Real Numbers: Class 10 Maths Chapter 1 Hinglish mein

Primes building blocks ki tarah, aur proofs ki kuch numbers fractions nahi hote

Mathematics · Hinglish notes, simulations aur practice · Official NCERT chapter PDF ↗

Published 12 Oct 2026 · Updated 12 Oct 2026

50 min
23 practice questions
14 MCQs

Is chapter mein aap kya seekhenge

Factor tree se kisi composite number ko prime powers ke product ke roop mein likhna, aur samjhana ki Fundamental Theorem of Arithmetic mein uniqueness ka matlab kya hai
Divisibility ke sawaal prime factorisations se solve karna, jaise kya koi power digit 0 par khatam ho sakti hai
Do ya teen numbers ka HCF aur LCM prime factorisation se nikaalna, aur do numbers ke liye HCF × LCM = product use karna
Samjhana ki jo prime kisi positive integer ke square ko divide karta hai, woh us integer ko bhi zaroor divide karta hai
Proof by contradiction se prove karna ki 2 ka, aur kisi bhi prime ka, square root irrational hai
Prove karna ki ek non-zero rational number aur aise square roots ke sums, differences, products aur quotients irrational hote hain

NCERT Mathematics Class 10, Chapter 1, pp. 1–9 par based original companion text. Yeh NCERT se juda nahi hai aur NCERT ne ise endorse nahi kiya hai.

1.1 Introduction

Paani hamesha do hydrogen atoms aur ek oxygen atom se milkar banta hai, kabhi kisi aur mix se nahi. Whole numbers bhi kuch aisa hi behave karte hain. Unke building blocks prime numbers hain, aur 1 se bade har whole number ki primes ki apni fixed recipe hoti hai: 60 hamesha do 2, ek 3 aur ek 5 se banta hai. Is chapter mein hum is fact ko ek theorem ke roop mein likhenge aur use kaam mein layenge.

Class 9 mein tum real numbers se mile the: rational numbers jaise $\frac{3}{4}$ aur $- 2.5,$ aur saath mein irrational numbers jaise $\sqrt{2}$ aur $π$ jinke barabar koi fraction nahi ho sakta (Fig. 1.1). Tumne $\sqrt{2}$ ko number line par mark kiya tha, lekin uska irrational hona bas maan liya gaya tha. Yeh chapter uska proof deta hai.

5/5Irrational numbers beech ke gaps bharte hain; rationals ke saath milkar woh real numbers banate hain.

Fig. 1.1Numbers ki har family agli family ke andar baithi hai; rationals aur irrationals milkar real numbers banate hain.

Gol kinaron wale boxes ek ke andar ek rakhe hain. Step 1 mein sirf natural numbers 1, 2, 3 dikhte hain. Step 2 uske chaaron taraf whole numbers ka ek bada box jodta hai jismein 0 bhi hai. Step 3 integers jodta hai, nayi band mein −5 aur −12 ke saath. Step 4 rational numbers jodta hai, three quarters, −2.5 aur 0.3 recurring ke saath. Step 5 rationals ke bagal mein irrational numbers ka ek dashed box jodta hai, jismein root 2, pi aur minus root 7 hain, aur sabke chaaron taraf ek bahari box hai jis par real numbers likha hai.

Ek hi result, Fundamental Theorem of Arithmetic, poore chapter ko chalata hai. Isse numbers ka HCF aur LCM nikaalne ka ek tez tareeka milta hai (HCF aur LCM). Isse hum yeh bhi prove kar paate hain ki $\sqrt{2},$ $\sqrt{3},$ $\sqrt{5}$ aur aise kai numbers irrational hain (Section 1.3).Tumhari textbook ka introduction Euclid's division algorithm aur decimal expansions ka bhi zikr karta hai. Woh sections ab ke syllabus se hata diye gaye hain, isliye yeh chapter unhe cover nahi karta.

Quick check

2 mein se 1

In numbers mein se kaunsa irrational hai?

\sqrt{49}

\frac{22}{7}

\sqrt{5}

0. \overline{6}

Answer dekho

Answer (C) $\sqrt{5}$

\sqrt{49} = 7,

\frac{22}{7}

pehle se hi fraction hai (yeh bas

π

ke kareeb hai), aur

0. \overline{6} = \frac{2}{3} .

Sirf

\sqrt{5}

ko fraction ke roop mein nahi likha ja sakta; yeh chapter ise prove karta hai.

1.2 The Fundamental Theorem of Arithmetic

Yaad karo ki prime number ke theek do factors hote hain, 1 aur khud woh number: 2, 3, 5, 7, 11, 13 waghairah. 12 ya 91 jaise composite number ka kam se kam ek aur factor hota hai. Number 1 dono mein se kuch nahi hai, kyunki uska sirf ek factor hai.

Har composite number do chhote factors mein toot jaata hai, aur woh factors phir se toot sakte hain. Yeh todna kahan rukta hai? Sirf un numbers par jinhe aur toda nahi ja sakta, yaani primes par. Ek factor tree poore process ko record karta hai: number ko upar rakho, use kinhi do factors mein todo, aur tab tak chalte raho jab tak har branch ek prime par khatam na ho jaaye. Fig. 1.2 13860 ke liye yahi karta hai.

5/5Gol ghere wale primes ko multiply karo: $13860 = 2^{2} \times 3^{2} \times 5 \times 7 \times 11.$

Fig. 1.2Pehla split koi bhi chalega; tree tab rukta hai jab har branch ek prime par khatam ho.NCERT p. 2

13860 ka ek factor tree. Composite numbers square boxes mein hain aur primes green circles mein, jaisa kone mein di gayi key dikhati hai. Step 1 mein sirf 13860 dikhta hai. Step 2 use 132 aur 105 mein todta hai. Step 3 132 ko 12 aur 11 mein, aur 105 ko 7 aur 15 mein todta hai; 11 aur 7 primes hain. Step 4 12 ko 4 aur 3 mein, 15 ko 3 aur 5 mein, aur 4 ko 2 aur 2 mein todta hai. Step 5 mein dashed rings branches ke ends par saat primes ko mark karte hain, aur neeche ki line kehti hai 13860 equals 2 squared times 3 squared times 5 times 7 times 11.

