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Answer:
1. let us assume that √7 is rational. it can be expressed in the form p/q where p and q are co-primes and q is not equal to zero.
i.e, √7 = p/q
q√7 = p
squaring both sides, we get
(q√7)^2 = (p)^2
7q^2 = p^2 - (eq^n 1)
7q = p
therefore, we can see that p divides 7q^2, it follows that p divides 7q also.
Now let us take p = 7r for some integer r - (eq^n 2)
combining equation 1 and 2, we get,
7q^2 = (7r)^2
7q^2 = 49r^2
q^2 = 49r^2/7
q^2 = 7r^2
q = 7r
therefore, we can see that q divides 7r^2, it follows that q divides 7r also.
therefore, p and q have at least 7 as a common factor. but this contradicts the fact that p and q are co-primes. this contradiction has arisen due to our incorrect assumption that √7 is rational
,so we conclude that √7 is irrational.
2. let us assume that 2-3√5 is rational. it can be expressed in the form p/q where p and q are integers and q is not equal to zero.
I.e., 2-3√5 = p/q
-3√5 = p/q -2
-3√5 = p-2q/q
√5 = p-2q/3q
therefore, p, 2q and 3q are integers, so p-2q/3q is rational. but this contradicts the fact that √5 is irrational.
so we conclude that 2-3√5 is irrational.
3. (c) three decimal places
explanation : if we write the denominator of this fraction in standard form, I.e., 33/200, and divide them, we get 0.165
now an interesting thing is that the power of 2 here in the exponential form of the denominator is also three, so we can conclude that the number which has the highest power of n in the exponential decimal expansion of any fraction, the places after which the decimal expansion of that fraction will terminate will be the same as the power of the higher exponent.