NCERT Exemplar Solutions - Class 10 - Mathematics - Unit 1: Real Numbers

Exercise 1.3

Step 1: Consider a positive integer.

Let \(n\) be any positive integer. Every integer is either even or odd.

Step 2: Case when \(n\) is even.

If \(n\) is even, we can write \(n = 2k\) for some integer \(k\).

Then,

\(n^2 = (2k)^2\)

\(= 4k^2\)

This is a multiple of 4, so it is of the form \(4q\), where \(q = k^2\).

Step 3: Case when \(n\) is odd.

If \(n\) is odd, we can write \(n = 2k + 1\) for some integer \(k\).

Then,

\(n^2 = (2k + 1)^2\)

\(= 4k^2 + 4k + 1\)

\(= 4(k^2 + k) + 1\)

This is of the form \(4q + 1\), where \(q = k^2 + k\).

Step 4: Conclusion.

Therefore, the square of any positive integer is either of the form \(4q\) (if the number is even) or \(4q+1\) (if the number is odd).

Idea. Every integer is either even or odd. Even numbers are multiples of 2, and odd numbers are of the form \(4t+1\) or \(4t+3\). We examine these cases and rewrite the cube as \(4\times(\text{integer})\) plus a small remainder.

Case 1: \(n\) is even.

Let \(n = 2k\).

Then \(n^3 = (2k)^3\).

So \(n^3 = 8k^3\).

Hence \(n^3 = 4\big(2k^3\big)\).

This is of the form \(4m\).

Case 2: \(n\) is odd of the form \(4t+1\).

Let \(n = 4t + 1\).

Compute \(n^3 = (4t+1)^3\).

Expand: \((4t+1)^3 = 64t^3 + 48t^2 + 12t + 1\).

Group multiples of 4: \(64t^3 = 4\cdot 16t^3\), \(48t^2 = 4\cdot 12t^2\), \(12t = 4\cdot 3t\).

Thus \(n^3 = 4\big(16t^3 + 12t^2 + 3t\big) + 1\).

This is of the form \(4m + 1\).

Case 3: \(n\) is odd of the form \(4t+3\).

Let \(n = 4t + 3\).

Compute \(n^3 = (4t+3)^3\).

Expand: \((4t+3)^3 = 64t^3 + 144t^2 + 108t + 27\).

Group multiples of 4: \(64t^3 = 4\cdot 16t^3\), \(144t^2 = 4\cdot 36t^2\), \(108t = 4\cdot 27t\).

Also write \(27 = 24 + 3 = 4\cdot 6 + 3\).

Therefore \(n^3 = 4\big(16t^3 + 36t^2 + 27t + 6\big) + 3\).

This is of the form \(4m + 3\).

Conclusion. For any positive integer \(n\), the cube \(n^3\) is congruent to \(0\), \(1\), or \(3\) modulo \(4\). Equivalently, \(n^3\) has the form \(4m\), \(4m+1\), or \(4m+3\) for some integer \(m\).

Step 1: Express an integer in terms of division by 5.

When any integer \(n\) is divided by 5, the remainder can only be 0, 1, 2, 3, or 4. So, every integer can be written in one of the following forms:

\(n = 5q\), or \(n = 5q + 1\), or \(n = 5q + 2\), or \(n = 5q + 3\), or \(n = 5q + 4\).

Step 2: Square each possibility.

If \(n = 5q\):

\(n^2 = (5q)^2 = 25q^2 = 5(5q^2)\).

This is clearly a multiple of 5, so the remainder is 0 when divided by 5.

If \(n = 5q + 1\):

\(n^2 = (5q + 1)^2 = 25q^2 + 10q + 1 = 5(5q^2 + 2q) + 1\).

So the remainder is 1 when divided by 5.

If \(n = 5q + 2\):

\(n^2 = (5q + 2)^2 = 25q^2 + 20q + 4 = 5(5q^2 + 4q) + 4\).

So the remainder is 4 when divided by 5.

If \(n = 5q + 3\):

\(n^2 = (5q + 3)^2 = 25q^2 + 30q + 9 = 25q^2 + 30q + 5 + 4\).

= \(5(5q^2 + 6q + 1) + 4\).

So the remainder is also 4 when divided by 5.

If \(n = 5q + 4\):

\(n^2 = (5q + 4)^2 = 25q^2 + 40q + 16 = 25q^2 + 40q + 15 + 1\).

= \(5(5q^2 + 8q + 3) + 1\).

So the remainder is 1 when divided by 5.

Step 3: Collect the results.

Therefore, depending on the form of \(n\), the possible remainders of \(n^2\) when divided by 5 are only:

0, 1, or 4.

Step 4: Compare with the given forms.

