Monday, November 29, 2021

Exponents and Radicals : Exercise (2.4) - Solution

  1. Simplify the following.

    (a) $3 \sqrt{5}+7 \sqrt{5}$


    $ \begin{aligned} &3\sqrt{5}+7\sqrt{5}\\\\ =&10\sqrt{5}\end{aligned}$


    (b) $\sqrt{75}-\sqrt{12}$


    $\begin{aligned} &\sqrt{{75}}-\sqrt{{12}}\\\\ =&\sqrt{{25\times 3}}-\sqrt{{4\times 3}}\\\\ =&5\sqrt{3}-2\sqrt{3}\\\\ =&3\sqrt{3} \end{aligned}$


    (c) $3 \cdot 3 \sqrt{3} \cdot 3 \sqrt{27}$


    $\begin{aligned} &3\cdot 3\sqrt{3}\cdot 3\sqrt{{27}}\\\\ =&27\cdot \sqrt{3}\cdot \sqrt{{27}}\\\\ =&27\cdot \sqrt{3}\cdot \sqrt{{9\times 3}}\\\\ =&27\cdot \sqrt{3}\cdot 3\sqrt{3}\\\\ =&81\cdot \sqrt{3}\cdot \sqrt{3}\\\\ =&81\cdot 3\\\\ =&243 \end{aligned}$


    (d) $2 \sqrt{5} \cdot 3 \sqrt{2}$


    $\begin{aligned} &2 \sqrt{5} \cdot 3 \sqrt{2}\\\\ =&6\sqrt{10} \end{aligned}$


    (e) $(4-\sqrt{3})^{2}$


    $\begin{aligned} &(4-\sqrt{3})^2\\\\ =&16-2\times 4\times \sqrt{3}+3\\\\ =&19-8\sqrt{3} \end{aligned}$


    (f) $(\sqrt{3}+2 \sqrt{2})(\sqrt{3}+\sqrt{2})$


    $\begin{aligned} &\left( {\sqrt{3}+2\sqrt{2}} \right)\left( {\sqrt{3}+\sqrt{2}} \right)\\\\ =&\sqrt{3}\left( {\sqrt{3}+2\sqrt{2}} \right)+\sqrt{2}\left( {\sqrt{3}+2\sqrt{2}} \right)\\\\ =&3+2\sqrt{6}+\sqrt{6}+4\\\\ =&7+3\sqrt{6} \end{aligned}$


    (g) $(\sqrt{7}-\sqrt{6})(\sqrt{7}+\sqrt{6})(\sqrt{x}+1)(\sqrt{x}-1)$


    $\begin{aligned} &\left( {\sqrt{7}-\sqrt{6}} \right)\left( {\sqrt{7}+\sqrt{6}} \right)\left( {\sqrt{x}+1} \right)\left( {\sqrt{x}-1} \right)\\\\ =&\left( {7-6} \right)\left( {x-1} \right)\\\\ =&x-1 \end{aligned}$


    (h) $\sqrt{75}-\dfrac{3}{4} \sqrt{48}-5 \sqrt{12}$


    $\begin{aligned} &\sqrt{{75}}-\frac{3}{4}\sqrt{{48}}-5\sqrt{{12}}\\\\ =&\sqrt{{25\times 3}}-\frac{3}{4}\sqrt{{16\times 3}}-5\sqrt{{4\times 3}}\\\\ =&5\sqrt{3}-\frac{3}{4}\times 4\sqrt{3}-5\times 2\sqrt{3}\\\\ =&5\sqrt{3}-3\sqrt{3}-10\sqrt{3}\\\\ =&-8\sqrt{3} \end{aligned}$


    (i) $\sqrt{2 x^{2}}+5 \sqrt{32 x^{2}}-2 \sqrt{98 x^{2}}$


    $\begin{aligned} &\sqrt{{2{{x}^{2}}}}+5\sqrt{{32{{x}^{2}}}}-2\sqrt{{98{{x}^{2}}}}\\\\ =&\sqrt{{2{{x}^{2}}}}+5\sqrt{{16\times 2{{x}^{2}}}}-2\sqrt{{49\times 2{{x}^{2}}}}\\\\ =&\left( {\sqrt{2}x} \right)+\left( {5\times 4\sqrt{2}x} \right)-\left( {2\times 7\sqrt{2}x} \right)\\\\ =&\sqrt{2}x+20\sqrt{2}x-14\sqrt{2}x\\\\ =&7\sqrt{2}x \end{aligned}$


    (j) $\sqrt{20 a^{3}}+a \sqrt{5 a}+\sqrt{80 a^{3}}$


    $\begin{aligned} &\sqrt{{20{{a}^{3}}}}+a\sqrt{{5a}}+\sqrt{{80{{a}^{3}}}}\\\\ =&\sqrt{{4\times 5{{a}^{3}}}}+a\sqrt{{5a}}+\sqrt{{16\times 5{{a}^{3}}}}\\\\ =&2a\sqrt{{5a}}+a\sqrt{{5a}}+4a\sqrt{{5a}}\\\\ =&7a\sqrt{{5a}} \end{aligned}$


  2. Rationalise the denominators and simplify.

    (a) $\dfrac{2}{\sqrt{5}}$


    $\begin{aligned} &\dfrac{2}{\sqrt{5}}\\\\ =&\dfrac{2}{\sqrt{5}}\times\dfrac{\sqrt{5}}{\sqrt{5}}\\\\ =&\dfrac{2\sqrt{5}}{5} \end{aligned}$