To 13860 primes 2, 2, 3, 3, 5, 7 aur 11 ka product hai. Ek bada number try karo: $999999 = 999 \times 1001,$ aur yeh aage aur tootte hain: $3^{3} \times 37$ aur $7 \times 11 \times 13.$ Bade numbers mein mushkil kaam yeh check karna hai ki aakhri factors sach mein prime hain. 37 ko test karne ke liye primes ki limit hai $\sqrt{37} \approx 6,$ isliye sirf 2, 3 aur 5 try karne hain.

Yeh process ulta bhi chalta hai. Koi bhi primes chuno, chaho to kuch repeat karo, aur unhe multiply karo:

$3 \times 5 \times 11 = 165$
$2 \times 2 \times 3 \times 11 = 132$
$2 \times 2 \times 2 \times 5 \times 5 = 200$

Har product ek whole number hai, aur kyunki primes infinitely many hain (Euclid ne ise lagbhag 300 BCE mein prove kiya tha), products kabhi khatam nahi hote. Kya is kabhi na khatam hone wali supply se koi composite number chhoot sakta hai? Nahi: har composite number waise hi tootta hai jaise 13860 toota, aur branches ke ends par primes hamesha wahi hote hain. Yahi Fundamental Theorem of Arithmetic (अंकगणित की आधारभूत प्रमेय) hai.

Theorem 1.1Fundamental Theorem of Arithmetic

Har composite number primes ka product hai. Yahi nahi, ek hi number ke aise kinhi do products mein bilkul wahi primes hote hain, aur har prime utni hi baar aata hai.

Factorisation kyun exist karta hai. Ek composite number $n$ is tarah tootta hai: $n = a \times b,$ jahan $a$ aur $b$ dono 1 se bade hain, lekin koi bhi utna bada nahi jitna $n .$ Kisi bhi composite factor ko phir se todo. Har split chhote numbers deta hai, aur whole numbers hamesha chhote nahi hote ja sakte, isliye todna kahin na kahin rukega. Yeh tabhi rukta hai jab har factor prime ho.

Yeh unique kyun hai. Yeh gehra hissa hai, aur iska proof is book ke level se aage ka hai (neeche history note dekho). Tum ise khud test kar sakte ho: ek hi number ke alag-alag factor trees hamesha wahi primes par khatam hote hain (Fig. 1.3).

Theorem ko order se koi fark nahi padta: $7 \times 2 \times 2$ aur $2 \times 7 \times 2$ 28 ka ek hi factorisation hain. Woh jo rule out karta hai woh hai primes ka alag collection. Fig. 1.3 mein 360 ko do alag pehle splits se factorise kiya gaya hai. Dono trees phir bhi teen 2, do 3 aur ek 5 par khatam hote hain.

3/3Dono ka result ek hi hai: $360 = 2^{3} \times 3^{2} \times 5.$ 360 ka koi bhi factor tree kuch aur nahi de sakta.

Fig. 1.3Alag trees, wahi primes: yahi Fundamental Theorem of Arithmetic ki uniqueness hai.

Step 1 tree A dikhata hai: 360 toot kar 8 aur 45 banta hai, 8 toot kar 2 aur 4, 4 toot kar 2 aur 2, 45 toot kar 5 aur 9, aur 9 toot kar 3 aur 3; neeche ek row uske primes ko order mein 2, 2, 2, 3, 3, 5 likhti hai. Step 2 uski jagah tree B dikhata hai: 360 toot kar 10 aur 36 banta hai, 10 toot kar 2 aur 5, 36 toot kar 6 aur 6, aur har 6 toot kar 2 aur 3; uske sorted primes 2, 2, 2, 3, 3, 5 doosri row mein aate hain. Step 3 dono ek jaisi rows ko ek bracket mein lekar same primes label lagata hai.

Factorisations ko compare karna aasaan banane ke liye hum primes ko increasing order mein likhte hain aur repeat hone wale primes ko powers ke roop mein ikattha karte hain. Ise prime factorisation ka standard form kehte hain:

\begin{aligned} n & = p_{1}^{a_{1}} \times p_{2}^{a_{2}} \\ \times \dots \times p_{k}^{a_{k}} \end{aligned}

jahan

$p_{1} < p_{2} < \dots < p_{k}$: woh alag-alag primes jo $n$ ko divide karte hain
$a_{1}, a_{2}, \dots, a_{k}$: har prime kitni baar aata hai (positive integers)

Jaise, $13860 = 2^{2} \times 3^{2} \times 5 \times 7 \times 11$ aur $360 = 2^{3} \times 3^{2} \times 5.$ Order fix ho jaane ke baad, ek hi number ko factorise karne wale do log bilkul ek hi cheez likhenge.

InteractivePrime factor lab

Do tabs wali ek interactive lab. Factor tree tab mein tum 2 se 999999 tak ka koi number type karte ho ya 360, 13860, 1001 ya 999999 jaisa koi preset chunte ho. Composite numbers square boxes ki tarah aur primes circles ki tarah dikhte hain. Kisi box par tap karo aur use todne ke liye uska koi factor pair chuno, ya smallest primes se tree poora karne ke liye ek button dabao. Jab har branch ek prime par khatam ho jaati hai, ek message standard form mein factorisation deta hai aur ginta hai ki tumne us number ke kitne alag trees banaye. HCF and LCM tab mein tum do ya teen numbers daalte ho; ek table har prime ko har number mein uski power ke saath list karti hai, lower aur higher powers mark karti hai, aur HCF, LCM dikhati hai aur yeh bhi ki HCF times LCM numbers ke product ke barabar hai ya nahi.

Fig. 1.4Apne tareeke se factor trees banao, phir do ya teen numbers ki prime powers compare karo.

Quick check

3 mein se 1

In mein se 588 ka prime factorisation kaunsa hai?

2^{2} \times 3 \times 7^{2}

2^{2} \times 147

4 \times 3 \times 49

2 \times 3^{2} \times 7^{2}

Answer dekho

Answer (A) $2^{2} \times 3 \times 7^{2}$

588 = 4 \times 147 = 2^{2} \times 3 \times 49 = 2^{2} \times 3 \times 7^{2} .

