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Real Numbers: Practice Worksheet

Prime factorisation, HCF and LCM, their applications, and irrational-number reasoning.

12 problems·20–25 min·★★★★☆
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  1. Problem 1
    Prime Factorization·★★☆☆☆
    Write $50$ as a product of prime powers.
    ▶Answer
    $50$ = $2 \times 5^{2}$.
    ▶Step-by-step solution
    1. Divide by prime numbers until only prime factors remain.
    2. The collected prime powers are $2 \times 5^{2}$.
    3. Thus $50$ = $2 \times 5^{2}$.
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  2. Problem 2
    HCF and LCM·★★★☆☆
    Use prime factorisation to find the HCF and LCM of $30$, $6$. Then verify the product relation.
    ▶Answer
    HCF = $6$ and LCM = $30$.
    ▶Step-by-step solution
    1. Write each number as a product of prime powers: $30$ = $2 \times 3 \times 5$; $6$ = $2 \times 3$.
    2. Take the smallest common exponents for the HCF: $2 \times 3$ = $6$.
    3. Take the greatest exponents present for the LCM: $2 \times 3 \times 5$ = $30$.
    4. Check: $6$ x $30$ = $180$, which equals $180$.
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  3. Problem 3
    HCF and LCM·★★★☆☆
    The HCF of $80$ and $150$ is $10$. Use the relation between HCF, LCM, and the two numbers to find their LCM.
    ▶Answer
    The LCM is $1200$.
    ▶Step-by-step solution
    1. For two positive integers, HCF x LCM equals the product of the integers.
    2. So LCM = ($80$ x $150$) / $10$.
    3. Therefore the LCM is $1200$.
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  4. Problem 4
    HCF and LCM·★★★★☆
    Use prime factorisation to find the HCF and LCM of $147$, $105$, $245$.
    ▶Answer
    HCF = $7$ and LCM = $735$.
    ▶Step-by-step solution
    1. Write each number as a product of prime powers: $147$ = $3 \times 7^{2}$; $105$ = $3 \times 5 \times 7$; $245$ = $5 \times 7^{2}$.
    2. Take the smallest common exponents for the HCF: $7$ = $7$.
    3. Take the greatest exponents present for the LCM: $3 \times 5 \times 7^{2}$ = $735$.
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  5. Problem 5
    Prime Factorization·★★★☆☆
    Establish that $36^n$ cannot have last digit 0 or 5 for any natural number n.
    ▶Answer
    It cannot end with 0 or 5.
    ▶Step-by-step solution
    1. The prime factors of $36$ are 2, 3; no factor 5 appears.
    2. Every power $36^n$ therefore has no factor 5.
    3. But any number ending in 0 or 5 is divisible by 5.
    4. So $36^n$ cannot end with 0 or 5.
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  6. Problem 6
    Divisibility·★★★☆☆
    Justify that $13 \times 11 \times 5 + 5$ is composite.
    ▶Answer
    It is composite because it equals $5 \times 144$.
    ▶Step-by-step solution
    1. Factor out the common factor $5$.
    2. This gives $13 \times 11 \times 5 + 5$ = $5(13 \times 11 + 1)$.
    3. Both factors $5$ and $144$ are greater than 1.
    4. So the number has a non-trivial factorization and is composite.
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  7. Problem 7
    Divisibility·★★★☆☆
    Show that $6! + 2$ is composite.
    ▶Answer
    $6! + 2$ is composite.
    ▶Step-by-step solution
    1. Since 2 is one of the factors in 6!, it divides 6!.
    2. Factor out 2: $6! + 2$ = $2\left(\frac{6!}{2} + 1\right)$.
    3. Both factors are integers greater than 1, so the number is composite.
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  8. Problem 8
    LCM·★★★☆☆
    For step lengths $30$, $32$, $50$ cm, determine the least distance that is an exact number of steps for all three walkers.
    ▶Answer
    The shortest common distance is $2400$ cm.
    ▶Step-by-step solution
    1. The distance must be a common multiple of the three step lengths.
    2. The shortest possible distance is therefore their LCM.
    3. LCM(30, 32, 50) = 2400.
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  9. Problem 9
    Irrational Numbers·★★★★☆
    Prove that $\sqrt{11}$ is irrational.
    ▶Answer
    $\sqrt{11}$ is irrational.
    ▶Step-by-step solution
    1. Assume $\sqrt{11}$ = $\frac{a}{b}$ for coprime positive integers $a$ and $b$.
    2. Squaring gives $a^2 = 11b^2$, so 11 divides $a^2$. By unique prime factorisation, 11 divides $a$.
    3. Write $a = 11c$. Substitution gives $b^2 = 11c^2$, so 11 also divides $b$.
    4. This contradicts that $a$ and $b$ are coprime. Therefore the square root is irrational.
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  10. Problem 10
    Irrational Numbers·★★★★☆
    Establish that $\sqrt{2} + \sqrt{3}$ is irrational.
    ▶Answer
    $\sqrt{2} + \sqrt{3}$ is irrational.
    ▶Step-by-step solution
    1. Suppose, for contradiction, that $\sqrt{2} + \sqrt{3} = r$, where r is a positive rational number.
    2. Then $\sqrt{2} = r - \sqrt{3}$.
    3. Squaring both sides gives $2 = r^2 + 3 - 2r\sqrt{3}$.
    4. So $\sqrt{3} = \frac{r^2 + 3 - 2}{2r}$, which would make $\sqrt{3}$ rational.
    5. Since 3 is prime, $\sqrt{3}$ is irrational. The contradiction proves the claim.
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  11. Problem 11
    Irrational Numbers·★★★☆☆
    Is $\frac{\sqrt{11}}{1}$ rational or irrational?
    ▶Answer
    It is irrational.
    ▶Step-by-step solution
    1. The divisor $1$ is a non-zero rational number.
    2. If $\frac{\sqrt{11}}{1}$ were rational, multiplying by $1$ would make $\sqrt{11}$ rational.
    3. This is impossible; hence the quotient is irrational.
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  12. Problem 12
    Irrational Numbers·★★★★☆
    Prove that $\frac{1}{\sqrt{5}}$ is irrational.
    ▶Answer
    $\frac{1}{\sqrt{5}}$ is irrational.
    ▶Step-by-step solution
    1. Suppose $\frac{1}{\sqrt{5}}$ were rational and non-zero.
    2. Its reciprocal would then also be rational, so $\sqrt{5}$ would be rational.
    3. But the square root of the prime 5 is irrational.
    4. This contradiction proves that the reciprocal is irrational.
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