Question

1. Prove that √7 is an irrational number. Hence 2 - 3√7 is also an irrational
number​

Answer 1

Answer:

please follow me and make this answer brainlest.

Step-by-step explanation:

let us assume that root 7 is an irrational number then

Let us assume that

√7

7

7 is rational. Then, there exist co-prime positive integers a and b such that

7 is rational. Then, there exist co-prime positive integers a and b such that7

7 is rational. Then, there exist co-prime positive integers a and b such that7

7 is rational. Then, there exist co-prime positive integers a and b such that7 =

7 is rational. Then, there exist co-prime positive integers a and b such that7 = b

7 is rational. Then, there exist co-prime positive integers a and b such that7 = ba

7 is rational. Then, there exist co-prime positive integers a and b such that7 = ba

7 is rational. Then, there exist co-prime positive integers a and b such that7 = ba

7 is rational. Then, there exist co-prime positive integers a and b such that7 = ba ⟹a=b

7 is rational. Then, there exist co-prime positive integers a and b such that7 = ba ⟹a=b 7

7 is rational. Then, there exist co-prime positive integers a and b such that7 = ba ⟹a=b 7

7 is rational. Then, there exist co-prime positive integers a and b such that7 = ba ⟹a=b 7

7 is rational. Then, there exist co-prime positive integers a and b such that7 = ba ⟹a=b 7 Squaring on both sides, we get

7 is rational. Then, there exist co-prime positive integers a and b such that7 = ba ⟹a=b 7 Squaring on both sides, we geta

7 is rational. Then, there exist co-prime positive integers a and b such that7 = ba ⟹a=b 7 Squaring on both sides, we geta 2

7 is rational. Then, there exist co-prime positive integers a and b such that7 = ba ⟹a=b 7 Squaring on both sides, we geta 2 =√7b

7b 2

7b 2

7b 2 Therefore, a

7b 2 Therefore, a 2

7b 2 Therefore, a 2 is divisible by 7 and hence, a is also divisible by7

7b 2 Therefore, a 2 is divisible by 7 and hence, a is also divisible by7so, we can write a=√7p, for some integer p.

7p, for some integer p.Substituting for a, we get 49p

7p, for some integer p.Substituting for a, we get 49p 2

7p, for some integer p.Substituting for a, we get 49p 2 =7b

7b 2

7b 2 ⟹b

7b 2 ⟹b 2

7b 2 ⟹b 2 =7p

7b 2 ⟹b 2 =7p 2

7b 2 ⟹b 2 =7p 2 .

7b 2 ⟹b 2 =7p 2 .This means, b

7b 2 ⟹b 2 =7p 2 .This means, b 2

7b 2 ⟹b 2 =7p 2 .This means, b 2 is also divisible by 7 and so, b is also divisible by 7.

7b 2 ⟹b 2 =7p 2 .This means, b 2 is also divisible by 7 and so, b is also divisible by 7.Therefore, a and b have at least one common factor, i.e., 7.

7b 2 ⟹b 2 =7p 2 .This means, b 2 is also divisible by 7 and so, b is also divisible by 7.Therefore, a and b have at least one common factor, i.e., 7.But, this contradicts the fact that a and b are co-prime.

7b 2 ⟹b 2 =7p 2 .This means, b 2 is also divisible by 7 and so, b is also divisible by 7.Therefore, a and b have at least one common factor, i.e., 7.But, this contradicts the fact that a and b are co-prime.Thus, our supposition is wrong.

7b 2 ⟹b 2 =7p 2 .This means, b 2 is also divisible by 7 and so, b is also divisible by 7.Therefore, a and b have at least one common factor, i.e., 7.But, this contradicts the fact that a and b are co-prime.Thus, our supposition is wrong.Hence,

7b 2 ⟹b 2 =7p 2 .This means, b 2 is also divisible by 7 and so, b is also divisible by 7.Therefore, a and b have at least one common factor, i.e., 7.But, this contradicts the fact that a and b are co-prime.Thus, our supposition is wrong.Hence, 7

7b 2 ⟹b 2 =7p 2 .This means, b 2 is also divisible by 7 and so, b is also divisible by 7.Therefore, a and b have at least one common factor, i.e., 7.But, this contradicts the fact that a and b are co-prime.Thus, our supposition is wrong.Hence, 7

7b 2 ⟹b 2 =7p 2 .This means, b 2 is also divisible by 7 and so, b is also divisible by 7.Therefore, a and b have at least one common factor, i.e., 7.But, this contradicts the fact that a and b are co-prime.Thus, our supposition is wrong.Hence, 7 is irrational.

Answer 2

Question:

Prove that √7 is an irrational number. Hence 2 - 3√7 is also an irrational

number

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Let us assume that √7 is a rational number.

So,

$\sqrt{7} \: = \frac{a}{b}$ (where a and b are co prime)

=> √7b = a ------(1)

Squaring both sides

=> a² = 7b²

Therefore, a² is divisible by 7 .

Hence, a is also divisible by 7.

So we can write a = 7c (for some integer c)

substituting value of a in eq(1)

=> (7c)² = 7b²

=> 49c² = 7b²

=> 7c² = b²

So, b² is divisible by 7.

Hence, b is also divisible by 7.

So both a and b have 7 as their common factor.

But this contradicts our assumption that a and b are co prime integers.

Thus, our assumption is wrong and √7 is an irrational number.

_______________________________

Now we have

2 - 3√7

Again let us assume that 2 - 3√7 is a rational number such that $2 - 3 \sqrt{7} = \frac{p}{q}$(where p and q are co prime integers)

$\implies \: 2 - 3 \sqrt{7} = \frac{p}{q} \\ \\ \implies \: - 3 \sqrt{7} = \frac{p}{q} - 2 \\ \\ \implies \: - 3 \sqrt{7} = \frac{p - 2q}{q} \\ \\ \implies \: \sqrt{7} = \frac{p- 2q}{ - 3q} \\ \\ \implies \: \sqrt{7} = \frac{2q - p}{3q}$

$\frac{2q- p}{3q} \: is \: a \: rational \: number.$

But in the above part we found that √7 is an irrational number.

Hence, 2 - 3√7 is also an irrational number.