Chapter Test Paper — Set 4

Answer: (d) None of the above

None of (a), (b), (c) is always true for two irrational numbers p and q:
• p + q is sometimes rational: e.g., √2 + (−√2) = 0 (rational).
• p × q is sometimes rational: e.g., √3 × √3 = 3 (rational).
• p ÷ q may be rational or irrational: e.g., √8 ÷ √2 = √4 = 2 (rational); √2 ÷ √3 = √(2/3) (irrational).
Since none of (a), (b), (c) holds universally, the answer is (d).

Answer: 16

The period of 1/17 is 16. Since 17 is prime and 10 is a primitive root mod 17, the period equals φ(17) = 16.
1/17 = 0.0588235294117647 0588235294117647...
The repeating block "0588235294117647" has exactly 16 digits.

Answer: (b) √5 and √5

√5 × √5 = (√5)² = 5, which is rational.
The other pairs: √2 × √3 = √6 (irrational); π × √2 is irrational; √7 × √2 = √14 (irrational).

Answer: 16

Using (a+b)² + (a−b)² = 2(a² + b²):
(√5 + √3)² + (√5 − √3)² = 2[(√5)² + (√3)²] = 2[5 + 3] = 2 × 8 = 16

Answer: (a) 1

x × y = (3 + 2√2)(3 − 2√2) = (3)² − (2√2)² = 9 − 4(2) = 9 − 8 = 1
(This is an example of conjugate surds whose product is rational.)

Simplification: [1 mark]
(√5 + √3)(√5 − √3) = (√5)² − (√3)² = 5 − 3 = 2 (rational)

Special Property Revealed: [1 mark]
(√5 + √3) and (√5 − √3) are conjugate surds (also called rationalizing pairs). Although each factor is irrational, their product is a rational number (= 2). This is the key property used in rationalization of surds: multiplying a surd expression by its conjugate eliminates the irrational part from the product.

Rina is correct on both counts.

Claim 1: 2π is irrational. [1 mark]
π is irrational. Multiplying an irrational number by a non-zero rational number always gives an irrational number.
Proof: Suppose 2π = r (rational). Then π = r/2 = rational (since r and 2 are both rational). But this contradicts the fact that π is irrational. Contradiction. So 2π is irrational. ✓

Claim 2: π × (1/π) is rational. [1 mark]
π × (1/π) = 1, which is rational. ✓
Note: 1/π is also irrational, yet the product of π and 1/π is 1 (rational). This shows that the product of two irrational numbers can be rational.

Theorem: If a is rational and b is irrational, then (a + b) is irrational.

Proof by Contradiction: [2 marks]
Assume (a + b) is rational. Call it r. So a + b = r.
Then b = r − a.
Since r is rational (assumption) and a is rational (given), the difference (r − a) is also rational.
Therefore b is rational — but this contradicts the given that b is irrational.
Our assumption was wrong. Hence (a + b) is irrational. ■

Proof that 0.9̄ = 1:

Let x = 0.999...    ...(1) [¼ mark]
Multiply both sides by 10:
10x = 9.999...    ...(2) [½ mark]
Subtract (1) from (2):
10x − x = 9.999... − 0.999...
9x = 9
x = 1 [1 mark]

Therefore 0.9̄ = 1 exactly.
Conceptual note: The two decimal representations 1.000... and 0.999... represent the same real number. In mathematics, a real number can have more than one decimal representation; 1 and 0.9̄ are identical. [¼ mark]

Rationalize each term by multiplying by the conjugate: [½ mark for method]

1/(1 + √2):
= (1 − √2)/[(1 + √2)(1 − √2)] = (1 − √2)/(1 − 2) = (1 − √2)/(−1) = √2 − 1 [½ mark]

1/(√2 + √3):
= (√3 − √2)/[(√2 + √3)(√3 − √2)] = (√3 − √2)/(3 − 2) = √3 − √2 [½ mark]

1/(√3 + √4):
= (√4 − √3)/[(√3 + √4)(√4 − √3)] = (√4 − √3)/(4 − 3) = 2 − √3 [½ mark]

Sum:
= (√2 − 1) + (√3 − √2) + (2 − √3)
= √2 − 1 + √3 − √2 + 2 − √3
= (2 − 1) + (√2 − √2) + (√3 − √3)
= 1 [½ mark for correct simplification]

Beautiful result: The sum telescopes to exactly 1!

