Question

Q 29​

sᴏʟᴠᴇ ____ᴋɪɴᴅʟʏ ᴅᴏɴ'ᴛ sᴘᴀᴍ !

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Answer 1

Question:

• Show that the relation R is in the set N × N defined by (a, b) R (c, d) if a² + d² = b² + c² ∀ a, b, c, d ∈ N, is an equivalence relation.

Basic concept:

❐ Let R be a relation defined on sample space S, then,

• R is reflexive if (a, b) R (a, b) where (a, b) ∈ S.

• R is symmetric if (a, b) R (c, d) then (c, d) R (a, b) where (a, b), (c, d) ∈ S.

• R is transitive if (a, b) R (c, d), (c, d) R (e, f) then (a, b) R (e, f) where (a, b), (c, d), (e, f) ∈ S.

❐ If R is reflexive, symmetric & transitive, then R is an equivalence relation. Now, let's solve.

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Reflexive:

• Let (a, b) ∈ N × N

Then,

⇒ a² + b² = b² + a²

⇒ (a, b) R (a, b)

• Hence, R is reflexive.

Symmetric:

• Let (a, b), (c, d) ∈ N × N

Then,

⇒ (a, b) R (c, d)

⇒ a² + d² = b² + c²

⇒ c² + b² = d² + a²

⇒ (c, d) R (a, b)

• Hence, R is symmetric.

Transitive:

• Let (a, b), (c, d) and (e, f) ∈ N × N

Then,

⇒ (a, b) R (c, d)

⇒ a² + d² = b² + c² ----(1)

Also,

⇒ (c, d) R (e, f)

⇒ c² + f² = d² + e² ----(2)

Adding equation (1) & (2),

⇒ a² + f² = b² + e²

⇒ (a, b) R (e, f)

• Hence, R is transitive.

Since,

• R is reflexive, symmetric & transitive.

Hence,

• R is an equivalence relation.

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Answer 2

$\huge \tt \underline \blue{Question \: - }$

─ ⚘ Show that the relation R in the set N x N defined by (a,b) R (c,d) if a2 + d2 =b2 + c2 ∀ a, b, c, d ∈ N, is an equivalence relation.

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$\huge \tt \underline \orange{ \alpha nswer}$

Here (a, b) R (c, d)

$\bf \: » a + d = b + c.$

(i) Now (a, b) R (a, b) if -

$\bf \ \: a + b = b + a,$

• which is true.

:: relation R is reflexive.

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(ii) Now (a, b) R (c, d)

$\bf \: - a +d = b + c = d +a \\ = c + b =$

$\bf \: → C+ b = d + \\ \bf → (c, d) R (a, b)$

:: relation R is symmetric.

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(iii) Now (a, b) R (c,d) and (c,d) R (e,f)

$\bf \: + a + d = b + c \\ \bf c +f = d + e$

$\bf \: (a + d) + (c + f) =$

$\bf \: (b + c) + (d + e)$

$\bf \: → a + f = b + =$

$\bf \: → (a, b) R (e, f)$

:: relation R is transitive.

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Now R is reflexive, symmetric and transitive

:: relation Ris an equivalence relation.

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I'm kinda busy! tht's why inactive ☘️

lobe uh too❤️ & mich uh more❤️❤️