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Real numbers

Lesson 87 of 113

1. Ye topic hai kya?

Number line pe jo bhi plot ho sake = real. Uske andar do khand: jo p/q likh sakte (rational), aur jo nahi (irrational). Is chapter ka kaam teen sawalon ka jawab dena hai:

  1. Ye number kis dabbe mein baitha — N, W, Z, Q, R, ya irrational?
  2. Iska decimal khatam hoga, block dohrayega, ya na-khatam na-cycle?
  3. Do numbers ka HCF Euclid se, aur unique prime toot se HCF / LCM / square?

Paper yahan kya maangta

Set-membership, terminating test (simplify pehle), repeating ko fraction, remainder lemma, integer ki form (2q, 4q+1…), Euclid algorithm, FTA, √n rational kab, √2-type proof, Q ± irrational.

Yahan kya nahi

Bells, tiles, rooms — HCF/LCM extra page. Binary, unit-digit, 3/9 se katna — Number system page. Polynomials agla chapter.

2. Sets — dabba andar dabba

Subset ka matlab: chhote dabbe ka har member bade mein bhi hai. Isliye chain yeh hai:

N ⊂ W ⊂ Z ⊂ Q ⊂ R
Natural ⊂ Whole ⊂ Integer ⊂ Rational ⊂ Real.

Ulta nahi chalta. 0 whole hai, natural nahi. −5 integer hai, whole nahi. 3/4 rational hai, integer nahi. √2 real hai, rational nahi.

−3 −2 −1 0 1 2 3 1/2 √2 π

Chhote dots = rational jagah (integers, 1/2). Doosre colour = irrational (√2 ≈ 1.41, π ≈ 3.14). Dono real line pe baithte hain — farq yeh ki irrational p/q nahi.

R · Real Jo is line pe ek point ho. Q ∪ irrationals. √−1 (imaginary) yahan nahi — wo line pe nahi baithta.
Q · Rational = p/q, p aur q integers, q ≠ 0 Same number kai fractions: 1/2 = 2/4 = 3/6. Lowest terms: HCF(p,q)=1.
Z · Integer = … −2, −1, 0, 1, 2 … Har integer rational hai kyunki n = n/1. q=1 allowed.
W · Whole = 0, 1, 2, 3 … Natural + 0. Negative nahi.
N · Natural = 1, 2, 3 … Is notes mein 0 natural nahi. “Smallest natural number?” → 1. Kuch books 0 ko N mein rakhte — paper ka convention dekho, yahan 1.

Membership — har number ko tick

Sawal aata hai: “0 kis-kis mein hai?” Ek-ek set pe haan/nahi. Jo chhote mein hai, uske baad ke bade mein bhi hoga — agar chain toot na jaye.

NumberNWZQRKyun
7haanhaanhaanhaanhaanNatural, aur 7/1
0nahihaanhaanhaanhaan0 = 0/1. Counting 1 se.
−5nahinahihaanhaanhaanNegative integer = −5/1
3/4nahinahinahihaanhaanFraction, poora nahi
−2/3nahinahinahihaanhaanNegative bhi Q mein
√9 = 3haanhaanhaanhaanhaanPehle simplify, phir set
√2nahinahinahinahihaanp/q nahi ban-ta
πnahinahinahinahihaan22/7 approximation hai, π nahi
Irrational = real minus rational. Line pe hai, p/q nahi.
Classic: √2, √3, √5, √p (p prime), π, 0.1010010001… (har baar extra 0, block same nahi).
√4 = 2 — radical dikha isliye irrational mat bolna. Andar perfect square ho to answer integer/rational.
√8 = 2√2 — 8 = 4×2, 4 nikal gaya, √2 reh gaya → irrational.
0.333… khatam nahi hota, lekin repeat karta hai → rational (1/3). “Khatam nahi = irrational” galat rule hai.
22/7 rational hai. π irrational. Paper inhe milata hai jaan-bujh ke.

Do rationals ke beech hamesha aur rationals milte hain (1/2 aur 1 ke beech 3/4, 5/8, 9/16…). Irrationals bhi. Isliye line “ghan” hai — exam mein rarely poochhte, idea yeh ki Q ke beech khali jagah nahi samajhna.