Options B aur C ka product 588 aata hai lekin woh composite factors (147, 4 aur 49) use karte hain. Option D to hai

2 \times 9 \times 49 = 882.

Kya koi power 0 par khatam ho sakti hai?

Koi number digit 0 par theek tab khatam hota hai jab 10 use divide karta hai. Kyunki $10 = 2 \times 5,$ yeh theek tab hota hai jab uske prime factorisation mein 2 aur 5 dono aate hain. Uniqueness hi asli baat hai: agar kisi number ke ek aur sirf ek factorisation mein koi 2 (ya koi 5) nahi hai, to koi bhi cheez use chupke se andar nahi la sakti. Fig. 1.5 powers ki teen families ko test karta hai.

4/4 $30^{n} = 2^{n} \times 3^{n} \times 5^{n}$ mein dono hain, isliye yeh hamesha 0 par khatam hota hai.

Fig. 1.5Sirf wahi number 0 par khatam ho sakta hai jiske prime factorisation mein 2 aur 5 dono hon.

Ek heading kehti hai ki 0 par khatam hone ke liye 2 aur 5 chahiye, uske neeche power, has 2, has 5 aur first values ke columns hain. Step 1 mein sirf heading dikhti hai. Step 2 15 ki powers jodta hai: 2 ka slot ek dashed, kata hua circle hai aur 5 ka slot bhara hua hai, aur uski values 15, 225 aur 3375 sab 5 par khatam hoti hain. Step 3 12 ki powers jodta hai: 2 maujood hai, 5 kata hua hai, aur 12, 144 aur 1728 2, 4 aur 8 par khatam hote hain. Step 4 30 ki powers jodta hai jismein 2 aur 5 dono maujood hain, aur 30, 900 aur 27000 sab 0 par khatam hote hain.

Example 1NCERT Example 1 · p. 4

15 ki powers aur digit 0

Prove karo: har natural number $n,$ chahe jo bhi ho, $15^{n}$ ka last digit 0 nahi hota.

Pehle khud try karo. Phir solution ek-ek step karke kholo.

Prime factorisation se HCF aur LCM

Do ya zyada numbers ka HCF (highest common factor) woh sabse bada number hai jo un sabko divide karta hai. Unka LCM (lowest common multiple) woh sabse chhota positive number hai jise woh sab divide karte hain. Pichhli classes mein tumne inhe factors aur multiples list karke, ya baar-baar divide karke nikaala tha. Prime factorisation dono ko ek nazar mein de deta hai, aur yeh bhi samjhata hai ki yeh method kaam kyun karta hai.

Example 2NCERT Example 2 · p. 4

60 aur 72 ka HCF aur LCM

Prime factorisation se 60 aur 72 ka HCF aur LCM nikaalo.

Pehle khud try karo. Phir solution ek-ek step karke kholo.

PropertyPrime factorisations se HCF aur LCM

Har number ko standard form mein likho. HCF un primes ka product hai jo numbers mein se har ek ko divide karte hain, aur har prime ki woh sabse chhoti power li jaati hai jismein woh aata hai. LCM un saare primes ka product hai jo numbers mein se kisi bhi ek ko divide karte hain, aur har prime ki woh sabse badi power li jaati hai jismein woh aata hai.

Ek prime ko follow karo, maan lo 2, jo 72 mein $2^{3}$ ki tarah aur 60 mein $2^{2}$ ki tarah aata hai. Common factor ko dono numbers ko divide karna hai, isliye usmein 2 ki power ki limit hai $2^{2},$ aur HCF yeh poora le leta hai. Common multiple ko dono se divisible hona hai, isliye use kam se kam chahiye $2^{3},$ aur LCM isse zyada nahi leta. Yahi reasoning baari-baari har prime ke liye kaam karti hai. Fundamental Theorem ki wajah se koi aur primes ya powers shaamil nahi ho sakte.

Fig. 1.6 is rule ko prime blocks ke stacks ki tarah dikhata hai. Jab do numbers mein koi bhi prime common nahi hota, to unka HCF 1 hota hai aur hum unhe coprime kehte hain, jaise 8 aur 15. Tab unka LCM bas unka product hota hai.

5/5Har column mein lower power + higher power = saare blocks. Isliye $12 \times 360 = 72 \times 60 = 4320.$

Fig. 1.6Har prime ke liye HCF chhota stack leta hai aur LCM lamba stack.

Teen columns, primes 2, 3 aur 5 ke liye. Step 1 72 ko square blocks ki tarah dikhata hai: teen 2 aur do 3. Step 2 unke bagal mein 60 ko gol blocks ki tarah jodta hai: do 2, ek 3 aur ek 5. Step 3 har column mein un blocks ko dashes se outline karta hai jo dono numbers mein common hain, do 2 aur ek 3, aur baaki ko halka kar deta hai; HCF hai 2 squared times 3, yaani 12. Step 4 har column ke lambe stack ko outline karta hai, teen 2, do 3 aur ek 5, jisse LCM 360 milta hai. Step 5 har column par label lagata hai ki HCF power plus LCM power uske saare blocks ke barabar hai: 2 + 3 = 5, 1 + 2 = 3 aur 0 + 1 = 1, isliye 12 times 360 equals 72 times 60, yaani 4320.

Do numbers ka product rule

\begin{aligned} HCF (a, b) \\ \times LCM (a, b) = a \times b \end{aligned}

(1.1)

jahan

$a, b$: koi bhi do positive integers

Yeh kaam kyun karta hai? Ek baar mein ek prime lo. Maan lo is prime ki power $a$ mein $m$ hai, aur doosre number mein $n$ , jo hai $b .$ Do powers hain, $m$ aur $n,$ jinmein se HCF chhoti leta hai, LCM badi, aur chhoti + badi $= m + n,$ jo theek utni hai jitni power is product mein hai: $a \times b .$ To ek baar do numbers ka HCF pata ho, to ek hi division se LCM mil jaata hai: $LCM = (a \times b) \div HCF .$

Example 3NCERT Example 3 · p. 4

HCF se LCM tak

Prime factorisation se 120 aur 405 ka HCF nikaalo, aur usse unka LCM nikaalo.

