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NCERT Exemplar · Class 10 Mathematics Real Numbers

39 questions · 39 still being checked

EXERCISE 1.3 1–10 (part 3 of 5)

  1. Exercise 1

    Show that the square of any positive integer is either of the form 4q\displaystyle 4 q or 4q+1\displaystyle 4 q+1 for some integer q\displaystyle q.

    Not cross-checked

    NCERT prints no numerical answer for this exercise, so this working has not been cross-checked against the book.

    \[a = 4m,\ 4m+1,\ 4m+2,\ 4m+3 \quad \text{(Euclid's division lemma, } b=4\text{)} \] \[(4m)^2 = 16m^2 = 4(4m^2) \] \[(4m+1)^2 = 16m^2+8m+1 = 4(4m^2+2m)+1 \] \[(4m+2)^2 = 16m^2+16m+4 = 4(4m^2+4m+1) \] \[(4m+3)^2 = 16m^2+24m+9 = 4(4m^2+6m+2)+1 \] Each case gives \(\displaystyle 4q\) or \(\displaystyle 4q+1\). Answer: \(\displaystyle a^2 = 4q\) or \(\displaystyle 4q+1\).

  2. Exercise 2

    Show that cube of any positive integer is of the form 4m,4m+1\displaystyle 4 m, 4 m+1 or 4m+3\displaystyle 4 m+3, for some integer m\displaystyle m.

    Not cross-checked

    NCERT prints no numerical answer for this exercise, so this working has not been cross-checked against the book.

    \[a = 4k,\ 4k+1,\ 4k+2,\ 4k+3 \quad \text{(Euclid's division lemma, } b=4\text{)} \] \[(4k)^3 = 64k^3 = 4(16k^3) \] \[(4k+1)^3 = 64k^3+48k^2+12k+1 = 4(16k^3+12k^2+3k)+1 \] \[(4k+2)^3 = 64k^3+96k^2+48k+8 = 4(16k^3+24k^2+12k+2) \] \[(4k+3)^3 = 64k^3+144k^2+108k+27 = 4(16k^3+36k^2+27k+6)+3 \] The four cases give \(\displaystyle 4m\), \(\displaystyle 4m+1\), \(\displaystyle 4m\), \(\displaystyle 4m+3\) -- only these three residues occur. Answer: \(\displaystyle a^3 = 4m\), \(\displaystyle 4m+1\), or \(\displaystyle 4m+3\).

  3. Exercise 3

    Show that the square of any positive integer cannot be of the form 5q+2\displaystyle 5 q+2 or 5q+3\displaystyle 5 q+3 for any integer q\displaystyle q.

    Not cross-checked

    NCERT prints no numerical answer for this exercise, so this working has not been cross-checked against the book.

    \[a = 5k,\ 5k+1,\ 5k+2,\ 5k+3,\ 5k+4 \quad \text{(Euclid's division lemma, } b=5\text{)} \] \[(5k)^2 = 25k^2 = 5(5k^2) \] \[(5k+1)^2 = 25k^2+10k+1 = 5(5k^2+2k)+1 \] \[(5k+2)^2 = 25k^2+20k+4 = 5(5k^2+4k)+4 \] \[(5k+3)^2 = 25k^2+30k+9 = 5(5k^2+6k+1)+4 \] \[(5k+4)^2 = 25k^2+40k+16 = 5(5k^2+8k+3)+1 \] The remainders are only \(\displaystyle 0,1,4\); \(\displaystyle 2\) and \(\displaystyle 3\) never occur. Answer: \(\displaystyle a^2\) is never of the form \(\displaystyle 5q+2\) or \(\displaystyle 5q+3\).

  4. Exercise 4

    Show that the square of any positive integer cannot be of the form 6m+2\displaystyle 6 m+2 or 6m+5\displaystyle 6 m+5 for any integer m\displaystyle m.

    Not cross-checked

    NCERT prints no numerical answer for this exercise, so this working has not been cross-checked against the book.

    \[a = 6k,\ 6k+1,\ 6k+2,\ 6k+3,\ 6k+4,\ 6k+5 \quad \text{(Euclid's division lemma, } b=6\text{)} \] \[(6k)^2 = 36k^2 = 6(6k^2) \] \[(6k+1)^2 = 36k^2+12k+1 = 6(6k^2+2k)+1 \] \[(6k+2)^2 = 36k^2+24k+4 = 6(6k^2+4k)+4 \] \[(6k+3)^2 = 36k^2+36k+9 = 6(6k^2+6k+1)+3 \] \[(6k+4)^2 = 36k^2+48k+16 = 6(6k^2+8k+2)+4 \] \[(6k+5)^2 = 36k^2+60k+25 = 6(6k^2+10k+4)+1 \] The remainders are only \(\displaystyle 0,1,3,4\); \(\displaystyle 2\) and \(\displaystyle 5\) never occur. Answer: \(\displaystyle a^2\) is never of the form \(\displaystyle 6m+2\) or \(\displaystyle 6m+5\).