3. Decimal — teen tarah, aur kyun

p/q ka matlab: p ko q se divide. Decimal point ke baad zeros laga ke chalte raho. Har step pe remainder bacha.

Remainder 0 → divide khatam → terminating (0.375).
Remainder wahi cycle mein wapas → digits dohrate → repeating (0.16).
Na 0, na cycle — yeh p/q se ho hi nahi sakta. Aisa decimal irrational (√2, π).

Kyun har rational ya to terminate ya repeat?

q se divide karte waqt remainder sirf 0, 1, 2, …, q−1 ho sakte — q possible values. Infinite steps, finite remainders. Do hi raaste:

  • Kisi step pe remainder 0 → khatam.
  • Koi remainder doosri baar aaya → uske baad wahi digits — cycle. Cycle ki lambai zyada se zyada q−1 (0 chhod ke).

Isliye: rational ⇔ terminating ya repeating. Ulta: non-terminating non-repeating ⇒ p/q nahi ⇒ irrational.

Long division — aankh se dekho

3/8 terminate

8 × 0 = 0, bacha 3. 30: 8×3=24, rem 6 → digit 3.

60: 8×7=56, rem 4 → digit 7.

40: 8×5=40, rem 0 → digit 5. Khatam.

3/8 = 0.375

1/6 repeat

6 × 0 = 0, bacha 1. 10: 6×1=6, rem 4 → digit 1.

40: 6×6=36, rem 4 → digit 6.

Wapas rem 4. Wahi 6,6,6…

1/6 = 0.16 (1 non-repeat, 6 cycle)

1/7 = 0.142857 — chhe digits ka block, phir wahi. 7 se remainder 1…6 hi ho sakte, isliye period ≤ 6. Yahan exactly 6.
TarahDecimalSetExample
TerminatingKhatamQ1/2=0.5, 3/8=0.375, 7/20=0.35
Non-term repeatingBlock dohrataQ1/3=0.3, 1/6=0.16, 2/7=0.142857
Non-term non-repeatNa 0 na cycleIrrational√2=1.414213…, π, 0.1010010001…

Terminating kab? Sabse important MCQ

Decimal khatam = fraction ko ? / 10, 100, 1000, … bana sakte. 10 = 2×5, isliye 10k = 2k5k. Denominator ke primes mein 2 aur 5 ke alawa kuch bacha to 10k nahi banega.

1) p/q ko lowest terms lao — HCF(p,q)=1. (Cancel pehle!)
2) Ab q ke prime factors dekho.
3) Sirf 2 aur/ya 5 → terminating. Matlab q = 2a5b (a,b ≥ 0, ek zero bhi chalega).
4) Koi 3, 7, 11, 13… bacha → non-terminating repeating.

Kitne digits pe khatam? q = 2a5b ho to decimal places = max(a, b). Extra 2 ya extra 5 se multiply karke dono powers barabar, denominator 10max.

Kaise likho terminating decimal — multiply wala rasta

3/8. 8 = 2³. 5 ki kami = 3. ×5³ dono taraf:

3/8 = 3×125 / 8×125 = 375/1000 = 0.375. Places = max(3,0) = 3.

7/20. 20 = 2²×5. Extra 2 ek. ×5: 35/100 = 0.35. Places = max(2,1) = 2.

9/25. 25 = 5². Extra 5. ×2²: 36/100 = 0.36.

7/80. 80 = 2⁴×5. Places = max(4,1) = 4 → 0.0875.

Pehle simplify. 21/56 dekh ke “56=7×8, 7 hai terminate nahi” — galat. 21 aur 56 ka HCF=7, 21/56=3/8, 8=2³, terminate. Numerator ke 7 ne denominator ka 7 kha liya.
“Denominator mein 2 hai to terminate” — nahi. 1/6 = 2×3, 3 extra → 0.16.
“Numerator prime hai to terminate nahi” — nahi. 7/8 terminate. Test q pe hota hai, p pe nahi (lowest terms ke baad).

Ulta: terminating decimal → p/q

Digits jitni, utna 10-power. Phir simplify.

0.375 = 375/1000. HCF 125 → 3/8. 0.6 = 6/10 = 3/5. 0.35 = 35/100 = 7/20.

