1.Revisiting Irrational Numbers Class 10

Revisiting Irrational Numbers Class 10

In this article Revisiting Irrational Numbers Class 10,we will prove that square root of 2,3,5 is irrational numbers.We use in our proof,the Fundamental Theorem of Arithmetic.
“Before moving to Revisiting Irrational Numbers Class 10,read our basic introduction to Fundamental Theorem of Arithmetic 10th.”

2.Revisiting Irrational Numbers Class 10 Illustrations

Illustration 1: prove that \sqrt{5} is irrational.
Proof: Let us assume, to the contrary, that \sqrt{5} is rational.
So, we can find integers r and s (s \neq 0) such that \sqrt{5} = \frac{r}{s}.
Suppose r and s have a common factor other than 1.
Then, we divide by the common factor to get \sqrt{5} = \frac{a}{b}, where a and b are coprime.
So, b\sqrt{5} = a.
Squaring on both sides and rearranging, we get 5b^2 = a^2.
Therefore, 5 divides a^2.
Now, by theorem FTA, it follows that 5 divides a.
So, we can write a = 5c for some inty integer c.
Substituting for a, we get 5b^2 = 25c^2, that is b^2 = 5c^2.
This means that 5 divides b^2, and so 5 divides b (again using Theorem FTA with p = 5).
But this contradicts the fact that a and b have no common factors other than 1.
This contradiction has arisen because of our incorrect assumption that \sqrt{5} is rational.
So, we conclude that \sqrt{5} is irrational.
Illustration 2: prove that 3 + 2\sqrt{5} is irrational.
Proof: Let us assume, to the contrary, that 3 + 2\sqrt{5} is rational.
That is, we can find co prime a and b (b \neq 0) such that 3 + 2\sqrt{5} = \frac{a}{b}.
Therefore, \frac{a}{2b} - \frac{3}{2} = \sqrt{5}.
Rearranging this equation, we get \sqrt{5} = \frac{a - 3b}{2b}.
Since a and b are integers, we get \frac{a - 3b}{2b} is rational, and so \sqrt{5} is rational.
But this contradicts the fact that \sqrt{5} is irrational.
This contradiction has arisen because of our incorrect assumption that 3 + 2\sqrt{5} is rational.
So, we conclude that 3 + 2\sqrt{5} is irrational.

Revisiting Irrational Numbers Class 10

llustration 3: Prove that the following are irrationals:
Illustration 3(i). \frac{1}{\sqrt{2}}
Proof: Let us assume, to the contrary, that \frac{1}{\sqrt{2}} is rational.
That is, we can find coprime a and b (b \neq 0) such that \frac{1}{\sqrt{2}} = \frac{a}{b}.
Therefore, \frac{b}{a} = \sqrt{2}.
Rearranging this equation, we get \sqrt{2} = \frac{b}{a}.
Since a and b are integers, we get \frac{b}{a} is rational, and so also \sqrt{2} is rational.
But this contradicts the fact that \sqrt{2} is irrational.
This contradiction has arisen because of our incorrect assumption that \frac{1}{\sqrt{2}} is rational.
So, we conclude that \frac{1}{\sqrt{2}} is irrational.
Illustration 3(ii). 7\sqrt{5}
Proof: Let us assume, to the contrary, that 7\sqrt{5} is rational.
That is, we can find coprime a and b (b \neq 0) such that 7\sqrt{5} = \frac{a}{b}.
Therefore, \frac{a}{7b} = \sqrt{5}.
Rearranging this equation, we get \sqrt{5} = \frac{a}{7b}.
Since a and b are integers, we get \frac{a}{7b} is rational, and so \sqrt{5} is irrational.
But this contradicts the fact that \sqrt{5} is irrational.
This contradiction has arisen because of our incorrect assumption that 7\sqrt{5} is rational.
So, we conclude that 7\sqrt{5} is irrational.
Illustration 3(iii). 6 + \sqrt{2}
Proof: Let us assume, to the contrary, that 6 + \sqrt{2} is rational.
That is, we can find coprime a and b (b \neq 0) such that 6 + \sqrt{2} = \frac{a}{b}.
Therefore, \frac{a}{b} - 6 = \sqrt{2}.
Rearranging this equation, we get \sqrt{2} = \frac{a-6b}{b}.
Since a and b are integers, we get \frac{a}{b} - 6 is rational, and so \sqrt{2} is rational.
But this contradicts the fact that \sqrt{2} is irrational.
This contradiction has arisen because of our incorrect assumption that 6 + \sqrt{2} is rational.
So, we conclude that 6 + \sqrt{2} is irrational.