Pehle khud try karo. Phir solution ek-ek step karke kholo.

Example 4NCERT Example 4 · p. 5

Ek saath teen numbers

Prime factorisation se 12, 30 aur 84 ka HCF aur LCM nikaalo.

Pehle khud try karo. Phir solution ek-ek step karke kholo.

Yahan HCF × LCM $= 6 \times 420 = 2520,$ lekin $12 \times 30 \times 84 = 30240.$ Product rule (1.1) sirf do numbers ke liye sahi hai.

Quick check

3 mein se 1

Do numbers ka HCF 18 hai aur unka product 6480. Unka LCM kya hai?

Answer dekho

Answer 360

LCM

= 6480 \div 18 = 360.

Jaise, 72 aur 90 ka HCF 18 aur LCM 360 hai.

Practice · 9 questions

Practice 1.1: answers ke saath

1
NumericalEasyNCERT Exercise 1.1, Q1
Har number ko standard form mein primes ke product ki tarah likho.
1. $252$
  Solution dikhao
  $252 = 2 \times 126 = 2^{2} \times 63,$ aur $63 = 3^{2} \times 7.$ To $252 = 2^{2} \times 3^{2} \times 7.$
2. $2730$
  Solution dikhao
  Baari-baari 2, 3 aur 5 se divide karo: 2730, 1365, 455, 91. Phir $91 = 7 \times 13.$ To $2730 = 2 \times 3 \times 5 \times 7 \times 13.$
3. $4675$
  Solution dikhao
  Yeh 75 par khatam hota hai, isliye 5 se do baar divide karo: $4675 = 5^{2} \times 187.$ Phir $187 = 11 \times 17.$ To $4675 = 5^{2} \times 11 \times 17.$
4. $9797$
  Solution dikhao
  97 se chhota koi prime ise divide nahi karta. 97 try karne par milta hai $9797 = 97 \times 101,$ aur 97 aur 101 dono prime hain.
2
NumericalMediumNCERT Exercise 1.1, Q2
Har pair ka HCF aur LCM prime factorisation se nikaalo. Phir check karo ki HCF × LCM pair ke product ke barabar hai.
1. $39$ aur $65$
  Solution dikhao
  $39 = 3 \times 13$ aur $65 = 5 \times 13.$ HCF $= 13$ aur LCM $= 3 \times 5 \times 13 = 195.$ Check: $13 \times 195 = 2535 = 39 \times 65.$
2. $154$ aur $210$
  Solution dikhao
  $154 = 2 \times 7 \times 11$ aur $210 = 2 \times 3 \times 5 \times 7.$ HCF $= 2 \times 7 = 14$ aur LCM $= 2 \times 3 \times 5 \times 7 \times 11 = 2310.$ Check: $14 \times 2310 = 32340 = 154 \times 210.$
3. $225$ aur $60$
  Solution dikhao
  $225 = 3^{2} \times 5^{2}$ aur $60 = 2^{2} \times 3 \times 5.$ HCF $= 3 \times 5 = 15$ aur LCM $= 2^{2} \times 3^{2} \times 5^{2} = 900.$ Check: $15 \times 900 = 13500 = 225 \times 60.$
3
NumericalMediumNCERT Exercise 1.1, Q3
Prime factorisation use karke teen numbers ke har set ka HCF aur LCM nikaalo.
1. $18,$ $24$ aur $30 :$ HCF $=$ aur LCM $=$ .
  Answer dekho
  Answer (1) 6, (2) 360
  $18 = 2 \times 3^{2},$ $24 = 2^{3} \times 3$ aur $30 = 2 \times 3 \times 5.$ HCF $= 2 \times 3 = 6$ aur LCM $= 2^{3} \times 3^{2} \times 5 = 360.$
2. $13,$ $19$ aur $31 :$ HCF $=$ aur LCM $=$ .
  Answer dekho
  Answer (1) 1, (2) 7657
  Teeno alag-alag primes hain, isliye unka akela common factor 1 hai, aur LCM unka product hai: $13 \times 19 \times 31 = 7657.$
3. $4,$ $15$ aur $49 :$ HCF $=$ aur LCM $=$ .
  Answer dekho
  Answer (1) 1, (2) 2940
  $4 = 2^{2},$ $15 = 3 \times 5$ aur $49 = 7^{2}$ mein koi prime common nahi hai, isliye HCF $= 1$ aur LCM $= 4 \times 15 \times 49 = 2940.$
NumericalEasyNCERT Exercise 1.1, Q4
345 aur 552 ka HCF 69 hai. Dobara factorise kiye bina unka LCM nikaalo.
Answer dekho
Answer 2760
LCM $= \frac{345 \times 552}{69} .$ Kyunki $345 = 5 \times 69,$ yeh hai $5 \times 552 = 2760.$
Short answerMediumNCERT Exercise 1.1, Q5
Kya koi natural number $n$ hai jiske liye $14^{n}$ ka last digit 0 ho? Reason do.
Haan, kuch even $n$ ke liyeHaan, har $n$ ke liyeNahi, kabhi nahi
Solution dikhao
0 par khatam hone wala number is number se divisible hota hai: $10 = 2 \times 5,$ isliye uske prime factorisation mein 5 aana hi chahiye. Lekin $14^{n} = 2^{n} \times 7^{n},$ aur uniqueness ke kaaran ismein koi aur prime nahi hai. 5 ke bina $14^{n}$ kabhi 0 par khatam nahi ho sakta.
6
Short answerMediumNCERT Exercise 1.1, Q6
Samjhao ki har number composite kyun hai.
1. $3 \times 5 \times 17 + 17$
  Solution dikhao
  Common factor 17 bahar nikaalo: number hai $17 \times (15 + 1) = 17 \times 16 = 272.$ Iske 1 aur khud ke alawa 16 aur 17 bhi factors hain, isliye yeh composite hai.
2. $8 \times 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1 + 7$
  Solution dikhao
  Pehla term hai $7 \times 5760,$ kyunki $8 \times 6 \times 5 \times 4 \times 3 \times 2 = 5760.$ To number hai $7 \times 5760 + 7 = 7 \times 5761.$ Iska ek factor 7 hai aur yeh 7 se bada hai, isliye yeh composite hai.
NumericalMediumNCERT Exercise 1.1, Q7
Do buses ek depot se saath-saath 7:00 a.m. par nikalti hain. Route A wali bus ek round trip poori karke har 40 minutes mein dobara nikalti hai; route B wali bus yahi har 75 minutes mein karti hai. Kitne minutes baad woh agli baar saath nikalengi, aur kis time par?
min
Solution dikhao
Woh 40-minute wale poore rounds aur 75-minute wale poore rounds, dono ke baad hi saath nikalti hain, isliye humein LCM chahiye.
$40 = 2^{3} \times 5$ aur $75 = 3 \times 5^{2},$ isliye LCM $= 2^{3} \times 3 \times 5^{2} = 600$ minutes, yaani 10 ghante. Woh agli baar 5:00 p.m. par saath nikalengi.
Quick answerMedium
Agar $p = x^{2} y^{3}$ aur $q = x^{4} y,$ jahan $x$ aur $y$ alag-alag primes hain, to LCM $(p, q)$ hai
$x^{2} y$ $x^{4} y^{3}$ $x^{6} y^{4}$ $x^{2} y^{3}$
Answer dekho
Answer (B) $x^{4} y^{3}$
Higher powers hain $x^{4}$ aur $y^{3} .$ Option A HCF hai, aur option C to product hai: $p q .$
Quick answerEasy
Do numbers ka HCF 16 aur product 3072 hai. Unka LCM hai
Answer dekho
Answer 192
LCM $= 3072 \div 16 = 192.$ Jaise, 48 aur 64 ka HCF 16 aur LCM 192 hai.