0.6 (ek 6, khatam) = 3/5. 0.6 (6 dohrata) = 2/3. Bar hai ya nahi — pehle dekho. 0.06 = 6/99 = 2/33, 6/99 mat chhodna bina simplify.

Repeating → fraction: x-trick (kyun 10, kyun 100)

Repeat ki length = k digits. 10k se multiply karo — decimal point k jagah aage, repeating tail line ho jati hai. Minus karo, tail cancel, bacha integer / (10k−1). 10k−1 = 9, 99, 999…

Example · 1 digit · 0.6

0.666… ko p/q banao.

Solution

  1. x rakho

    x = 0.666…

  2. ×10

    1 digit repeat → 10x = 6.666… Tail same.

  3. Minus

    10x − x = 6.666… − 0.666… = 6. 9x = 6, x = 6/9 = 2/3.

Answer2/3
Example · 2 digit · 0.27

0.2727… ko p/q banao.

Solution

  1. ×100

    2 digit → 100x = 27.2727…

  2. Minus

    99x = 27, x = 27/99. HCF 9 → 3/11.

Answer3/11

Mixed — pehle non-repeating digits, phir cycle. Pehle 10m se non-repeat ko point ke left lao. Phir 10k aur, taaki dono copies ki tail same ho. Minus.

Example · mixed · 0.16

0.1666… = ? (yeh 1/6 hai — steps se nikalo, yaad mat karo.)

Solution

  1. m=1, k=1

    x = 0.1666… Non-repeat ek (1), repeat ek (6).

  2. ×10

    10x = 1.666… Ab decimal ke baad sirf cycle.

  3. ×100

    100x = 16.666… Tail 666… dono mein same.

  4. Minus

    100x − 10x = 16.666… − 1.666… = 15. 90x = 15, x = 15/90 = 1/6.

Answer1/6

Shortcut yaad: pure repeat of k digits = (block) / (k nines), phir simplify. Mixed mein nines ke saath zeros bhi aate (90, 990…) — shortcut ratne se steps safer.

0.1010010001… alag jaanwar hai: zeros badhte hain, koi fixed block nahi. Repeating nahi → irrational. 0.101001000100001… ko 0.10 mat padhna.

4. Euclid lemma — remainder aur “form”

Lemma division hi hai, bas remainder ki boundary lock: r kabhi negative nahi, kabhi divisor se bada ya barabar nahi.

Positive integers a, b (b > 0): unique integers q, r exist —
a = bq + r · 0 ≤ r < b
q = kitni baar b aata (quotient), r = bacha (remainder).

r < b kyun? Agar r ≥ b, ek aur b nikaal sakte, naya remainder chhota. Isliye r = b allowed nahi. r = 5, b = 5 galat — wo to agla quotient hai, remainder 0 hona chahiye.

17 ko 5 se: 5×3=15, bacha 2. 17 = 5×3 + 2. Agar 15 exactly: r=0, matlab b, a ko divide karta hai.

q=2, r=7 for 17÷5 mat likhna. 7 < 5 false. q=4, r=−3 bhi nahi — r ≥ 0.

Har integer kisi form mein

b fix, r ghoomta 0 se b−1. Yehi “n kis form ka hai?” MCQ.

bHar integerBolchaal
22q ya 2q+1even / odd
33q, 3q+1, 3q+23 se rem 0,1,2
44q, 4q+1, 4q+2, 4q+3odd = 4q+1 ya 4q+3
66q … 6q+56 se coprime odds: 6q+1, 6q+5

Odd hamesha 4q+1 ya 4q+3 — kyun?

Lemma b=4: sirf chaar buckets. 4q even (4 se). 4q+2 = 2(2q+1) even (2 se, 4 se nahi — jaise 6, 10, 14). Bachhe do: 4q+1 aur 4q+3 — dono odd. Koi odd inke bahar nahi.

Example · form

47 ko 4q + r, 0 ≤ r < 4. Odd isliye kaun si form?

Solution

  1. Divide

    4×11=44, r=3. 47=4×11+3.

  2. Odd check

    r=1 ya 3 hona chahiye. 3 aaya → 4q+3, q=11.

Answer4q+3

Square ki form — even / odd

Even n=2k → n²=4k²=4m. Even ka square 4 se kat’ta. 6²=36, 36/4=9.

