Question

Prove that the following is irrational 7 root3 - 3

Answer 1

Answer:

for √3:

Let us assume that √3 is a rational number.

then, as we know a rational number should be in the form of p/q  where p and q are co- prime number.

√3 = p/q { where p and q are co- prime}

√3q = p

Now, by squaring both the side gives:

(√3q)² = p²

3q² = p² ........ ( i )

if 3 is the factor of p² , then, 3 is also a factor of p

Let p = 3m { where m is any integer }

squaring both sides gives:

p² = (3m)²

p² = 9m²

putting the value of p² in equation ( i )

3q² = p²

3q² = 9m²

q² = 3m²

if 3 is factor of q², then, 3 is also factor of q

Since, 3 is factor of p & q both

So, our assumption that p & q are co- prime is wrong

hence,. √3 is an irrational number

Step-by-step explanation:

let us suppose that 7√3 - 3 is rational.

then there exists two positive integers a and b which are co primes such that:

7√3 - 3 = a/b

7√3 = a/b - 3

7√3 = a-3b/b

√3 = a-3b/7b

a-3b/7b = rational number but we know that √3 is irrational( proved above)

rational number ≠ irrational number

hence our supposition is wrong

therefore, 7√3 - 3 is irrational

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Answer 2

Answer:

Step-by-step explanation:

Let assume that 7√3-3 as a rational number if possible ....

Let 7√3-3 =a/b

7√3=a-3b/b

√3=a-3b/7b

Integer-integer/integer

= A rational number ...

Here,√3 is a rational no.but we contradicts that √3 is an irrational no.

So, 7√3-3 is an. Irrational number.......

Hope I help u