Also Read This Article:- Euclid Division Algorithm Class 10 and Important Questions with Solution

3.Practice Questions of Revisiting Irrational Numbers Class 10 for students

Revisiting Irrational Numbers Class 10

Prove that the following are irrationals

(1.)\sqrt{2}+\sqrt{3}  (2.)4-5\sqrt{2}
By solving the above questions,you can understand the Revisiting Irrational Numbers Class 10 well because the concept is well understood when you solve it practically.

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Also Read This Article:- Irrational number

4.Frequently Asked Questions Related to Revisiting Irrational Numbers Class 10

**छात्र-छात्राओं से आज का प्रश्न**

*”सेना की एक परेड में 24 सदस्यों वाले बैंड के पीछे 308 जवानों का एक समूह कदमताल करता हुआ चलता है।दोनों समूहों को समान पंक्तियों में कदमताल करना है।बताइये,अधिकतम कितनी पंक्तियों में कदमताल करते हुए चल सकते हैं?”*
**Today’s Question to Students**
*”In an army parade, a group of 308 soldiers walk behind a band of 24 members. Both groups have to move in equal lines. Tell me, how many rows can you walk in the maximum number of rows?”*
*पिछली प्रश्नोत्तरी का उत्तर*
माना चार भाग क्रमशः x,y,z तथा 80-(x+y+z)
प्रश्नानुसार: x+3 = y-3 = 3z = \frac{80-(x+y+z)}{3} \\ \Rightarrow x+3 = y-3  \\ \Rightarrow x - y = -6 \cdots (1)\\ y-3 = 3z \Rightarrow y - 3z = 3 \cdots (2)\\ 3z = \frac{80-(x+y+z)}{3} \Rightarrow x + y + 10z = 80 \cdots (3)
समीकरण (2) को 10 से तथा (3) को 3 से गुणा करने पर:
\begin{array}{c}10y - 30z = 30 \cdots (4) \\ 3x + 3y + 30z = 240 \cdots(5) \\ \hline \end{array}
जोड़ने परः
3x + 13y = 270 \cdots (6)
समीकरण (1) को 13 से गुणा करने पर:
\begin{array}{c}13x - 13y = -78 \cdots(7)\\ 3x + 13y = 270 \cdots(6) \\ \hline \end{array}
जोड़ने परः
16x=192 \Rightarrow x=12
(1) से:y=18,(2) से:z=5
तथा (3) से:80-(x+y+z)=45
अतः 80 के चार भाग:12,18,5,45
*Previous Quiz Solution*
Let the four parts be x, y, z and 80 respectively (x+y+z)
According to the question:
x+3 = y-3 = 3z = \frac{80-(x+y+z)}{3} \\ \Rightarrow x+3 = y-3 \\ \Rightarrow x - y = -6 \cdots (1)\\ y-3 = 3z \Rightarrow y - 3z = 3 \cdots (2)\\ 3z = \frac{80-(x+y+z)}{3} \Rightarrow x + y + 10z = 80 \cdots (3)
Multiplying equation (2) by 10 and (3) by 3:
\begin{array}{c}10y - 30z = 30 \cdots (4) \\ 3x + 3y + 30z = 240 \cdots(5) \\ \hline \end{array}
On adding:
3x + 13y = 270 \cdots (6)
Multiplying equation (1) by 13:
\begin{array}{c}13x - 13y = -78 \cdots(7)\\ 3x + 13y = 270 \cdots(6) \\ \hline \end{array}
On adding :
16x=192 \Rightarrow x=12
(1) from:y=18,(2) from:z=5
and (3) from:80-(x+y+z)=45
Hence, four parts of 80:12,18,5,45
*”This article has been prepared by **Satyam Coaching Centre** on the **Satyam Mathematics** blog.”*