Odd n=2k+1 → n²=4k²+4k+1=4k(k+1)+1. k aur k+1 consecutive — ek even, isliye k(k+1) even, 4×(even)=8 se: odd square = 8m+1 (saath 4m+1 bhi). 5²=25=8×3+1. 7²=49=8×6+1.

Exam: “odd integer ka square 8 se divide karke remainder?” → 1. “4 se?” → 1. Even square 4 se remainder 0; 8 se 0 ya 4 ho sakta (6²=36=8×4+4).

NCERT application — teen consecutive

n, n+1, n+2. Lemma b=2: inme se kam se kam ek even (asli mein ek ya do). Lemma b=3: teen consecutive mein exactly ek 3 se kat’ta. Even × (3-se-katne wala) → product 6 se kat’ta.

n(n+1)(n+2) hamesha 6 se divide. n³−n = n(n−1)(n+1) bhi teen consecutive, same — 6 se.

Bells/tile type word-problem yahan nahi. Form se “hamesha divisible” wale proof yahan.

5. Euclid algorithm — HCF kyun chal-ta hai

HCF = sabse bada number jo dono ko divide kare. Euclid ka rasta: divide, remainder lo, remainder naya divisor.

a = bq + r ho to HCF(a,b) = HCF(b,r).
r=0 aate hi: HCF = last non-zero divisor (jo b abhi tha).
Special: HCF(a,0) = a — a, a ko bhi 0 ko bhi divide karta (0 = a×0).

Kyun equal HCF? (yeh missing piece tha)

Jo d, a aur b dono ko divide kare: d | a, d | b. Phir d, a − bq ko bhi — matlab d | r. Jo common divisor (a,b) ka, wahi (b,r) ka.

Ulta: d | b aur d | r → d | (bq+r) → d | a. Common divisors ka set same → sabse bada bhi same.

Har step pair chhota hota jaata, remainder < divisor, kabhi infinite nahi. Last non-zero remainder = HCF.

Example · algorithm

HCF(405, 126) — har remainder ka matlab.

Solution

  1. Bada ÷ chhota

    405 = 126×3 + 27. (126×3=378, 405−378=27). Ab HCF(126, 27).

  2. Dobara

    126 = 27×4 + 18. (108, bacha 18). HCF(27, 18).

  3. Dobara

    27 = 18×1 + 9. HCF(18, 9).

  4. r = 0

    18 = 9×2 + 0. Stop. HCF = 9.

  5. Check

    405÷9=45, 126÷9=14. 45 aur 14 coprime — 9 se bada common nahi.

Answer9

Teen numbers: HCF(a,b,c) = HCF( HCF(a,b), c ). Pehle do, phir teesra. LCM bhi chain: LCM( LCM(a,b), c ).

Chhote numbers: listing factors bhi chal-ta (12, 18). Bade: Euclid tez. Prime-factor tab jab FTA/LCM saath chahiye. Word problems (sabse badi tile, bells saath) alag page — yahan tool.

6. FTA — unique prime toot

Prime: exactly do distinct positive divisors — 1 aur khud (2, 3, 5, 7, 11…). Composite: 1 ke alawa aur khud ke alawa koi divisor (4, 6, 9, 15, 21…).

1 na prime na composite. Prime ki definition “exactly two divisors” — 1 ke paas sirf ek (khud). 2 hi even prime; baaki even 2 se toot-te, composite.
Har integer n > 1 ko primes ke product mein likh sakte, aur order chhod ke unique. Yeh Fundamental Theorem of Arithmetic.

360 = 2³ × 3² × 5. 8 × 9 × 5 FTA form nahi — 8=2³, 9=3² primes nahi. Powers ko primes ke upar likho.

Euclid lemma (prime wala) — √2 proof ka engine

Agar p prime aur p, product ab ko divide kare, to p, a ko ya b ko (ya dono ko) divide karega. Composite pe yeh toot-ta: 4, 2×6=12 ko kat-ta hai, lekin 4 na 2 ko kat-ta na 6 ko. Prime pe aisa nahi.

p prime, p | a² ⇒ p | a. Kyunki a² = a×a, lemma se p | a ya p | a — matlab p | a.
Odd ka square odd, even ka even — isi se √2 proof mein “p² even ⇒ p even”.

