Irrational numbers and real numbers

A rational number has a decimal expansion that is eventually periodic (repeating). In 9.0202202220... the zeros occur at positions 1, 3, 6, 10, ... which are the triangular numbers \\(T_n=\frac{n(n+1)}{2}\\). The blocks of consecutive 2's between zeros have lengths 1, 2, 3, 4, ... which grow without bound, so no fixed repeating block exists. Hence the decimal is non-terminating and non-repeating, and the number is irrational. Exam tip: check whether gaps between recognizable markers (here zeros) stabilize; if they keep changing indefinitely, the decimal cannot be periodic and the number is irrational.

Conjugate multiplication gives (pq=7^2-(\sqrt{11})^2=49-11=38). Use (a^2-b^2) in such questions.

The governing idea is expansion of a binomial containing surds. Substitute x = \sqrt{5}+\sqrt{2} and square it: x^2 = (\sqrt{5}+\sqrt{2})^2 = 5+2+2\sqrt{5}\sqrt{2} = 7+2\sqrt{10}. Thus x^2-2\sqrt{10} = (7+2\sqrt{10})-2\sqrt{10}=7. Since 7 is an integer, it is also a rational real number. Option B is incorrect because the irrational surd terms cancel exactly. Option C is impossible because the expression is formed from real numbers, and option D does not describe 7. Therefore option A is the unique correct answer.

On adding the two squares the radical terms cancel and the result is (32). Identify cancelling terms in conjugates.

By identity the difference is (4ab), where (a=\sqrt{11}) and (b=\sqrt{5}). So the answer is (4\sqrt{55}).

(\sqrt{18}=3\sqrt{2}), (\sqrt{50}=5\sqrt{2}), and (\sqrt{8}=2\sqrt{2}), so (y=6\sqrt{2}). Its square is (72).

The governing concept is rationalizing a denominator containing a surd. The conjugate of 3+√5 is 3−√5, so multiply both numerator and denominator by this conjugate: 1/(3+√5)=(3−√5)/[(3+√5)(3−√5)]. Applying the difference-of-squares identity makes the denominator 3²−(√5)²=9−5=4. Therefore the rationalized form is (3−√5)/4, which is option A. Option B uses the original denominator expression rather than its conjugate, so it does not produce the required cancellation. Option C has an incorrect denominator, and option D omits the irrational part of the numerator and is not equal to the original fraction. Thus A is the unique correct answer.

Multiplying by the conjugate makes the denominator (7-3=4). Hence we get (\frac{2(\sqrt{7}+\sqrt{3})}{4}).

Since ((2+\sqrt{3})(2-\sqrt{3})=1), the reciprocal is (2-\sqrt{3}). Recognizing conjugates is a fast method.

The first product is (25-6=19) and (\sqrt{24}=2\sqrt{6}) is irrational. A rational plus an irrational is irrational.

Rationalize the denominator to find the reciprocal.
\(\dfrac{1}{\sqrt{2}+\sqrt{3}}=\dfrac{\sqrt{3}-\sqrt{2}}{(\sqrt{3})^2-(\sqrt{2})^2}=\sqrt{3}-\sqrt{2}\).
Add to \(r\): \(\sqrt{2}+\sqrt{3}+\sqrt{3}-\sqrt{2}=2\sqrt{3}\). Hence the simplified result is \(2\sqrt{3}\).
Why other options are wrong: option C is just \(r\) (not the sum), option D equals \(1/r\), and option B is an incorrect arithmetic result. Exam tip: use conjugates to rationalize denominators whenever square roots appear in the denominator.

Convert each radical to a common √3 factor: \(2\sqrt{12}=2\sqrt{4\cdot3}=4\sqrt{3}\), \(-3\sqrt{27}=-3\sqrt{9\cdot3}=-9\sqrt{3}\), \(\sqrt{75}=\sqrt{25\cdot3}=5\sqrt{3}\). Summing gives \(4\sqrt{3}-9\sqrt{3}+5\sqrt{3}=(4-9+5)\sqrt{3}=0\). Hence the expression simplifies to \(0\). The common distractor \(10\sqrt{3}\) arises from ignoring the negative sign on the middle term, so it is incorrect. Exam tip: factor out perfect squares inside radicals first, then combine like radical terms.

(\sqrt{8}=2\sqrt{2}) and (\sqrt{18}=3\sqrt{2}), so (a=5\sqrt{2}). Its square is (50), a rational number.

Simplify first: \(\sqrt{45}=3\sqrt{5}\), \(\sqrt{80}=4\sqrt{5}\), \(\sqrt{125}=5\sqrt{5}\). So the expression equals \(3\sqrt{5}+4\sqrt{5}-5\sqrt{5}=2\sqrt{5}\). Since \(\sqrt{5}\) is irrational, multiplying by the nonzero rational 2 yields an irrational number, so \(2\sqrt{5}\) is irrational. Why others are wrong: it cannot be rational because \(\sqrt{5}\) is irrational; it is not an integer; it is not zero because the coefficient 2 is nonzero. Exam tip: factor and combine like surd terms (same \(\sqrt{\;\;}\)) to simplify quickly.

If it were rational, its square (3+\sqrt{5}) would be rational, which it is not. Hence it is real irrational.

Use the identity \((a-b)^2=a^2+b^2-2ab\). With \(a=\sqrt6\) and \(b=\sqrt2\), we get \(a^2+b^2=6+2=8\) and \(2ab=2\sqrt{12}=4\sqrt3\). Thus \(x^2=8-4\sqrt3\). Option B corresponds to a common halving error in the \(2ab\) term; option C has the sign of the middle term wrong. Exam tip: compute \(a^2+b^2\) and \(2ab\) separately to avoid sign or factor mistakes.

The two binomials have the form \((u+v)(u-v)\), which is the difference-of-squares identity \(u^2-v^2\). Using this identity avoids multiplying every term separately and also shows why the square-root terms cancel. The result is rational even though the individual factors contain irrational numbers.

Take \(u=\sqrt{13}\) and \(v=\sqrt{12}\). Then \((\sqrt{13}+\sqrt{12})(\sqrt{13}-\sqrt{12})=(\sqrt{13})^2-(\sqrt{12})^2=13-12=1\). Thus option A is correct. The expression is not 25, and \(\sqrt{156}\) would arise from multiplying only the radicals, not from the complete conjugate product.

Multiplying by the conjugate makes the denominator (5-2=3). So the rationalized form is (\frac{\sqrt{5}-\sqrt{2}}{3}).

(\sqrt{27}=3\sqrt{3}) and (\sqrt{12}=2\sqrt{3}), so the numerator is (5\sqrt{3}). Dividing gives (5).

Actually (x=\sqrt{2}+\sqrt{3}) satisfies (x^4-10x^2+1=0), not a simple quadratic here. Read powers carefully in such trick questions.