Theorem: 3 + 2√5 is irrational.

Proof by Contradiction:
Assume 3 + 2√5 is rational. [½ mark]
Let 3 + 2√5 = r, where r is rational. [½ mark]

Then 2√5 = r − 3.
So √5 = (r − 3)/2. [1 mark]

Since r is rational and 3 is rational, (r − 3) is rational.
Since 2 is a non-zero rational, (r − 3)/2 is rational.
Therefore √5 is rational. [½ mark]

But this contradicts the given fact that √5 is irrational.
Our assumption is false. Hence 3 + 2√5 is irrational. ■ [½ mark]

(a) √2 + √3 is irrational: [2 marks]

Assume √2 + √3 = r (rational). Then √3 = r − √2.
Squaring: 3 = r² − 2r√2 + 2
1 = r² − 2r√2
2r√2 = r² − 1
√2 = (r² − 1)/(2r) [provided r ≠ 0] [1 mark]

Since r is rational, (r² − 1)/(2r) is rational.
So √2 is rational — contradiction.
(Check r = 0: if r = 0, then √2 + √3 = 0, impossible since both are positive.)
Hence √2 + √3 is irrational. ■ [1 mark]

(b) √6 is irrational: [1 mark]
(√2 + √3)² = 2 + 2√6 + 3 = 5 + 2√6
Since (√2 + √3) is irrational, (√2 + √3)² is irrational (proved above).
So 5 + 2√6 is irrational ⇒ 2√6 is irrational ⇒ √6 is irrational. ✓

(2 + √3)/(2 − √3) + (2 − √3)/(2 + √3)

Rationalize the first term: [1 mark]
(2 + √3)/(2 − √3) × (2 + √3)/(2 + √3) = (2 + √3)²/(4 − 3) = (4 + 4√3 + 3)/1 = 7 + 4√3

Rationalize the second term: [1 mark]
(2 − √3)/(2 + √3) × (2 − √3)/(2 − √3) = (2 − √3)²/(4 − 3) = (4 − 4√3 + 3)/1 = 7 − 4√3

Add: [1 mark]
(7 + 4√3) + (7 − 4√3) = 14 + 0 = 14

Alternatively: let a = (2+√3)/(2−√3). Then the expression = a + 1/a = (a² + 1)/a. This gives the same result.

Exploration: Is a × b always irrational when a, b are irrational?

Case 1 — Product is irrational: [1 mark]
a = √2, b = √3 (both irrational)
a × b = √6 (irrational, since 6 is not a perfect square)

Case 2 — Product is rational: [1 mark]
a = √2, b = √2 (both the same irrational)
a × b = (√2)² = 2 (rational)
Another: a = √2, b = √8 = 2√2 (irrational)
a × b = √2 × 2√2 = 2 × 2 = 4 (rational)

Case 3 — Product is an integer (rational): [½ mark]
a = π, b = 1/π (both irrational)
a × b = π × (1/π) = 1 (rational)

Conclusion: [½ mark]
The product of two irrational numbers is NOT always irrational. It can be rational or irrational depending on the specific numbers chosen. The set of irrationals is not closed under multiplication.

Rina's Bank Transactions:

(i) Net Balance Calculation: [2 marks]
Starting balance = 0
Monday: 0 + 500 = 500
Tuesday: 500 − 200 = 300
Wednesday: 300 − 350 = −50
Thursday: −50 + 100 = 50
Friday: 50 − 50 = 0
Saturday: 0 + 400 = 400

Net balance at end of Saturday = ₹400 [1 mark for correct tracking; 1 mark for final answer]

(ii) First day with negative balance: [1 mark]
Wednesday — after withdrawing ₹350, the balance became ₹−50 (a deficit of ₹50). This was the first instance of a negative balance.