HCF / LCM primes se

Dono (ya teeno) ko primes mein likho. Jo prime ek mein nahi, uski power 0 samjho. Phir:

HCF = har prime ki minimum power (jo common hai, uski chhoti power; jo missing, min=0, wo prime HCF mein nahi).
LCM = har prime ki maximum power (kisi ek mein bhi ho to LCM mein poori max power).

Min kyun? Common divisor us prime ko utni hi baar le sakta jitni dono allow karein — bottleneck = chhoti power. LCM multiple hai — har number ko cover karna, isliye sabse lambi power chahiye.

235Value
122102²×3
181202×3²
301112×3×5
HCF min1106
LCM max221180

Do numbers: HCF × LCM = a × b — kyun, aur teen pe kyun nahi

HCF ko d bolo. a = d·m, b = d·n, aur m, n coprime (warna d aur bada hota). LCM ko d, m, n sab chahiye: LCM = d·m·n. Product: HCF×LCM = d·(d m n) = (d m)(d n) = a b.

Sirf do numbers: HCF(a,b) × LCM(a,b) = a × b.
Teen pe galat: upar HCF=6, LCM=180, 6×180=1080. 12×18×30=6480. Barabar nahi — teen ke coprime-hisse overlap karte.

HCF=1 (coprime) → LCM = product. 8 aur 9 coprime, LCM=72. 8, 9, 6 teeno coprime nahi (6 aur 8 mein 2) — teeno ka LCM product nahi.

Ek number, HCF, LCM pata: doosra = (HCF × LCM) / diya hua. Formula do numbers ka hi.

Example · doosra number

HCF=12, LCM=180, ek number 36. Doosra?

Solution

  1. Formula

    Doosra = 12×180 / 36 = 2160/36 = 60.

  2. Check

    36=2²×3², 60=2²×3×5. HCF=2²×3=12, LCM=2²×3²×5=180.

Answer60

Perfect square / cube

Square = k×k. Prime toot mein har exponent double ho jata. Isliye square tabhi jab har exponent even (0, 2, 4…).

Cube = k×k×k → har exponent 3 ka multiple (0, 3, 6…).

Square banana: jo exponent odd hai, us prime se ek baar multiply (odd→even). Jo already even, chhodo.
Cube banana: har exponent ko agle 3-multiple tak le jao (1→3 ×p², 2→3 ×p, 4→6 ×p²).

180 = 2² × 3² × 5¹. Square nahi (5 odd). Chhota multiply: ×5 → 2²3²5² = 900 = 30².

Chhota divide karke square: odd powers hatao — 180÷5=36=6².

Cube ke liye 180: 2 ko 3 banana ×2, 3 ko 3 ×3, 5 ko 3 ×5². Multiply 2×3×25=150. (Paper square zyada poochhta.)

Factors kitne? (ginti, list nahi)

n = pa qb rc. Ek factor = px qy rz jahan x = 0 se a (a+1 choices), y = 0 se b, z = 0 se c.

Factors ki ginti = (a+1)(b+1)(c+1). 1 aur n khud shamil.

360 = 2³ × 3² × 5¹ → (3+1)(2+1)(1+1)=4×3×2=24 factors.

7. Irrational — proof aur operations

Positive integer n ke liye: √n tab rational jab n perfect square (0, 1, 4, 9, 16, 25, 36…). Warna √n irrational. √p, p prime — hamesha irrational (p square nahi).

√4=2 rational. √8=√(4×2)=2√2 irrational. √12=2√3 irrational. Pehle andar se square factor nikaalo, phir decide.

√2 irrational — contradiction (poora)

Assume opposite, to collision nikaalo. Collision = assumption galat.

  1. Assume

    √2 = p/q, p,q integers, q ≠ 0, fraction already lowest terms — HCF(p,q)=1. (Hamesha simplify karke shuru.)

  2. Square

    2 = p²/q² → p² = 2q². Left even, isliye p² even. Odd ka square odd (upar form), isliye p even. p = 2k.

  3. q bhi even

    (2k)² = 2q² → 4k² = 2q² → q² = 2k². q² even → q even.