If \(n^2\) were of the form \(5q + 2\), its remainder upon division by 5 would be 2. Similarly, if \(n^2\) were of the form \(5q + 3\), its remainder would be 3.

But we have just shown that the remainder can only be 0, 1, or 4.

Conclusion. Hence, the square of any positive integer can never be of the form \(5q + 2\) or \(5q + 3\).

Idea. Every integer, when divided by 6, leaves a remainder from \(0\) to \(5\). We will check the square in each case and look at the remainder (mod 6).

Step 1: Write the number by division algorithm.

Let \(n\) be any positive integer. Then \(n = 6q + r\).

Here \(q\) is an integer and \(r\in\{0,1,2,3,4,5\}\).

Step 2: Square in each remainder case and reduce modulo 6.

Case \(r=0\): \(n = 6q\), so \(n^2 = 36q^2\).

Thus \(n^2\equiv 0\pmod{6}\).

Case \(r=1\): \(n = 6q + 1\), so \(n^2 = 36q^2 + 12q + 1\).

Thus \(n^2\equiv 1\pmod{6}\).

Case \(r=2\): \(n = 6q + 2\), so \(n^2 = 36q^2 + 24q + 4\).

Thus \(n^2\equiv 4\pmod{6}\).

Case \(r=3\): \(n = 6q + 3\), so \(n^2 = 36q^2 + 36q + 9\).

Since \(9\equiv 3\pmod{6}\), we get \(n^2\equiv 3\pmod{6}\).

Case \(r=4\): \(n = 6q + 4\), so \(n^2 = 36q^2 + 48q + 16\).

Since \(16\equiv 4\pmod{6}\), we get \(n^2\equiv 4\pmod{6}\).

Case \(r=5\): \(n = 6q + 5\), so \(n^2 = 36q^2 + 60q + 25\).

Since \(25\equiv 1\pmod{6}\), we get \(n^2\equiv 1\pmod{6}\).

Step 3: Collect possible square remainders.

From the cases we have \(n^2\equiv 0, 1, 3, 4\pmod{6}\).

Therefore \(n^2\) can never have remainder \(2\) or \(5\) upon division by 6.

Conclusion. A perfect square cannot be of the form \(6m + 2\) or \(6m + 5\) for any integer \(m\).

Step 1: Express an odd integer.

Any odd integer can be written as \(n = 2k + 1\), where \(k\) is some integer. This form ensures that \(n\) is always one more than an even number, hence odd.

Step 2: Square the expression.

Now square both sides:

\(n^2 = (2k + 1)^2\)

Step 3: Expand the square.

Using the formula \((a + b)^2 = a^2 + 2ab + b^2\), we get:

\(n^2 = (2k)^2 + 2 \cdot (2k) \cdot 1 + 1^2\)

\(n^2 = 4k^2 + 4k + 1\)

Step 4: Factorize.

Take 4 common from the first two terms:

\(n^2 = 4(k^2 + k) + 1\)

Step 5: Identify the form.

Let \(q = k^2 + k\), which is an integer (since \(k\) is an integer).

Thus, \(n^2 = 4q + 1\).

Conclusion. The square of any odd integer is always of the form \(4q + 1\) for some integer \(q\).

Step 1: Express an odd integer.

Any odd integer can be written as

\(n = 2k + 1\), where \(k\) is an integer.

Step 2: Expand \(n^2 - 1\).

\(n^2 - 1 = (2k + 1)^2 - 1\)

\(= 4k^2 + 4k + 1 - 1\)

\(= 4k^2 + 4k\)

\(= 4k(k + 1)\).

Step 3: Analyze the product \(k(k+1)\).

The numbers \(k\) and \(k+1\) are consecutive integers.

In any two consecutive integers, one is always even.

Therefore, their product \(k(k+1)\) is always even.

Step 4: Multiply by 4.

Since \(k(k+1)\) is even, we can write

\(k(k+1) = 2m\), for some integer \(m\).

So, \(n^2 - 1 = 4 \times 2m = 8m\).

Conclusion.

This shows that \(n^2 - 1\) is a multiple of 8. Hence, it is divisible by 8.

Step 1: Write odd numbers in standard form.

Since \(x\) and \(y\) are odd, there exist integers \(a\) and \(b\) such that

\(x = 2a + 1\)

\(y = 2b + 1\)

Step 2: Square each expression.

\(x^2 = (2a + 1)^2\)

\(\quad = 4a^2 + 4a + 1\)

\(\quad = 4\big(a^2 + a\big) + 1\)

Similarly, \(y^2 = (2b + 1)^2\)

\(\quad = 4b^2 + 4b + 1\)

\(\quad = 4\big(b^2 + b\big) + 1\)

Step 3: Add the squares.

