Question

Can somebody please help as soon as possible with this? I need the step by step answer...thanks

Answer 1

Answer:

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1) √3 is irrational

Proof :

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.

So,

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

√3q = p

Now, by squaring both the side

we get,

(√3q)² = p²

3q² = p² ........ ( i )

So,

If 3 is the factor of p²,

then, 3 is also a factor of p ..... ( ii )

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

Squaring both sides

p² = (3m)²

p² = 9m²

Putting the value of p² in equation ( i )

3q² = p²

3q² = 9m²

q² = 3m²

So,

If 3 is factor of q²,

then, 3 is also factor of q.

Since,

3 is a factor of p & q both,

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

Hence,

√3 is an irrational number

2) 7+2√3 is irrational

Proof :

Let, if possible, 7 + 2 under root 3 be a rational no.

This will give rise to 2 unique integers a and b such that,

7+2 under root 3, where a and b are co primes and b is not equal to 0. 

Now,

7+2√3 = a/b

2√3= (a/b)-7

√3 = (a/2b)-(7/2)

Here, as a , b , 7 and 2 are all integers, (a/2b)-(7/2) is real. 

This implies that √3 is also a rational no. 

However, we have just proved above that √3 is an irrational no. 

So, This is a contradiction.

Therefore, our assumption was wrong. 

Hence,

7+2 under root 3 is an irrational no. 

HOPE IT HELPS,

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