Chapter 1
The Real Numbers & Completeness
In this chapter the logical flow is: \(\sqrt2\) exposes a gap in \(\mathbb Q\) โ supremum describes the edge of a set โ completeness says that edge exists in \(\mathbb R\) โ Archimedean property gives control using natural numbers โ density places rational/irrational numbers between any two reals.
Hinglish: Pehle hum dekhenge \(\mathbb Q\) mein problem kya hai. Phir supremum se set ki exact upper edge define karenge. Fir completeness bolega ki \(\mathbb R\) mein aisi edge missing nahi hoti. Isi foundation par aage sequences ke limits khade honge.
Irrationality of \( \sqrt{2} \) & The "Gap" in \( \mathbb{Q} \)
To prove \( \sqrt{2} \notin \mathbb{Q} \), we use proof by contradiction: assume \( \sqrt{2} = p/q \) in lowest terms. Then \( p^2 = 2q^2 \), so \( p \) must be even. Write \( p = 2k \), then \( 4k^2 = 2q^2 \implies q^2 = 2k^2 \), so \( q \) is also even. But both being even contradicts "lowest terms"!
Hinglish: Maano \( \sqrt{2} = p/q \) hai simplified form mein. Square karo: \( 2 = p^2/q^2 \), so \( p^2 = 2q^2 \). Iska matlab \( p^2 \) even hai, toh \( p \) bhi even hoga (kyunki odd ร odd = odd). Likhte hain \( p = 2k \). Daal ke solve karo: \( q^2 = 2k^2 \), toh \( q \) bhi even nikla! Dono even? Toh \( p/q \) simplified nahi tha โ virodhabhas (contradiction)! Matlab \( \sqrt{2} \) rational ho hi nahi sakta.
There is no rational number whose square is 2. i.e., \( \sqrt{2} \notin \mathbb{Q} \).
Hinglish: Hum sochte hain ki do rational numbers ke beech ek aur rational hamesha mil jaata hai (density). Phir bhi, number line pe "holes" (gaps) hain. Jab hum \( \sqrt{2} \) ko rationally approximate karne ki koshish karte hain, sequence us gap ke paas jaake atak jaata hai.
In the proof that \(\sqrt2\) is irrational, why do we assume \(\frac pq\) is in lowest terms?
Kyunki proof ke end mein hume milta hai ki \(p\) bhi even hai aur \(q\) bhi even hai.
Matlab dono 2 se divide hote hain.
Lekin agar fraction already lowest terms mein tha, toh dono ka common factor 2 hona impossible hai.
Yahi contradiction hai.
Least Upper Bounds (LUB Property)
For \( A \subseteq \mathbb{R} \), \( s = \sup A \) if (i) \( s \) is an upper bound, and (ii) every upper bound of \( A \) is \( \geq s \) (i.e. \( s \) is the least upper bound).
Hinglish: \( \max A \) tabhi exist karta hai jab set ka sabse bada element khud A ke andar ho. \( \sup A \) ko A ke andar hone ki zaroorat nahi hai. Jaise \( A=(0,1) \): \( \sup A = 1 \), lekin \( 1 \notin A \), isliye \( \max A \) exist hi nahi karta! \( \sup A \) hamesha exist karta hai agar set bounded above hai (\( \mathbb{R} \) mein).
Let \[ A=\left\{1-\frac1n:n\in\mathbb N\right\}. \] Find \(\sup A\).
Pehle terms dekho:
\[
0,\frac12,\frac23,\frac34,\frac45,\dots
\]
Ye 1 ke paas ja rahe hain, lekin kabhi 1 ke equal nahi hote.
Har term \(<1\), so 1 upper bound hai.
Aur 1 se chhota koi number permanent upper bound nahi ban sakta,
kyunki terms 1 ke arbitrarily close aa jaate hain.
Isliye:
\[
\sup A=1.
\]
Archimedean Property
For any \( x > 0 \) and any \( y \in \mathbb{R} \), there exists \( n \in \mathbb{N} \) with \( nx > y \).
Hinglish: Chahe \( x \) kitna bhi chhota ho, aur \( y \) kitna bhi bada โ \( x \) ko baar-baar (n times) jodte jaao, ek din wo \( y \) se aage nikal jaayega. \( \mathbb{R} \) mein koi "infinitely small" number nahi hota jo Archimedean property ko defy kare.
Claim: Let \( x = 0.001 \) and \( y = 5 \). Find \( n \) such that \( nx > y \).
Solution: We need \( n \times 0.001 > 5 \implies n > 5000 \). So \( n = 5001 \) works: \( 5001 \times 0.001 = 5.001 > 5 \). โ
Hinglish: Chahe \( x = 0.001 \) bahut chhota hai, 5001 baar jodne pe 5 se bada ho jaata hai! Yahi Archimedean property ka matlab hai โ chhota + chhota + chhota + ... = bada. Koi number itna chhota nahi hota ki kabhi bade number tak na pahunch sake.
Between any two distinct real numbers \( a < b \), there exists a rational number \( q \in \mathbb{Q} \) such that \( a < q < b \). Similarly, there exists an irrational number between them.
Hinglish: Chahe do numbers kitne bhi paas hon (jaise 3.1415926 aur 3.1415927), unke beech mein ek rational aur ek irrational number zaroor milega! Number line pe rationals aur irrationals dono har jagah bhare hue hain โ koi "khaali patch" nahi hai.
Example: \( \sqrt{2} \approx 1.414 \) aur \( \sqrt{2} + 0.001 \approx 1.415 \) ke beech \( q = 1.4145 \) (rational) mil jayega.