  4. Collision

    p aur q dono even → 2 common factor, HCF ≥ 2. Lowest terms ke khilaaf. Isliye koi aisa p/q nahi. √2 irrational.

√3, √5, √7 same dhang: p² = 3q² ⇒ 3 | p² ⇒ 3 | p (prime lemma), p=3k, phir 3 | q, HCF toot-ta. Har prime p ke √p ke liye yahi.

3 + 2√5 bhi irrational — pattern yaad rakho

Agar 3+2√5 = r rational hota, to 2√5 = r−3 rational, √5 = (r−3)/2 rational. √5 irrational hai — collision. Isliye 3+2√5 irrational.

Non-zero rational + irrational = irrational. (Proof: agar sum rational, to irrational = sum − rational = rational. Collision.)
Non-zero rational × irrational = irrational. (Agar product rational r, to irrational = r / (wo rational). 0 se divide nahi.)

Exception jo log bhoolte: 0 × √2 = 0 rational. Zero is rational. Isliye “non-zero” zaroori. 0 + √2 = √2 ab bhi irrational — 0 add exception nahi.

Operations table — lock, phir trap

KaamResultKyun / example
Q + Qhamesha Q1/2+1/3=5/6. Sum of fractions = fraction.
Q × Qhamesha Q(2/3)×(9/4)=3/2. 0×kuch=0, Q.
Q − Q, Q÷Q (÷0 nahi)hamesha QQ closed in +, −, ×, ÷ (q≠0).
Q + irrhamesha irr2+√3, π+2/3. 0+√3=√3 irr.
Non-zero Q × irrhamesha irr5√2, −√2, π×2. Trap: 0×irr=0 Q.
irr + irrQ ya irr√2+(−√2)=0 Q; √2+√3 irr
irr × irrQ ya irr√2×√8=4 Q; √2×√3=√6 irr
Example · irr × irr = Q

√2 × √8 = ? Rational ya nahi?

Solution

  1. Root jod

    √2 × √8 = √(2×8) = √16 = 4. (Positive roots, Class-X.)

  2. Ya toot

    √8 = √(4×2)=2√2. √2 × 2√2 = 2×2 = 4.

Answer4 · rational

√2 + √3 irrational (short): maan lo = r rational. Square: 2 + 2√6 + 3 = r² → 5 + 2√6 = r² → √6 = (r²−5)/2 rational. √6 irrational (6 perfect square nahi). Collision.

“Irrational + irrational = hamesha irrational” — galat. Cancel ho sakte.
√2 + √8 = √2 + 2√2 = 3√2, ab bhi irrational. √8 ko 2√2 banana bhoolna mat, warna √2+√8 ko “do irr ka sum, cannot say” pe atak jaoge — yahan simplify pehle.
(√2+√3)(√3−√2) = 3−2 = 1, rational. Product of irrationals bhi Q ho sakta.
π + 2/3 irr. π × 0 = 0 Q. 22/7 + 1/3 Q — kyunki 22/7 π nahi.

8. Sawal — basic se pro

Upar ke rules yahan lagao. Solution step-card mein hai — pehle khud socho.

Q1 · Basic

0.333… (3 repeat) kaun sa?

  1. Irrational
  2. Rational
  3. Natural nahi isliye irrational
  4. Integer

Solution

  1. Decimal type

    3 dohrata hai — repeating. Repeating = p/q. Yahan x=0.333…, 10x−x=3, 9x=3, x=1/3.

  2. Set

    1/3 natural/integer nahi, lekin Q mein hai. C galat rule hai (“N nahi to irr”). D integer nahi.

AnswerB · Rational
Q2 · Basic

1/8 ka decimal terminating hai?

  1. Haan
  2. Nahi, 8 bada hai
  3. Sirf 1/2 aur 1/5 terminate
  4. Irrational

Solution

  1. Lowest terms

    Pehle se 1/8. q=8=2³. Sirf 2, koi 3/7 nahi.

  2. Likho

    ×5³: 125/1000 = 0.125. Places = max(3,0)=3. B size se nahi hota. C galat — 1/8, 1/4, 7/20 bhi terminate.

AnswerA · Haan
Q3 · Basic

1/6 terminating?