1.3 Irrational Numbers par Dobara Nazar

Tumhara calculator kehta hai $\sqrt{2} = 1.414213562.$ Kya yeh fraction hai? Har terminating decimal ek fraction hota hai: $1.414213562 = \frac{1414213562}{10^{9}} .$ Lekin iska square karne par aata hai $1.99999999894 \dots,$ theek 2 nahi. Screen sirf ek approximation dikhati hai, aur koi bhi screen itne digits nahi dikha sakti ki tay ho jaaye ki $\sqrt{2}$ fraction hai ya nahi. Humein proof chahiye, aur saara bhaari kaam Fundamental Theorem of Arithmetic karega.

Definition

Irrational number

Ek real number tab irrational hota hai jab koi bhi fraction $\frac{p}{q},$ jismein numerator integer $p$ ho aur denominator non-zero integer $q,$ uske barabar na ho.

Example $\sqrt{11},$ $π$ aur $0.123456789101112 \dots$ (ginti ke numbers ek ke baad ek likhe hue) irrational hain.

Square ko divide karne wala prime

$\sqrt{2}$ ke proof ke liye primes aur squares ke baare mein ek fact chahiye. Kisi number ka square karne se uske prime factorisation ki har power double ho jaati hai, lekin koi naya prime nahi aata. Jaise, $30 = 2 \times 3 \times 5$ aur $30^{2} = 900 = 2^{2} \times 3^{2} \times 5^{2} .$ To 900 ko divide karne wala koi bhi prime 2, 3 ya 5 hi hoga, aur inmein se har ek 30 ko divide karta hai (Fig. 1.7). Short mein, $p ∣ a$ ka matlab hai $p$ , $a$ ko divide karta hai, aur $p ∤ a$ ka matlab hai ki nahi karta.

Theorem 1.2Square ko divide karne wala prime

Maan lo $p$ ek prime hai aur $a$ ek positive integer. Agar $p$ divide karta hai $a^{2},$ to $p$ divide karta hai $a .$

Tumhari textbook is proof ko examination ke liye nahi maanti; lekin theorem khud baar-baar use hota hai.

$a$ ko primes ke product ki tarah likho, $a = p_{1} p_{2} \dots p_{n}$ (repeats allowed). Tab $a^{2} = p_{1}^{2} p_{2}^{2} \dots p_{n}^{2},$ isliye $a^{2}$ ke primes theek yahi hain: $p_{1}, \dots, p_{n},$ har ek do baar use hua.

Ab maan lo $p$ divide karta hai $a^{2},$ yaani $a^{2} = p \times m .$ $m$ ko primes mein factorise karo aur aage $p$ lagao: yeh $a^{2}$ ka ek aisa prime factorisation hai jismein aata hai $p .$ Uniqueness se yeh wahi hai jo $p_{1}^{2} p_{2}^{2} \dots p_{n}^{2},$ isliye $p$ inmein se ek hai: $p_{1}, \dots, p_{n} .$

To $p$ un primes mein se ek hai jinse banta hai $a,$ aur isliye $p$ divide karta hai $a .$

4/44 prime nahi hai: $4 ∣ 36$ lekin $4 ∤ 6,$ kyunki uske do 2 6 ki alag-alag copies se aate hain.

Fig. 1.7Square karne se wahi primes repeat hote hain; isiliye

a^{2}

ko divide karne wala prime zaroor divide karega

a .

Step 1 a = 30 ko teen tiles ki tarah dikhata hai: 2, 3 aur 5. Step 2 a squared = 900 ko unhi teen tiles ki do copies ki tarah jodta hai. Step 3 chhe tiles ko bracket karta hai, note karta hai ki sirf 2, 3 aur 5 aate hain, aur kehta hai ki agar koi prime a squared ko divide kare to woh a ko divide karta hai. Step 4 mein, ek dashed line ke neeche, sawaal hai ki prime kyun: 6 squared tiles 2 aur 3 ki do copies hai, aur ek dashed link dono 2-tiles ko jodta hai, har copy se ek, kyunki 4 ko dono 2 chahiye. To 4, 6 squared ko divide karta hai lekin 4, 6 ko divide nahi karta.

Proof by contradiction: $\sqrt{2}$ irrational hai

Koi kaise prove kar sakta hai ki kuch impossible hai, ki infinitely many fractions mein se ek bhi aisa nahi jiska square theek 2 ho? Hum proof by contradiction use karte hain. Jo prove karna hai uska ulta maan lo. Us assumption se sahi-sahi reasoning karo. Agar tum kisi aisi baat par pahunchte ho jo sach ho hi nahi sakti, to assumption galat tha.