  5. Exercise 5

    Show that the square of any odd integer is of the form 4q+1\displaystyle 4 q+1, for some integer q\displaystyle q.

    Not cross-checked

    NCERT prints no numerical answer for this exercise, so this working has not been cross-checked against the book.

    \[a = 2k+1 \quad \text{(odd integer)} \] \[a^2 = 4k^2+4k+1 = 4(k^2+k)+1 \] Answer: \(\displaystyle a^2 = 4q+1\), where \(\displaystyle q = k^2+k\).

  6. Exercise 6

    If n\displaystyle n is an odd integer, then show that n2−1\displaystyle n^2-1 is divisible by 8.

    Not cross-checked

    NCERT prints no numerical answer for this exercise, so this working has not been cross-checked against the book.

    \[n = 2k+1 \quad \text{(odd integer)} \] \[n^2-1 = 4k^2+4k = 4k(k+1) \] \[k(k+1) = 2r \quad \text{(product of consecutive integers is even)} \] \[n^2-1 = 8r \] Answer: \(\displaystyle n^2-1\) is divisible by 8.

  7. Exercise 7

    Prove that if x\displaystyle x and y\displaystyle y are both odd positive integers, then x2+y2\displaystyle x^2+y^2 is even but not divisible by 4.

    Not cross-checked

    NCERT prints no numerical answer for this exercise, so this working has not been cross-checked against the book.

    \[x = 2m+1,\ y = 2n+1 \quad \text{(odd positive integers)} \] \[x^2+y^2 = 4m(m+1)+1+4n(n+1)+1 \] \[x^2+y^2 = 4\big[m(m+1)+n(n+1)\big]+2 \] This is \(\displaystyle 4k+2\) for an integer \(\displaystyle k\): even, remainder \(\displaystyle 2\) mod \(\displaystyle 4\). Answer: \(\displaystyle x^2+y^2\) is even but not divisible by 4.

  8. Exercise 8

    Use Euclid's division algorithm to find the HCF of 441\displaystyle 441, 567\displaystyle 567, 693.

    Matches the book, not yet reviewed

    This working reaches the answer NCERT prints. It has not yet been read through by hand.

    NCERT’s answer
    $\displaystyle 63$
    Apply Euclid's algorithm to $\displaystyle 441$ and $\displaystyle 567$ first, then to the result and 693. \[567 = 441 \times 1 + 126 \] \[441 = 126 \times 3 + 63 \] \[126 = 63 \times 2 + 0 \] So \(\displaystyle \text{HCF}(441,567) = 63\). \[693 = 63 \times 11 + 0 \] So \(\displaystyle \text{HCF}(63,693) = 63\).Answer: \(\displaystyle 63\).

  9. Exercise 9

    Using Euclid's division algorithm, find the largest number that divides 1251\displaystyle 1251, 9377\displaystyle 9377 and 15628\displaystyle 15628 leaving remainders 1\displaystyle 1, 2\displaystyle 2 and 3\displaystyle 3, respectively.

    Matches the book, not yet reviewed

    This working reaches the answer NCERT prints. It has not yet been read through by hand.

    NCERT’s answer
    $\displaystyle 625$
    The required number divides \(\displaystyle 1251-1\), \(\displaystyle 9377-2\), \(\displaystyle 15628-3\) exactly, so it is their HCF. \[1251-1=1250, \quad 9377-2=9375, \quad 15628-3=15625 \] \[9375 = 1250 \times 7 + 625 \] \[1250 = 625 \times 2 + 0 \] So \(\displaystyle \text{HCF}(1250,9375)=625\). \[15625 = 625 \times 25 + 0 \]Answer: \(\displaystyle 625\).

  10. Exercise 10

    Prove that 3+5\displaystyle \sqrt{3}+\sqrt{5} is irrational.

    Not cross-checked

    NCERT prints no numerical answer for this exercise, so this working has not been cross-checked against the book.

    Suppose, for contradiction, \(\displaystyle \sqrt3+\sqrt5=r\) with \(\displaystyle r\) rational; \(\displaystyle r\neq0\) since \(\displaystyle \sqrt3+\sqrt5>0\). \[\sqrt3 = r-\sqrt5 \] \[3 = r^2 - 2r\sqrt5 + 5 \] \[\sqrt5 = \frac{r^2+2}{2r} \] The right side is rational, so \(\displaystyle \sqrt5\) would be rational, contradicting that \(\displaystyle \sqrt5\) is irrational. So the assumption is false.Answer: \(\displaystyle \sqrt3+\sqrt5\) is irrational.