  1. Haan, 0.16
  2. Nahi — 6=2×3, 3 extra prime
  3. Haan kyunki 2 hai
  4. Cannot say

Solution

  1. q ke primes

    HCF(1,6)=1. 6=2×3. 3, 2×5 ke bahar. Terminate nahi.

  2. Kya hota

    0.1666… = 0.16 repeating. A “0.16” two-digit truncate hai, poora decimal nahi. C: 2 hona kaafi nahi.

AnswerB
Q4 · Basic · Euclid

a=17, b=5. Lemma: 17 = 5q + r, 0 ≤ r < 5. q aur r?

  1. q=3, r=2
  2. q=2, r=7
  3. q=4, r=−3
  4. q=3, r=5

Solution

  1. Divide

    5×3=15, bacha 2. 0 ≤ 2 < 5. Check: 5×3+2=17.

  2. Options maar

    B: r=7 ≮ 5. C: r negative. D: r=5, r < b toot-ta — wo 5×4+0 hona chahiye, 17 nahi.

AnswerA · q=3, r=2
Q5 · Medium

12 aur 18. Prime se HCF, phir LCM formula se?

  1. HCF 6, LCM 36
  2. HCF 3, LCM 216
  3. HCF 12, LCM 18
  4. HCF 6, LCM 30

Solution

  1. Factors

    12=2²×3, 18=2×3². HCF min = 2¹×3¹=6. (12 HCF nahi — 12, 18 ko nahi kat-ta.)

  2. LCM

    Max = 2²×3²=36. Check do-number formula: 6×36=216, 12×18=216. D: 30=2×3×5, extra 5 kahan se?

AnswerA
Q6 · Medium · FTA

360 ka prime factor form?

  1. 2³ × 3² × 5
  2. 2² × 3³ × 5
  3. 8 × 9 × 5
  4. 2 × 3 × 5 × 12

Solution

  1. Toot

    360=36×10=(2²×3²)×(2×5)=2³×3²×5. Check: 8×9×5=360 value theek, lekin 8,9 prime nahi — FTA form nahi (C). D mein 12 composite + primes double-count.

  2. B

    2²×3³×5=4×27×5=540, 360 nahi.

AnswerA
Q7 · Medium

Kaun sa irrational?

  1. √4 + √9
  2. √8
  3. 0.25
  4. 22/7

Solution

  1. Simplify pehle

    A: √4=2, √9=3, sum=5 Q. C: 0.25=1/4 Q. D: 22/7 = p/q, π nahi, Q.

  2. B

    √8=√(4×2)=2√2. 8 perfect square nahi → irrational.

AnswerB · √8
Q8 · Medium

π + 2/3 kaun sa?

  1. Rational (π ≈ 22/7)
  2. Irrational
  3. Integer
  4. Terminating decimal

Solution

  1. Rule

    Irrational + non-zero rational = irrational. Agar π+2/3 = r Q hota, π = r−2/3 Q — galat.

  2. Trap A

    22/7 ≈ π, barabar nahi. 22/7 + 2/3 rational hota; π + 2/3 nahi.

AnswerB
Q9 · Pro · simplify pehle

21/56 ka decimal terminating hai?

  1. Nahi, 56 mein 7 hai
  2. Haan — simplify 3/8, 8=2³
  3. Nahi, numerator 21
  4. Irrational

Solution

  1. HCF

    21=3×7, 56=7×8. HCF=7. 21/56=3/8.

  2. Ab q

    8=2³. Sirf 2 → terminating 0.375. Trap A: 7 cancel ho chuka. Test lowest terms ke baad hota hai.

AnswerB
Q10 · Pro · Euclid

HCF(135, 225) Euclid se?

  1. 15
  2. 45
  3. 75
  4. 5

Solution

  1. Bada pehle

    225 = 135×1 + 90. HCF(135, 90).

  2. Dobara

    135 = 90×1 + 45. HCF(90, 45).

  3. r=0

    90 = 45×2 + 0. Last non-zero divisor = 45. 15 common hai lekin sabse bada 45 (135=3×45, 225=5×45).

AnswerB · 45

Practice

Pehle khud, phir neeche kholo.

12345 678910
P1

0.10100100010000… (har baar ek extra 0) kaun sa?