\(x^2 + y^2 = \big[4(a^2 + a) + 1\big] + \big[4(b^2 + b) + 1\big]\)

\(\quad = 4\big[(a^2 + a) + (b^2 + b)\big] + 2\)

Step 4: Conclude parity and divisibility by 4.

The expression has the form \(4N + 2\) for some integer \(N\).

Therefore, \(x^2 + y^2\) is even (since it is \(2\) more than a multiple of \(4\))

but it is not divisible by \(4\) (a multiple of \(4\) cannot leave remainder \(2\)).

Optional check (modulo 4 view).

From the expansions above, \(x^2 \equiv 1 \pmod{4}\) and \(y^2 \equiv 1 \pmod{4}\).

Hence, \(x^2 + y^2 \equiv 1 + 1 = 2 \pmod{4}\), which matches \(4N + 2\).

Step 1: Recall Euclid’s Division Algorithm.

For any two positive integers \(a\) and \(b\), with \(a > b\), we can write:

\(a = bq + r\), where \(0 \leq r < b\).

If \(r = 0\), then \(b\) is the HCF of \(a\) and \(b\). Otherwise, repeat the process with \(b\) and \(r\).

Step 2: Find HCF of 567 and 441.

Divide 567 by 441:

\(567 = 441 \times 1 + 126\)

Now divide 441 by 126:

\(441 = 126 \times 3 + 63\)

Now divide 126 by 63:

\(126 = 63 \times 2 + 0\)

Since remainder is 0, the HCF of 567 and 441 is 63.

Step 3: Find HCF of 693 and 63.

Divide 693 by 63:

\(693 = 63 \times 11 + 0\)

So, the HCF of 693 and 63 is 63.

Step 4: Combine results.

Hence, the HCF of 441, 567, and 693 is 63.

Step 1: Understand the problem.

We want a number that divides each of the three given numbers, but leaves a remainder: 1 when dividing 1251, 2 when dividing 9377, and 3 when dividing 15628.

Step 2: Subtract the remainders.

If a number leaves remainder \(r\), then it divides the number minus \(r\).

So, the required number must divide:

\(1251 - 1 = 1250\)

\(9377 - 2 = 9375\)

\(15628 - 3 = 15625\)

Step 3: Find the HCF of these adjusted numbers.

We now need the HCF (highest common factor) of 1250, 9375, and 15625.

Step 4: Apply Euclid’s division algorithm.

First, find HCF(9375, 1250).

Divide 9375 by 1250:

\(9375 = 1250 \times 7 + 625\)

Now divide 1250 by 625:

\(1250 = 625 \times 2 + 0\)

So HCF(9375, 1250) = 625.

Step 5: Include the third number.

Now check HCF(15625, 625).

Divide 15625 by 625:

\(15625 = 625 \times 25 + 0\)

So the HCF is 625.

Final Answer: The largest required number is 625.

Goal. Show that \(\sqrt{3}+\sqrt{5}\) cannot be a rational number.

Step 1: Assume the opposite.

Let \(s = \sqrt{3} + \sqrt{5}\).

Assume \(s\) is rational.

Step 2: Isolate one radical.

From \(s = \sqrt{3} + \sqrt{5}\), we get

\(\sqrt{5} = s - \sqrt{3}\).

Step 3: Square to remove \(\sqrt{5}\).

Square both sides:

\(5 = (s - \sqrt{3})^2\).

Expand the right-hand side:

\(5 = s^2 + 3 - 2s\sqrt{3}\).

Step 4: Solve for \(\sqrt{3}\).

Rearrange:

\(2s\sqrt{3} = s^2 - 2\).

Therefore,

\(\sqrt{3} = \dfrac{s^2 - 2}{2s}\).

Step 5: Get the contradiction.

Since we assumed \(s\) is rational, both \(s^2\) and \(2s\) are rational.

Hence \(\dfrac{s^2 - 2}{2s}\) is rational.

That means \(\sqrt{3}\) is rational, which is false.

Step 6: Conclude.

Our assumption that \(s\) is rational leads to a contradiction.

Therefore, \(\sqrt{3}+\sqrt{5}\) is irrational.

Note. We divided by \(2s\) in Step 4. This is valid because \(s = \sqrt{3}+\sqrt{5} > 0\).

Step 1: Prime factorisation of \(12^n\).

We know that

\(12^n = (3 \times 4)^n\).

This can be written as

\(12^n = 3^n \times 4^n\).

And since \(4^n = (2^2)^n = 2^{2n}\), we finally get

\(12^n = 3^n \times 2^{2n}\).

So \(12^n\) has only the prime factors 2 and 3.

Step 2: Condition for a number to end with 0 or 5.