Tum pehle se aise argue karte ho. Ek dost kehti hai ki woh tez baarish mein bina chhatri ke paidal ghar aayi, phir bhi woh bilkul sookhi pahunchti hai. Agar uski kahani sach hoti, to woh geeli hoti. Woh sookhi hai, isliye kahani jhoothi hai. Proof mein impossible baat aksar ek aisa statement hoti hai jo hamare hi kisi starting fact se takraata hai.

Theorem 1.3

\sqrt{2}

irrational hai

Do integers ka koi aisa fraction nahi hai jiska square 2 ho; yaani $\sqrt{2}$ irrational hai.

Ulta maan lo. Maan lo $\sqrt{2}$ rational hai, to $\sqrt{2} = \frac{a}{b}$ kuch integers $a$ aur $b$ ke liye, jahan $b \neq 0.$ Pehle saare common factors cancel kar do, taaki $a$ aur $b$ coprime ho jaayein.

2 b^{2} = a^{2}

To 2 divide karta hai $a .$ Left side 2 ka multiple hai, isliye 2 divide karta hai $a^{2} .$ Kyunki 2 prime hai, Theorem 1.2 kehta hai ki 2 divide karta hai $a .$ Ab $a = 2 c$ likho, jahan koi integer hai $c .$

b^{2} = 2 c^{2}

To 2 $b$ ko bhi divide karta hai. Ab 2 divide karta hai $b^{2},$ aur Theorem 1.2 phir se batata hai ki 2 divide karta hai $b .$

Contradiction. 2 dono ko divide karta hai, $a$ aur $b,$ jabki humne unhe coprime chuna tha. Assumption ke baad ka har step sahi tha, isliye assumption khud galat hai. Isliye $\sqrt{2}$ irrational hai.

5/52 dono ko divide karta hai, $a$ aur $b,$ lekin woh coprime the. To $\sqrt{2}$ irrational hai.

Fig. 1.8Assumption ke baad ka har step sahi hai, isliye end ka takraav sirf assumption se hi aa sakta hai.

Paanch boxes neeche ki taraf arrows se jude hain. Step 1 pehla box dikhata hai: maan lo root 2 equals a over b, lowest terms mein. Step 2 jodta hai dono sides ka square, jisse a squared equals 2 b squared milta hai. Step 3 jodta hai to 2 divides a, aur a = 2c. Step 4 jodta hai phir b squared equals 2 c squared, isliye 2 divides b. Step 5 ek cross wala dashed box jodta hai jo kehta hai ki 2 divides a aur b, jo lowest terms se takraata hai, aur conclusion ki root 2 irrational hai.

Example 5NCERT Example 5 · p. 7

Root ke neeche ek aur prime

Prove karo ki $\sqrt{7}$ irrational hai.

Pehle khud try karo. Phir solution ek-ek step karke kholo.

Example 5 ko phir se dekho: 7 ke baare mein humne sirf yeh fact use kiya ki woh prime hai. 7 ki jagah koi bhi prime $p$ rakho aur har line phir bhi kaam karti hai. To $\sqrt{3},$ $\sqrt{5},$ $\sqrt{11},$ aur general mein $\sqrt{p}$ , chahe jo bhi prime ho $p,$ sab irrational hain. Yeh argument 9 jaise perfect square ke liye fail hota hai, aur yeh sahi bhi hai, kyunki $\sqrt{9} = 3$ rational hai.

Quick check

3 mein se 1

Us proof ke steps ko order mein rakho jo dikhata hai ki

\sqrt{5}

irrational hai.

To $5 ∣ a;$ rakho $a = 5 c$
5 dono ko divide karta hai: contradiction
Maan lo $\sqrt{5} = \frac{a}{b},$ coprime
$5 b^{2} = a^{2}$ milta hai
Milta hai $b^{2} = 5 c^{2},$ to $5 ∣ b$

Answer dekho

Answer

Maan lo $\sqrt{5} = \frac{a}{b},$ coprime
$5 b^{2} = a^{2}$ milta hai
To $5 ∣ a;$ rakho $a = 5 c$
Milta hai $b^{2} = 5 c^{2},$ to $5 ∣ b$
5 dono ko divide karta hai: contradiction

Maan lo, square karo, nikaalo ki 5 divide karta hai

a,

substitute karke dikhao ki 5 divide karta hai

b,

phir coprime ke saath takraav pakdo.

Rational aur irrational numbers ko milana

Kya $2 + \sqrt{7}$ irrational hai? Kya $\frac{\sqrt{7}}{3}$ hai? Dono sawaalon ka jawaab ek hi hai aur chhota sa proof bhi wahi. Maan lo number rational hai, phir rational part ko undo karo. Rational number wale step ko undo karne par tum rationals ke beech hi rehte ho, isliye $\sqrt{7}$ khud rational nikal aayega, jo impossible hai. Fig. 1.9 number line par yeh plan dikhata hai.

3/5Agar $2 + \sqrt{7} = \frac{a}{b},$ to $\sqrt{7} = \frac{a}{b} - 2$ rational hota. Impossible.

Fig. 1.9Non-zero rational se add, subtract, multiply ya divide karna ek rational se hi undo ho sakta hai, isliye yeh kabhi kisi irrational number ko rational nahi banata.

0 se 5 tak ek number line jismein root 7 2.6 se thoda aage mark hai. Step 1 sirf root 7 dikhata hai. Step 2 root 7 se 2 + root 7 tak, lagbhag 4.65 ke paas, +2 label wala ek solid arc banata hai. Step 3 −2 label wala ek dashed arc wapas root 7 tak aur ek cross banata hai: agar 2 + root 7, a over b hota, to root 7, a over b minus 2 ke barabar hota, jo ek rational number hai. Step 4 iski jagah root 7 se root 7 over 3 tak, lagbhag 0.88 ke paas, divide by 3 label wala arc banata hai. Step 5 times 3 label wala ek dashed arc wapas root 7 tak aur ek cross banata hai, kyunki tab root 7, 3a over b ke barabar hota.

