No R(Q) is correct answer.. R(Q) is not a vector space, while R(C) is. R(Q) represents the set of rational numbers, which does not form a vector space over the field of real numbers (R) because it is not closed under scalar multiplication. In other words, multiplying a rational number by a real number can result in an irrational number, which is not in R(Q). On the other hand, R(C) represents the set of complex numbers, which form a vector space over the field of real numbers (R). This is because complex numbers are closed under addition and scalar multiplication by real numbers. So, the correct answer is R(Q).
@@umairktk2091 mery pas original Paper ha....or question same asy he pucha gaya...jesy mn ny write kia....mn ny all sources sy confirm kr k ye ans btaya ha....phir b "To err is human"....
Very helpful
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Question number 33.the correct answer is R(C) not R(Q).
No R(Q) is correct answer..
R(Q) is not a vector space, while R(C) is.
R(Q) represents the set of rational numbers, which does not form a vector space over the field of real numbers (R) because it is not closed under scalar multiplication. In other words, multiplying a rational number by a real number can result in an irrational number, which is not in R(Q).
On the other hand, R(C) represents the set of complex numbers, which form a vector space over the field of real numbers (R). This is because complex numbers are closed under addition and scalar multiplication by real numbers.
So, the correct answer is R(Q).
@@mathtips96 how uh can write the superset set in bracket as compared to small set bcz real number is the subset of complex number??
@@umairktk2091 mery pas original Paper ha....or question same asy he pucha gaya...jesy mn ny write kia....mn ny all sources sy confirm kr k ye ans btaya ha....phir b "To err is human"....
@@mathtips96 well but mene is mcqs ka answer kahi aur dekha h isi leye
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