Real Numbers

The governing concept is the quotient property of square roots: for non-negative radicands and a non-zero denominator, √a ÷ √b = √(a/b). Applying it gives √48 ÷ √3 = √(48/3) = √16 = 4, so option B is correct. A second method confirms the result. Since 48 = 16 × 3, √48 = 4√3; therefore (4√3) ÷ √3 = 4 because √3 is non-zero. Options 2, 8, and 16 do not equal the quotient. They may arise from treating the radicands or the radicals incorrectly, such as dividing 48 by an unsuitable number or forgetting that √16 equals 4. Both valid simplifications independently produce 4, establishing B as the unique answer.

Option B is correct. In \(0.272727\ldots\), the block 27 repeats indefinitely, so it is a recurring decimal and can be expressed as a ratio of two integers; therefore, it is rational. The student's mistake is assuming that every non-terminating decimal is irrational. Non-terminating non-recurring decimals are irrational, whereas recurring decimals are rational. Exam tip: a terminating or recurring decimal represents a rational number.

Using the Pythagorean theorem for the right triangle formed by two sides of the square, \(d^2=6^2+6^2=72\). Hence, \(d=\sqrt{72}=6\sqrt{2}\) metres, so option C is correct. Option A is only the side length, while option B is the sum of two sides, not the diagonal. Exam tip: the diagonal of a square is side \(\times\sqrt{2}\); therefore, this diagonal is also an irrational real number.

Since \(27=9\times3\) and \(12=4\times3\), we get \(\sqrt{27}=3\sqrt{3}\) and \(\sqrt{12}=2\sqrt{3}\). Therefore, \(\sqrt{27}+\sqrt{12}=3\sqrt{3}+2\sqrt{3}=5\sqrt{3}\), so option A is correct. Option C simplifies only \(\sqrt{27}\) and does not include the second term. In an exam, first factor out perfect-square factors from each radical, then combine like surds.

The student's statement is incorrect. In \(0.101001000100001\ldots\), the number of zeros between successive 1s keeps increasing, so no fixed block of digits repeats. Therefore, its decimal expansion is non-terminating and non-recurring, which makes the number irrational. Option A is wrong because only terminating or recurring decimals are rational. Exam tip: For an infinite decimal, always check whether a fixed digit pattern repeats.

Since 5 and 6 are positive numbers, compare their squares: 5² = 25 and 6² = 36. Because 25 < 30 < 36, we get 5 < \(\sqrt{30}\) < 6. Therefore, option B is correct. Exam tip: to locate a square root, compare the number under the radical with the squares of the surrounding integers. \(\sqrt{24}\) is less than 5, while \(\sqrt{37}\) and \(\sqrt{50}\) are greater than 6.

\(\sqrt{9}=3\) because the principal square root is taken as non-negative. The negative sign outside the radical then gives \( -\sqrt{9}=-3 \). Option B ignores the outside negative sign. Exam tip: evaluate the square root first, then apply any sign written outside the radical.

Absolute value represents a number’s distance from 0 on the number line, so it is positive for every non-zero number. Therefore, \( |-12|=12 \). Option A incorrectly retains the negative sign. Exam tip: when finding an absolute value, remove the sign of a negative number.

First subtract inside the absolute-value bars: \(4-9=-5\). Absolute value is a number’s distance from zero, so \(|-5|=5\). Option A is the value obtained before applying absolute value. Exam tip: The absolute value of a real number is never negative.

The decimal 1.25 has two digits after the decimal point, so write it over 100: \\(1.25=\frac{125}{100}\\). Now simplify by dividing numerator and denominator by their common factor 25: \\(\frac{125}{100}=\frac{5}{4}\\). Therefore option A is correct. The denominator 100 is used because there are two decimal places; it is then reduced to the simplest form.

A useful check is that \\(\frac{5}{4}=1.25\\), since 5 divided by 4 equals 1.25. Option \\(\frac{25}{10}\\) also has numerical value 2.5, not 1.25, while \\(\frac{1}{25}\\) and \\(\frac45\\) are different values. Thus the correct fraction is \\(\frac54\\), and the supplied answer A is mathematically consistent. Writing a terminating decimal over a power of 10 is the standard conversion method.

( \sqrt{2} ) and ( \sqrt{8}=2\sqrt{2} ) are both irrational. Square roots of perfect squares can be rational.