If a number ends with digit 0, it must be divisible by 10. That means it must have both 2 and 5 as prime factors.

If a number ends with digit 5, it must be divisible by 5. So it must contain 5 as a prime factor.

Step 3: Check divisibility by 5.

From Step 1, we saw that \(12^n\) contains only factors of 2 and 3.

It does not have any factor of 5.

Conclusion.

Since \(12^n\) has no factor of 5, it can never end with 0 or 5.

Therefore, for any natural number \(n\), the number \(12^n\) cannot end with the digit 0 or 5.

Step 1: Understand the problem.

Each person takes steps of different lengths: 40 cm, 42 cm, and 45 cm. We want a distance that is a common multiple of all three step lengths. The minimum such distance will be the Least Common Multiple (LCM) of 40, 42, and 45.

Step 2: Prime factorisation of each number.

Factorise 40:

\(40 = 2^3 \times 5\)

Factorise 42:

\(42 = 2 \times 3 \times 7\)

Factorise 45:

\(45 = 3^2 \times 5\)

Step 3: Take the highest power of each prime factor.

From 40: we need \(2^3\) and \(5\).

From 42: we need \(3\) and \(7\).

From 45: we need \(3^2\) (this is larger than the single \(3\) from 42).

Step 4: Multiply them together.

LCM = \(2^3 \times 3^2 \times 5 \times 7\)

= \(8 \times 9 \times 5 \times 7\)

= \(2520\)

Step 5: Interpret the result.

The minimum distance that each can walk in complete steps is 2520 cm.

Convert into metres:

\(2520\,\text{cm} = 25.2\,\text{m}\).

Final Answer: The minimum distance is 2520 cm or 25.2 m.

Step 1: Prime factorisation of the denominator.

We start with the denominator \(5000\).

Break it down: \(5000 = 5 \times 1000\).

Now, \(1000 = 10^3 = (2 \cdot 5)^3 = 2^3 \cdot 5^3\).

So, \(5000 = 5 \times (2^3 \cdot 5^3) = 2^3 \cdot 5^4\).

Thus, in the required form, the denominator is \(2^3 \cdot 5^4\).

Step 2: Find the number of decimal places.

The highest power among \(2^3\) and \(5^4\) is \(5^4\).

This means the decimal expansion will terminate after 4 places.

Step 3: Write the decimal without division.

The fraction is \(\dfrac{257}{5000}\).

We can think of this as dividing numerator and denominator in steps.

First, divide \(257\) by \(5\):

\(\dfrac{257}{5} = 51.4\).

Now we still need to divide by \(1000\) (since \(5000 = 5 \times 1000\)).

So, \(\dfrac{257}{5000} = \dfrac{51.4}{1000} = 0.0514\).

Final Answer:

The denominator in prime form is \(2^3 \cdot 5^4\). The decimal expansion is \(0.0514\).

Recall. For a prime \(p\), the number \(\sqrt{p}\) is irrational. If \(\sqrt{p}\) were rational, say \(\sqrt{p}=\dfrac{m}{n}\) in lowest terms, then squaring would give \(p=\dfrac{m^2}{n^2}\). Hence \(pn^2=m^2\), so \(p\) divides \(m^2\), and therefore divides \(m\), contradicting lowest terms unless \(n=1\) and \(p\) were a perfect square. But a prime is not a perfect square. So \(\sqrt{p}\) is irrational.

Goal. Show that the sum \(\sqrt{p}+\sqrt{q}\) cannot be rational.

Step 1. Assume, for contradiction, that the sum is rational. Let \(s\in\mathbb{Q}\) with

\(\sqrt{p}+\sqrt{q}=s.\)

Step 2. Isolate one square root and square. From

\(\sqrt{q}=s-\sqrt{p}\)

square both sides to get

\(q=s^2+p-2s\,\sqrt{p}.\)

Step 3. Solve for \(\sqrt{p}\). Rearranging gives

\(2s\,\sqrt{p}=s^2+p-q.\)

If \(s\neq 0\) (the sum of two positive square roots cannot be \(0\)), then

\(\sqrt{p}=\dfrac{s^2+p-q}{2s}.\)

Step 4. Reach the contradiction. The right-hand side is a quotient of rationals, hence rational. Thus \(\sqrt{p}\) is rational, which contradicts the recall above.

Special case \(p=q\). Then \(\sqrt{p}+\sqrt{q}=2\sqrt{p}.\) Since \(\sqrt{p}\) is irrational and \(2\) is a nonzero rational, the product \(2\sqrt{p}\) is irrational.

Conclusion. In all cases, the assumption that \(\sqrt{p}+\sqrt{q}\) is rational leads to a contradiction. Therefore \(\sqrt{p}+\sqrt{q}\) is irrational.