbers and n ≠ 0. But, later we find out that exist no coprime integers m and n, so our assumption was wrong. This was the one way to prove that √2 is an irrational number. The other method that could be used is the long division method. In this, we find the decimal expansion of √2 and check whether it is nonterminating nonrepeating, or not. If it is, then it is an irrational number as per the properties of irrational numbers.
Is 2 times the square root of 2 Irrational?
Yes, 2 times √2 is irrational. When we multiply any rational number to an irrational number, the product is always an irrational number. So, we know that √2 is irrational, and therefore, 2√2 is also an irrational number.
How to Prove Root 2 is Irrational by Contradiction?
To prove that √2 is irrational by the contradiction method, we first assume that √2 is a rational number. Now, if it is a rational number, there exist two coprime integers x and y, such that √2 = x/y, where x and y have no other common factors except 1 and y ≠ 0. So, our equation is √2 = x/y. By squaring both sides, we get, 2 = x^{2}/y^{2}, which can be rewritten as 2y^{2} = x^{2}. It states that x^{2} is a multiple of 2, which also means that x is also a multiple of 2 [as when a prime number is a factor of a number, let's say, p^{2}, it is also a factor of p]. Now, we can write x as 2c, as we just found that x is a multiple of 2. So, our new equation is, x = 2c. Substituting this value in the equation 2y^{2} = x^{2}, we get, 2y^{2} = (2c)^{2}.
⇒ 2y^{2} = 4c^{2}
⇒ y^{2} = 2c^{2}
It means that y is also a multiple of 2. When both x and y have a common multiple 2, it means they are not coprime numbers. So, our assumption was wrong. √2 is not a rational number. Therefore, √2 is an irrational number.
Is 3 Times the Square Root of 2 Irrational?
Yes, 3 times √2 is irrational, as the product of a rational and an irrational number is always an irrational number. So, we know that √2 is irrational, therefore, 3√2 is also an irrational number.
How to Prove that 1 by Root 2 is irrational?
Let us prove that 1/√2 is irrational by the contradiction method. So, first, let us assume that 1/√2 is a rational number. It means that there exist two coprime integers p and q, such that 1/√2 = p/q. By squaring both sides, we get, 1/2= p^{2}/q^{2}, which can be rewritten as 2p^{2} = q^{2}. It states that q^{2} is a multiple of 2, which also means that q is a multiple of 2 [as when an integer is a multiple of a prime number, then its square root is also a multiple of that prime number]. So, we can form a new equation where we can express q as, q= 2c, where c ∈ Z. Now, substitute this value in the previous equation 2p^{2} = q^{2} to get, 2p^{2} = (2c)^{2}
⇒ 2p^{2} = 4c^{2}
⇒ p^{2} = 2c^{2}
It means that p is also a multiple of 2, which contradicts our assumption. Therefore, 1/√2 is an irrational number.
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