prove that difference and quotient of (3+2√3) and 3-2√3) are irrational.?

Difference
= (3 + 2√3) - (3 - 2√3)
= 3 + 2√3 - 3 + 2√3
= 4√3 is a irrational number


Quotient
= (3 + 2√3) / (3 - 2√3)
= (3 + 2√3)² / [(3 - 2√3)(3 + 2√3)]
= [(3)² + 2(3)(2√3) + (2√3)²] / [(3)² - (2√3)²]
= [9 + 12√3 + 12] / (-3)
= -7 - 4√3 is an irrational number

First let's simplify both expressions:

(3+2√3) - (3-2√3) = 4√3,

(3+2√3) / (3-2√3)
= ((3+2√3)(3+2√3)) / ((3-2√3)(3+2√3))
= (9+12+12√3) / (9-12)
= -7–4√3.

Note that both expressions simplify to the form M+N√3 with M, N integers and N≠0. Therefore it suffices to show that this form is irrational. Ad absurdum, assume otherwise. Then, subtracting M, which is rational, the difference N√3 must also be rational. Since N is nonzero and rational, we can divide by it so that the quotient √3 must still be rational. The rest of the proof is now standard: for some coprime integers p and nonzero q, we have

√3 = p/q.

Multiplying by q then squaring,

3q²= p².

So 3 | p². Therefore 3 | p. So

9 | p² = 3q².

Hence 3 | q². Therefore 3 | q as well, which contradicts the assumption that p and q are coprime because both are divisible by 3, q.e.d

Difference is 3+ 2√3 - 3 + 2√3 = 4√3 is irrational

Quotient:
(3+2√3)/(3-2√3)
multiply and divide by (3+2√3)
= (3+2√3)(3+2√3)/((3-2√3)(3+2√3))

The denominator is (a-b)(a+b) = a^2-b^2
a=3
a^2=9
b=2√3
b^2=12
a^2-b^2 = 9-12=-3
=(3+2√3)(3+2√3)/ -3
= (9+6√3+6√3+12) / -3
= (21+12√3)/ (-3)
= -7 -4√3 ----> irrational number

The difference:

[3 + 2sqrt(3)] - [3 - 2sqrt(3)]
3 + 2sqrt(3) - 3 + 2sqrt(3)
= 4sqrt(3) (which is irrational)
--------
The quotient:

[3 + 2sqrt(3)] / [3 - 2sqrt(3)]

Rationalise the denominator by multiplying the top and bottom by the conjugate of the denominator [3 + 2sqrt(3)]:

[3 + 2sqrt(3)][3 + 2sqrt(3)] / [3 + 2sqrt(3)][3 - 2sqrt(3)]
[9 + 12sqrt(3) + 12] / [9 - 12]
[21 + 12sqrt(3)] / -3
= -7 - 4sqrt(3) (which is also irrational)

The sum and product of these terms would be rational as the radicals would cancel out.

The difference of (3+2√3) and(3-2√3) is: 3+2√3-3+2√3=4√3 which is irrational.
The quotient of (3+2√3) and(3-2√3) is:
(3+2√3)/(3-2√3)
=(3+2√3)(3+2√3)/(3-2√3)(3+2√3)
=(3+2√3)²/[(3)²-(2√3)²]
={9+2.3.2√3+(2√3)²}/(9-12)
=(9+12√3+12)/(-3)
=(21+12√3)/(-3)
=3(7+4√3)/(-3)
=7+4√3 is an irrational number
= - (7+4√3) is an irrational number
The quotient of (3+2√3) and(3-2√3) is an irrational number.

Proff by negation might be the easiest way to do it.

the difference of ( 3+ 2 x 3^1/2 ) and ( 3 - 2x 3^1/2 ) = ( 3 + 2x 3^1/2 ) - ( 3- 2x 3^1/2 ) = 3 - 3 + 2 x 3^1/2 + 2 x 3^1/2 = 4x 3^1/2 , a irrational number.
the quotient of ( 3 + 2x 3^1/2 ) and ( 3 - 2 x 3^1/2 ) = ( 3 + 2 x 3^1/2 ) / ( 3 - 2 x 3^1/2 ) , and multiply both nominator and denominator by ( 3 + 2x 3^1/2 ) it becomes ; 3^2 + 2 x3 x2 x 3^1/2 + 2^2 x 3 / 3^2 - 2^2 x 3 = 21 + 12 x 3^1/2 / -3 = - 7 - 4 x 3^1/2 . an irrational number.

(3 + 2*sqrt(3)) - (3 - 2*sqrt(3)) = 4*sqrt(3)

For the quotient (3 + 2*sqrt(3)) / (3 - 2*sqrt(3)), multiply both sides by (3 + 2*sqrt(3)) because we hate square roots in denominators. You should end up with (9 + 12*sqrt(3) + 12) / (9 - 12), which simplifies to -7 - 4*sqrt(3).

Both the difference and quotient result in an answer that includes the sqrt(3), which is an irrational number (I'm assuming you don't need to prove that, but if you do there are multiple answers online).