Example 6NCERT Example 6 · p. 8

Irrational part wala sum

Dikhao ki $2 + \sqrt{7}$ irrational hai.

Pehle khud try karo. Phir solution ek-ek step karke kholo.

Example 7NCERT Example 7 · p. 8

Irrational part wala quotient

Dikhao ki $\frac{\sqrt{7}}{3}$ irrational hai.

Pehle khud try karo. Phir solution ek-ek step karke kholo.

Examples 6 aur 7 do aise facts ke special cases hain jo tumne Class 9 mein dekhe the. Maan lo $r$ ek rational number hai aur $s$ ek irrational number. Tab:

$r + s$ aur $r - s$ irrational hain;
$r \times s$ aur $\frac{s}{r}$ irrational hain, bas shart yeh hai: $r \neq 0.$

Proofs bilkul in do examples jaise hain: $r$ replace karta hai 2 ya 3 ko, aur $s$ replace karta hai $\sqrt{7} .$

Practice · 5 questions

Practice 1.2: answers ke saath

ProofMediumNCERT Exercise 1.2, Q1
Prove karo ki $\sqrt{11}$ irrational hai.
Solution dikhao
Maan lo $\sqrt{11} = \frac{a}{b}$ , jahan $a,$ $b$ coprime integers hain aur $b \neq 0.$
Tab $11 b^{2} = a^{2},$ isliye 11 divide karta hai $a^{2} .$ Kyunki 11 prime hai, 11 divide karta hai $a;$ likho $a = 11 c .$
Tab $11 b^{2} = 121 c^{2},$ isliye $b^{2} = 11 c^{2} .$ To 11 divide karta hai $b^{2},$ aur isliye 11 divide karta hai $b .$
To 11 dono ko divide karta hai, $a$ aur $b,$ jo unke coprime hone ke khilaaf hai. Isliye $\sqrt{11}$ irrational hai.
ProofMediumNCERT Exercise 1.2, Q2
Prove karo ki $4 - 5 \sqrt{3}$ irrational hai. Tum yeh fact use kar sakte ho ki $\sqrt{3}$ irrational hai.
Solution dikhao
Maan lo $4 - 5 \sqrt{3} = \frac{a}{b}$ , jahan $a,$ $b$ integers hain aur $b \neq 0.$
Tab $5 \sqrt{3} = 4 - \frac{a}{b},$ isliye $\sqrt{3} = \frac{4 b - a}{5 b} .$
Right side integers ka ratio hai aur $5 b \neq 0,$ isliye yeh rational hai. Lekin $\sqrt{3}$ irrational hai (Example 5, 7 ki jagah 3 lekar). Yeh contradiction dikhata hai ki $4 - 5 \sqrt{3}$ irrational hai.
3
ProofEasyNCERT Exercise 1.2, Q3
Prove karo ki har number irrational hai.
1. $\frac{1}{\sqrt{3}}$
  Solution dikhao
  Agar $\frac{1}{\sqrt{3}} = \frac{a}{b}$ , jahan $a,$ $b$ integers hain (dono non-zero, kyunki number 0 nahi hai), to $\sqrt{3} = \frac{b}{a}$ rational hota. Yeh impossible hai, isliye $\frac{1}{\sqrt{3}}$ irrational hai.
2. $- 2 \sqrt{5}$
  Solution dikhao
  Agar $- 2 \sqrt{5} = \frac{a}{b},$ to $\sqrt{5} = - \frac{a}{2 b},$ ek rational number. Lekin $\sqrt{5}$ irrational hai (5 prime hai), isliye $- 2 \sqrt{5}$ irrational hai.
3. $\sqrt{2} - 8$
  Solution dikhao
  Agar $\sqrt{2} - 8 = \frac{a}{b},$ to $\sqrt{2} = \frac{a + 8 b}{b},$ jo rational hai. Yeh Theorem 1.3 ke khilaaf hai, isliye $\sqrt{2} - 8$ irrational hai.
ProofHard
Prove karo ki $\sqrt{2} + \sqrt{3}$ irrational hai.
Solution dikhao
Maan lo $\sqrt{2} + \sqrt{3} = r,$ ek rational number. Tab $r > 0,$ isliye $r \neq 0.$
Tab $\sqrt{3} = r - \sqrt{2} .$ Square karne par, $3 = r^{2} - 2 r \sqrt{2} + 2.$
To $2 r \sqrt{2} = r^{2} - 1,$ jisse milta hai $\sqrt{2} = \frac{r^{2} - 1}{2 r} .$
Right side rational hai, isliye $\sqrt{2}$ rational ho jaata, jo Theorem 1.3 ke khilaaf hai. Isliye $\sqrt{2} + \sqrt{3}$ irrational hai.
Quick answerEasy
In mein se kaunsa number rational hai?
$\sqrt{2} + \sqrt{8}$ $(\sqrt{5})^{3}$ $(2 + \sqrt{3}) (2 - \sqrt{3})$ $\frac{3}{\sqrt{6}}$
Answer dekho
Answer (C) $(2 + \sqrt{3}) (2 - \sqrt{3})$
$(2 + \sqrt{3}) (2 - \sqrt{3}) = 4 - 3 = 1.$ Baaki simplify hokar bante hain $3 \sqrt{2},$ $5 \sqrt{5}$ aur $\frac{\sqrt{6}}{2},$ jo sab irrational hain.

1.4 Summary

Recap

Summary aur key points

Ek factor tree kisi bhi composite number ko primes mein tod deta hai (Section 1.2).
Fundamental Theorem of Arithmetic: order ko chhod kar, ek composite number ka sirf ek hi prime factorisation hota hai (Theorem 1.1).
Uniqueness divisibility ke sawaal suljha deti hai: $15^{n}$ kabhi 0 par khatam nahi hota, kyunki ismein factor 2 nahi hai (Example 1).
HCF: common primes, unki lowest powers par. LCM: saare primes, unki highest powers par (HCF aur LCM).
Do numbers ke liye HCF × LCM $= a \times b$ ((1.1)); teen ke liye yeh aksar fail hota hai (Example 4).
Agar prime $p$ divide karta hai $a^{2},$ to $p$ divide karta hai $a$ (Theorem 1.2).
$\sqrt{2}$ aur $\sqrt{p},$ chahe jo bhi prime ho $p,$ dono irrational hain: proof by contradiction (Theorem 1.3, Example 5).
Ek rational number ko irrational ke saath $+,$ $-,$ $\times$ ya $\div$ (non-zero rational) se jodne par irrational milta hai (numbers ko milana).

