This mathematics problem involves applying core mathematical principles and formulas. Below you will find a complete step-by-step solution with detailed explanations for each step, helping you understand not just the answer but the method behind it.
ScanSolve AI Answer
1. State whether the following statements are true or false. Justify your answers. i)* Every irrational number is a real number. True. The set of real numbers includes both rational and irrational numbers. ii)* Every point on the number line is of the form sqrt(m), where m is a natural number. False. Justification 1: Negative numbers exist on the number line, but sqrt(m) (where m is a natural number) is always non-negative. For example, -2 is on the number line but cannot be written as sqrt(m). Justification 2: Numbers like (1)/(2) are on the number line, but (1)/(2) cannot be expressed as sqrt(m) for any natural number m. iii)* Every real number is an irrational number. False. Justification: Real numbers include rational numbers as well. For example, 2 is a real number but it is a rational number, not an irrational number. 2. Are the square roots of all positive integers irrational? If not, give an example of the square root of a number that is a rational number. No, the square roots of all positive integers are not irrational. Example: Consider the positive integer 4. Its square root is sqrt(4) = 2. Since 2 can be written as (2)/(1), it is a rational number. Therefore, No, and an example is sqrt(4) = 2. 3. Show how sqrt(5) can be represented on the number line. Step 1: Draw a number line and mark a point O as 0. Step 2: From O, mark a point A at 2 units to the right (representing the number 2). So, OA = 2 units. Step 3: At point A, draw a line segment AB perpendicular to OA, with length 1 unit. Step 4: Join O to B. Triangle OAB is a right-angled triangle. Step 5: Using the Pythagorean theorem, the length of the hypotenuse OB is: OB = sqrt(OA^2 + AB^2) OB = sqrt(2^2 + 1^2) OB = sqrt(4 + 1) OB = sqrt(5) Step 6: With O as the center and OB as the radius, draw an arc that intersects the number line at a point P. Step 7: The point P on the number line represents sqrt(5). 4. Classroom activity (Constructing the 'square root spiral'): Take a large sheet of paper and construct the 'square root spiral' in the following fashion. Start with a point O and draw a line segment OP_1 of unit length. Draw a line segment P_1P_2 perpendicular to OP_1 of unit length (see Fig. 1.9). Now draw a line segment P_2P_3 perpendicular to OP_2. Then draw a line segment P_3P_4 perpendicular to OP_3. Continuing in this manner, you can get the line segment P_n-1P_n by drawing a line segment of unit length perpendicular to OP_n-1. In this manner, you will have created the points P_2, P_3,..., P_n,..., and joined them to create a beautiful spiral depicting sqrt(2), sqrt(3), sqrt(4), ... To construct the square root spiral: Step 1: Start with a point O on a large sheet of paper. Draw a line segment OP_1 of 1 unit length. Step 2: At P_1, construct a line segment P_1P_2 of 1 unit length, perpendicular to OP_1. Join O to P_2. By the Pythagorean theorem, OP_2 = sqrt(OP_1^2 + P_1P_2^2) = sqrt(1^2 + 1^2) = sqrt(2) units. Step 3: At P_2, construct a line segment P_2P_3 of 1 unit length, perpendicular to OP_2. Join O to P_3. By the Pythagorean theorem, OP_3 = sqrt(OP_2^2 + P_2P_3^2) = sqrt((2))^2 + 1^2 = sqrt(2+1) = sqrt(3) units. Step 4: At P_3, construct a line segment P_3P_4 of 1 unit length, perpendicular to OP_3. Join O to P_4. By the Pythagorean theorem, OP_4 = sqrt(OP_3^2 + P_3P_4^2) = sqrt((3))^2 + 1^2 = sqrt(3+1) = sqrt(4) = 2 units. Step 5: Continue this process. For each subsequent step, draw a line segment P_n-1P_n of 1 unit length, perpendicular to the previous hypotenuse OP_n-1. Then join O to P_n. The length of the new hypotenuse OP_n will be sqrt(n) units. This sequence of hypotenuses (OP_1, OP_2, OP_3, ...) forms a spiral, with lengths sqrt(1), sqrt(2), sqrt(3), sqrt(4), .