In 0.101001000100001…, the number of zeros between successive 1s keeps increasing, so the digits do not follow a fixed repeating pattern and the decimal expansion is non-terminating. Therefore, the number is irrational. Merely using the digits 0 and 1 does not make a number rational; a rational number has a terminating or repeating decimal expansion. Exam tip: a non-terminating, non-repeating decimal is irrational.

Multiply numerator and denominator by ( \sqrt{5} ). This gives ( \frac{1}{\sqrt{5}}=\frac{\sqrt{5}}{5} ).

To rationalise the denominator, multiply numerator and denominator by ( \sqrt{3} ). The answer is ( \frac{2\sqrt{3}}{3} ).

Direct answer: Option B, \(\sqrt3\). Both 2 and 3 are positive, and the square-root function preserves order for positive numbers. Since \(3>2\), we get \(\sqrt3>\sqrt2\). A numerical check also helps: \(\sqrt2\) is about 1.41, while \(\sqrt3\) is about 1.73. Option A, \(\sqrt2\), is not correct because it is the smaller root. Option B is correct because the number under its root, 3, is larger than the number under the other root, 2. Option C, both are equal, is wrong because equal non-negative square roots would have equal squares, but their squares are 2 and 3, which are different. Option D, both are not real, is wrong because the square roots of positive numbers are real numbers. A useful safe method is to square both positive quantities: comparing \(\sqrt2\) and \(\sqrt3\) is equivalent to comparing 2 and 3. Remember: for non-negative radicands, the larger radicand has the larger square root.

( \sqrt{5}\approx2.236 ), so ( -\sqrt{5}\approx -2.236 ). ( -2 ) is closer to zero, so it is greater.

sqrt{1}=1, whereas frac{3}{4}=0.75. Since 1>0.75, sqrt{1} is the greater number. As an exam tip, simplify the square root and convert the fraction to a decimal when comparing such numbers.

The direct answer is C, infinitely many. Rational numbers have the density property: between any two distinct real numbers, no matter how close they are, there are infinitely many rational numbers. To understand it, let the two numbers be a and b with a<b. Their average \\(\frac{a+b}{2}\\) lies between them. We can then find more numbers between a and that average, and between the average and b; repeating this process never ends. In this question the endpoints are real numbers, and rational numbers are densely placed throughout the real number line, so infinitely many rational numbers occur between them. Option A, only 1, is wrong because the average already gives one, and more can be found. Option B, only 2, is similarly too small. Option C is correct. Option D, none, is wrong because at least one rational number can be found between any two distinct real numbers, and in fact there are infinitely many. Exam cue: “between any two distinct real numbers” plus “rational numbers” usually tests density, whose answer is infinitely many.

Since (2^2<6<3^2), ( \sqrt{6} ) lies between (2) and (3). Since (6) is not a perfect square, it is irrational.

A square root is the number whose square equals the given number. Since \(0.2^2=0.04\), \(\sqrt{0.04}=0.2\). The principal square root is always non-negative; checking the decimal places helps avoid choosing 0.02 or 0.4.

Since \(0.5 \times 0.5 = 0.25\), \(\sqrt{0.25}=0.5\). The principal square root is always non-negative, so \(-0.5\) is not taken as the answer. In an exam, verify a decimal square root by squaring the obtained value.

Taking the square root of the numerator and denominator gives \(\sqrt{\frac{49}{64}}=\frac{\sqrt{49}}{\sqrt{64}}=\frac{7}{8}\). The principal square root is taken as positive, so \(-\frac{7}{8}\) would not be the answer. In exams, simplify the numerator and denominator separately under the square root.

Using the square-root property, \(\sqrt{\frac{16}{81}}=\frac{\sqrt{16}}{\sqrt{81}}\). Since \(\sqrt{16}=4\) and \(\sqrt{81}=9\), the value is \(\frac{4}{9}\). The principal square root is taken as positive, so \(-\frac{4}{9}\) is not the answer. Exam tip: For a fraction made of perfect squares, take the square root of the numerator and denominator separately.

Adding irrational ( \sqrt{3} ) to rational (2) gives an irrational number. It is also a real number.

Subtracting an irrational number from a rational number gives an irrational result. Thus (5-\sqrt{2}) is an irrational real number.