Flash cards · 8

Quick recall

1 / 8

Board-exam practice

Class 10 board papers is chapter ko multiple-choice aur assertion–reason questions, short answers, case studies aur irrationality ke proofs se test karte hain. Neeche ke questions isi style mein hain.

Apna NCERT exercise question dhoondo

Hamare practice questions ncert.ai ke liye likhe gaye hain aur NCERT exercises ko follow karte hain. Apni textbook ke question se match karne wala question dhoondo; uska answer uske saath hai.

Tumhari NCERT textbook mein	Yahan practice question
NCERT Exercise 1.1, Q1	Practice 1.1 · Q1
NCERT Exercise 1.1, Q2	Practice 1.1 · Q2
NCERT Exercise 1.1, Q3	Practice 1.1 · Q3
NCERT Exercise 1.1, Q4	Practice 1.1 · Q4
NCERT Exercise 1.1, Q5	Practice 1.1 · Q5
NCERT Exercise 1.1, Q6	Practice 1.1 · Q6
NCERT Exercise 1.1, Q7	Practice 1.1 · Q7
NCERT Exercise 1.2, Q1	Practice 1.2 · Q1
NCERT Exercise 1.2, Q2	Practice 1.2 · Q2
NCERT Exercise 1.2, Q3	Practice 1.2 · Q3

Quick check: MCQs

1.1 Introduction2 questions
1.2 The Fundamental Theorem of Arithmetic3 questions
Prime factorisation se HCF aur LCM3 questions
Proof by contradiction: $\sqrt{2}$ irrational hai3 questions
Rational aur irrational numbers ko milana3 questions

Glossary · 12

Key terms (English + Hindi)

composite numberभाज्य संख्याbhajya sankhya: 1 se bada woh whole number jiska 1 aur khud ke alawa bhi koi factor ho, jaise 9 ya 91. Number 1 na prime hai na composite.Chapter mein pehli baar yahan aaya →
coprimeसहअभाज्यsah-abhajya: Do integers coprime hote hain jab unka akela positive common factor 1 ho, jaise 8 aur 15.Chapter mein pehli baar yahan aaya →

factor treeगुणनखंड वृक्षgunankhand vriksh: Ek diagram jo number ko baar-baar do factors mein todta hai, jab tak har branch ek prime par khatam na ho.Chapter mein pehli baar yahan aaya →
Fundamental Theorem of Arithmeticअंकगणित की आधारभूत प्रमेयankganit ki aadharbhoot pramey: Har composite number primes ka product hai, aur factors ke order ko chhod kar yeh product unique hai.

HCF (highest common factor)महत्तम समापवर्तकmahattam samapavartak: Woh sabse bada whole number jo diye gaye har number ko poora divide karta hai.Chapter mein pehli baar yahan aaya →

irrational numberअपरिमेय संख्याaparimey sankhya: Aisa real number jiske barabar do integers ka koi fraction nahi ho sakta, jaise $\sqrt{2}$ ya $π .$

LCM (lowest common multiple)लघुत्तम समापवर्त्यlaghuttam samapavartya: Woh sabse chhota positive whole number jo diye gaye har number ka multiple ho.Chapter mein pehli baar yahan aaya →

prime factorisationअभाज्य गुणनखंडनabhajya gunankhandan: Primes ke product ki tarah likha gaya number, jaise $60 = 2^{2} \times 3 \times 5.$ Chapter mein pehli baar yahan aaya →
prime numberअभाज्य संख्याabhajya sankhya: 1 se bada woh whole number jiske factors sirf 1 aur khud woh number hain, jaise 2, 7 ya 13.Chapter mein pehli baar yahan aaya →
proof by contradiction: Aisa proof jo claim ka ulta maanta hai, sahi reasoning se kisi impossible baat tak pahunchta hai, aur isliye assumption ko reject kar deta hai.Chapter mein pehli baar yahan aaya →

rational numberपरिमेय संख्याparimey sankhya: Aisa number jo do integers ke fraction $\frac{p}{q}$ ke barabar ho, jahan $q \neq 0,$ jaise $\frac{3}{4}$ ya $- 5.$
real numberवास्तविक संख्याvastavik sankhya: Koi bhi rational ya irrational number; dono milkar poori number line bhar dete hain.Chapter mein pehli baar yahan aaya →

Real Numbers: Class 10 Maths Chapter 1 Hinglish mein

Is chapter mein aap kya seekhenge

1.1 Introduction

1.2 The Fundamental Theorem of Arithmetic

Kya koi power 0 par khatam ho sakti hai?

Prime factorisation se HCF aur LCM

Practice 1.1: answers ke saath

1.3 Irrational Numbers par Dobara Nazar

Square ko divide karne wala prime

Proof by contradiction: $\sqrt{2}$ irrational hai

Rational aur irrational numbers ko milana

Practice 1.2: answers ke saath

1.4 Summary

Summary aur key points

Board-exam practice

Practice questions answers ke saath (NCERT “Board-exam practice”)

Apna NCERT exercise question dhoondo

Quick check: MCQs

Key terms (English + Hindi)

Tum chapter ke end tak pahunch gaye

1.1 Introduction

1.2 The Fundamental Theorem of Arithmetic

Kya koi power 0 par khatam ho sakti hai?

Prime factorisation se HCF aur LCM

1.3 Irrational Numbers par Dobara Nazar

Square ko divide karne wala prime

Proof by contradiction: 2 irrational hai

Rational aur irrational numbers ko milana

1.4 Summary

Board-exam practice

Apna NCERT exercise question dhoondo

Quick check: MCQs

Tum chapter ke end tak pahunch gaye

Proof by contradiction: $\sqrt{2}$ irrational hai