Question

prove that 2 root 3 minus 7 as irrational​

Answer 1

Step-by-step explanation:

Let us assume that 2√3 - 7 is a rational number.

So, 2√3 - 7 = r

where, r is rational

2√3 = r + 7

√3 = (r + 7)/2

As we can see that RHS is purely rational but LHS is irrational.

So it contradicts to what we assume earlier. Hence, our assumption is wrong.

Therefore,

2√3 - 7 is an irrational number.

Hence proved !!

Answer 2

[Start with proving √3 as an irrational number]

Let us assume √3 to be a rational number,

√3 = p/q (where p and q are co-primes)

(√3)² = (p/q)²

3 = p²/q²

p² = 3q²

(hence, 3 is a factor of p) -------------(i)

let p = 3k (where k∈any integer)

p² = (3k)²

3q² = 9k²

q² = 9k²/3

q² = 3k²

(hence 3 is a factor of q) ---------(ii)

From (i) and (ii),

3 is a factor of both p and q.

However, this shouldn't be possible as p and q are co-primes.

Hence our assumption was wrong.

∴√3 is an irrational number

Now, solving for 2√3-7

let n = any integer

2√3-7 = n

2√3 = n+7

√3 = n+7/2

However, this cannot be possible as √3 is an irrational number whereas n+7/2 is a rational number,

And Irrational numbers cannot be equal to rational numbers.

∴2√3-7 is an irrational number.

Hope it helped...

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