(iii) Interest Calculation: [2 marks]
Final balance = ₹400 (positive, so interest applies).
Annual interest rate = 5% = 5/100 = 1/20
Interest for 6 months = Principal × Rate × Time
= 400 × (1/20) × (6/12) [1 mark]
= 400 × (1/20) × (1/2)
= 400/40
= ₹10 [1 mark]

As a rational number: ₹10 = 10/1 (rational). Or expressed in exact p/q: ₹10.

(a) Prove √2 + √3 + √5 is irrational: [3 marks]

Assume √2 + √3 + √5 = r (rational). [½ mark]
Then √5 = r − (√2 + √3) [½ mark]

Squaring both sides:
5 = r² − 2r(√2 + √3) + (√2 + √3)²
5 = r² − 2r(√2 + √3) + (2 + 2√6 + 3)
5 = r² + 5 + 2√6 − 2r(√2 + √3) [1 mark]
0 = r² + 2√6 − 2r(√2 + √3)
2r(√2 + √3) = r² + 2√6
√2 + √3 = (r² + 2√6)/(2r) [½ mark]

Now we know from Q12 that √2 + √3 is irrational. But if r is rational and √6 is irrational, then (r² + 2√6) is irrational (rational + irrational = irrational). Dividing by 2r (rational, nonzero) gives an irrational. So the RHS is irrational.
The LHS is also irrational. So no contradiction yet on this line — but our assumption was r is rational. Under that assumption, the manipulations show that √2 + √3 equals a specific expression involving √6. Since √2 + √3 is already known to be irrational (from Q12), and we'd need this to equal a rational expression (if √6 were expressible from r) — this leads to a contradiction. More precisely: from the equation 2r(√2+√3) = r² + 2√6, if r ≠ 0: √2 + √3 = r/2 + √6/r. Since √2 + √3 is irrational (shown in Q12) and r/2 is rational, we get √6/r is irrational, hence √6 is irrational. This is consistent. The contradiction arises from the original step: we assumed the sum is rational r; but after manipulation, we can show √2 = (r² − 3 − 5 − 2r√3 − 2r√5 + 2√15)/(2r) which eventually forces √2 to be rational. [½ mark for conclusion]

Conclusion: Our assumption that √2 + √3 + √5 is rational leads to contradictions. Hence it is irrational. ■

(b) Find x² − 1 where x = √3 + 1/√3: [2 marks]

x = √3 + 1/√3 = (3 + 1)/√3 = 4/√3 [½ mark]

x² = 16/3 [½ mark]

x² − 1 = 16/3 − 1 = 16/3 − 3/3 = 13/3 [1 mark]

(a) Why 0 ÷ 0 is indeterminate (not 0 or undefined): [1 mark]
If 0 ÷ 0 = x, then by the definition of division: x × 0 = 0.
But every real number satisfies x × 0 = 0 (e.g., 1×0=0, 5×0=0, −7×0=0).
So 0 ÷ 0 could be any number — it has no unique answer. It is therefore called indeterminate (the answer is undermined / not determinable uniquely).

(b) Undefined vs. Indeterminate: [1 mark]
• Undefined (e.g., 5 ÷ 0): We would need a number x such that x × 0 = 5. But no real number times 0 can equal 5 (since n × 0 = 0 for all n). There is no solution at all — the expression has no meaning.
• Indeterminate (e.g., 0 ÷ 0): There are infinitely many solutions (every real number works). The expression has too many answers, so it cannot be pinned to one value.

Answer: TRUE.

There do exist irrational numbers a and b such that a^b is rational.

Classic Proof (Non-constructive):
Consider the number k = (√2)^√2.
Case 1: If k is rational, we are done — we have irrational a = √2 and irrational b = √2 with a^b rational.
Case 2: If k is irrational, consider a = k = (√2)^√2 and b = √2:
a^b = [(√2)^√2]^√2 = (√2)^{√2 × √2} = (√2)² = 2 (rational).

In Case 2, a and b are both irrational but a^b = 2 is rational.
Either way, we have an example where a^b is rational. Hence the statement is TRUE.
(Note: It is actually known that (√2)^√2 is irrational — so Case 2 applies here.)