  1. Rational, repeating
  2. Irrational — pattern repeat nahi (zeros badhte)
  3. Integer
  4. Terminating

Solution

  1. Cycle?

    10, 100, 1000 — zeros ki ginti badh rahi. Koi fixed block 10-10-10 nahi. Non-term non-repeat → irrational.

  2. A kyun nahi

    Repeating hota to same length ka block. Yahan length badal rahi.

AnswerB
P2

180 ko perfect square banana. Sabse chhota multiply?

  1. 2
  2. 5
  3. 10
  4. 45

Solution

  1. FTA

    180=2² × 3² × 5. Even: 2 aur 3. Odd sirf 5.

  2. Chhota

    ×5 → 2²3²5²=900=30². ×10=×2×5 extra 2 ki zaroorat nahi. ×45 bada.

AnswerB · 5
P3

LCM(8, 9, 6) primes se?

  1. 24
  2. 72
  3. 48
  4. 36

Solution

  1. Max power

    8=2³, 9=3², 6=2×3. LCM=2³×3²=8×9=72.

  2. Trap

    Teen numbers pe HCF×LCM=product mat lagana. 8×9=72 yahan isliye kyunki 6 ke primes already 8 aur 9 mein cover. A 24=2³×3, 9 miss.

AnswerB · 72
P4

0.6 = ?

  1. 6/10
  2. 2/3
  3. 1/6
  4. 6/99

Solution

  1. x-trick

    x=0.666…, 10x=6.666…, 9x=6, x=6/9=2/3.

  2. Options

    A=0.6 terminating = 3/5. C=0.16. D=0.06 (do-digit, 06/99).

AnswerB · 2/3
P5

125 = 8q + r, 0 ≤ r < 8. r?

  1. 5
  2. 7
  3. 4
  4. 8

Solution

  1. Divide

    8×15=120, r=5. 0 ≤ 5 < 8. D: r=8 allowed nahi (r < 8), wo q=16, r=0 hota.

AnswerA · 5
P6

√3 + √12 simplify ke baad?

  1. √15 rational
  2. 3√3 irrational
  3. 2√3 rational
  4. 15

Solution

  1. Toot

    √12=√(4×3)=2√3. Sum: √3+2√3=3√3.

  2. Set

    3≠0, √3 irr → 3√3 irr. A: √3+√12 ≠ √15. Roots tab jodte jab same andar, ya product √a√b=√(ab).

AnswerB
P7

HCF(24,36)=12. LCM?

  1. 48
  2. 72
  3. 96
  4. 6

Solution

  1. Do-number formula

    LCM = 24×36 / 12. 24/12=2, 2×36=72.

  2. Primes

    24=2³×3, 36=2²×3². LCM=2³×3²=8×9=72. D 6 to HCF jaisa soch, ulta.

AnswerB · 72
P8

7/80 terminating?

  1. Haan, 80=2⁴×5
  2. Nahi, 80 bada
  3. Nahi, 7 prime
  4. Irrational

Solution

  1. q

    HCF(7,80)=1. 80=16×5=2⁴×5. Sirf 2,5 → terminate. Places=max(4,1)=4 → 0.0875.

  2. C

    Numerator prime hona rok-ta nahi. Test denominator pe.

AnswerA
P9

13/210 terminating?

  1. Haan, 2 aur 5 hain
  2. Nahi — 210=2×3×5×7, 3 aur 7 bache
  3. Haan, 13 prime
  4. Cannot say

Solution

  1. Cancel?

    13, 210 ko nahi kat-ta (210=2×3×5×7). Lowest terms yahi.

  2. Extra primes

    2 aur 5 hain, lekin 3 aur 7 bhi. “2 aur 5 hain” kaafi nahi — sirf 2 aur 5 chahiye. Repeating, terminate nahi.

AnswerB
P10

Euclid: HCF(26, 91)?

  1. 1
  2. 13
  3. 7
  4. 26

Solution

  1. 91 ÷ 26

    26×3=78, rem 13. HCF(26, 13).

  2. 26 ÷ 13

    13×2 + 0. HCF=13. 26, 91 ko divide karta (91=7×13), isliye D nahi — 26, 91 ko nahi kat-ta.

AnswerB · 13

← Index · Input–output

Agle topic: Polynomials.