Question

Q.1. Prove that √2 is irrational.

Q.2. Prove that 3√2 is irrational.

Q.3. Prove that √2 + √5 is irrational.

Answer 1

Hey there!

We will prove whether √2 is irrational by contradiction method.
Let √2 be rational 
It can be expressed as √2 = a/b ( where a, b are integers and co-primes. 
√2 = a/b
2= a²/b² 
2b² = a²
2 divides a²
By the Fundamental theorem of Arithmetic
so, 2 divides a .

a = 2k (for some integer) 

a² = 4k² 
2b² = 4k² 
b² = 2k² 

2 divides b²
2 divides b. 

Now 2 divides both a & b this contradicts the fact that they are co primes. 
this happened due to faulty assumption that √2 is rational. Hence, √2 is irrational. 

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We will prove whether 3√2 is irrational by contradiction method.
Let 3√2 be rational 
It can be expressed as 3√2 = a/b ( where a, b are integers and co-primes. 
3√2 = a/b
2= a²/3b² 
6b² = a²
6 divides a²
By the Fundamental theorem of Arithmetic
so, 6 divides a .

a = 6k (for some integer) 

a² = 36k² 
6b² = 36k² 
b² = 6k² 

6 divides b²
6 divides b. 

Now 6 divides both a & b this contradicts the fact that they are co primes. 
this happened due to faulty assumption that 3√2 is rational. Hence, 3√2 is irrational. 

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We will prove whether √2+ √5 is irrational by contradiction method.
Let √2+√5 be rational 
It can be expressed as √2 + √5 = a/b ( where a, b are integers and co-primes. 
√2 + √5 = a/b
2 + 5 + 2 √10 = a/b
2√10 = a/b - 7
2√10 = a - 7b/b
√10 = a - 7b/2b


Now √10 is irrational , where we a - 7b/2b is rational as a, b are integers.
this happened due to faulty assumption that √2+√5 is rational. Hence, √2+√5 is irrational. 

Answer 2

hy

its literally easy !!
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$3 \sqrt{2} \: \: \: \: = rational \: \: no \:$
$3 \sqrt{2} \: = q \:$
$( 3{ \sqrt{2)} }^{2} = {q}^{2}$

$9 \times 2 = {q}^{2}$
$\sqrt{18 } \: \: \: = q$

$\sqrt{18 \: \: \: doesnot} \: comes \: \: out \\ of \: \: square \: root \: \: so \: it \: is \: \: an \\ irrational \: no$
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$let \: \sqrt{2 + \sqrt{5 \: \: \: be} } \: \: a \: \: rational \\ \: number$
$\sqrt{2} + \sqrt{5} = \: q$
squaring both sides

$( \sqrt{2} + \sqrt{5} )^{2} = \: (q) ^{2}$

$\sqrt{2 + 5 } = q$
$q \: = \sqrt{7}$
$\sqrt{7} \: \: \: \: doesnot \: \: come \: out \: \: \: of \: \\ \: square \: \: root \: \: so \: its \: \: an \: irration \\ al \: \: number$
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$hope \: helped$