    (b) $\dfrac{5}{2+\sqrt{3}}$


    $\begin{aligned} &\dfrac{5}{{2+\sqrt{3}}}\\\\ =&\dfrac{{5\left( {2-\sqrt{3}} \right)}}{{\left( {2+\sqrt{3}} \right)\left( {2-\sqrt{3}} \right)}}\\\\ =&\dfrac{{5\left( {2-\sqrt{3}} \right)}}{{4-3}}\\\\ =&5\left( {2-\sqrt{3}} \right) \end{aligned}$


    (c) $\dfrac{12}{\sqrt{5}-\sqrt{3}}$


    $\begin{aligned} &\dfrac{{12}}{{\sqrt{5}-\sqrt{3}}}\\\\ =&\dfrac{{12\left( {\sqrt{5}+\sqrt{3}} \right)}}{{\left( {\sqrt{5}-\sqrt{3}} \right)\left( {\sqrt{5}+\sqrt{3}} \right)}}\\\\ =&\dfrac{{12\left( {\sqrt{5}+\sqrt{3}} \right)}}{{5-3}}\\\\ =&\dfrac{{12\left( {\sqrt{5}+\sqrt{3}} \right)}}{2}\\\\ =&6\left( {\sqrt{5}+\sqrt{3}} \right) \end{aligned}$


    (d) $\dfrac{\sqrt{2}+1}{2 \sqrt{2}-1}$


    $\begin{aligned} &\dfrac{{\sqrt{2}+1}}{{2\sqrt{2}-1}}\\\\ =&\dfrac{{\left( {\sqrt{2}+1} \right)\left( {2\sqrt{2}+1} \right)}}{{\left( {2\sqrt{2}-1} \right)\left( {2\sqrt{2}+1} \right)}}\\\\ =&\dfrac{{\sqrt{2}\left( {2\sqrt{2}+1} \right)+1\left( {2\sqrt{2}+1} \right)}}{{8-1}}\\\\ =&\dfrac{{4+\sqrt{2}+2\sqrt{2}+1}}{7}\\\\ =&\dfrac{{5+3\sqrt{2}}}{7} \end{aligned}$


    (e) $\dfrac{\sqrt{7}+3 \sqrt{2}}{\sqrt{7}-\sqrt{2}}$


    $\begin{aligned} &\dfrac{{\sqrt{7}+3\sqrt{2}}}{{\sqrt{7}-\sqrt{2}}}\\\\ =&\dfrac{{\left( {\sqrt{7}+3\sqrt{2}} \right)\left( {\sqrt{7}+\sqrt{2}} \right)}}{{\left( {\sqrt{7}-\sqrt{2}} \right)\left( {\sqrt{7}+\sqrt{2}} \right)}}\\\\ =&\dfrac{{\sqrt{7}\left( {\sqrt{7}+\sqrt{2}} \right)+3\sqrt{2}\left( {\sqrt{7}+\sqrt{2}} \right)}}{{7-2}}\\\\ =&\dfrac{{7+\sqrt{{14}}+3\sqrt{{14}}+6}}{5}\\\\ =&\dfrac{{13+4\sqrt{{14}}}}{5} \end{aligned}$


    (f) $\dfrac{\sqrt{17}-\sqrt{11}}{\sqrt{17}+\sqrt{11}}$


    $\begin{aligned} &\dfrac{{\sqrt{{17}}-\sqrt{{11}}}}{{\sqrt{{17}}+\sqrt{{11}}}}\\\\ =&\dfrac{{\left( {\sqrt{{17}}-\sqrt{{11}}} \right)\left( {\sqrt{{17}}-\sqrt{{11}}} \right)}}{{\left( {\sqrt{{17}}+\sqrt{{11}}} \right)\left( {\sqrt{{17}}-\sqrt{{11}}} \right)}}\\\\ =&\dfrac{{17-2\sqrt{{17}}\sqrt{{11}}+11}}{{17-11}}\\\\ =&\dfrac{{28-2\sqrt{{187}}}}{6}\\\\ =&\dfrac{{2\left( {14-\sqrt{{187}}} \right)}}{6}\\\\ =&\dfrac{{14-\sqrt{{187}}}}{3} \end{aligned}$


    (g) $\dfrac{1}{2 \sqrt{2}-\sqrt{3}}$


    $\begin{aligned} &\dfrac{1}{{2\sqrt{2}-\sqrt{3}}}\\\\ =&\dfrac{{1\cdot \left( {2\sqrt{2}+\sqrt{3}} \right)}}{{\left( {2\sqrt{2}-\sqrt{3}} \right)\left( {2\sqrt{2}+\sqrt{3}} \right)}}\\\\ =&\dfrac{{2\sqrt{2}+\sqrt{3}}}{{8-3}}\\\\ =&\dfrac{{2\sqrt{2}+\sqrt{3}}}{5} \end{aligned}$


    (h) $\dfrac{\sqrt{6}+1}{3-\sqrt{5}}$


    $\begin{aligned} &\dfrac{{\sqrt{6}+1}}{{3-\sqrt{5}}}\\\\ =&\dfrac{{\left( {\sqrt{6}+1} \right)\left( {3+\sqrt{5}} \right)}}{{\left( {3-\sqrt{5}} \right)\left( {3+\sqrt{5}} \right)}}\\\\ =&\dfrac{{\sqrt{6}\left( {3+\sqrt{5}} \right)+1\left( {3+\sqrt{5}} \right)}}{{9-5}}\\\\ =&\dfrac{{3\sqrt{6}+\sqrt{{30}}+3+\sqrt{5}}}{4}\\\\ =&\dfrac{{3+\sqrt{5}+3\sqrt{6}+\sqrt{{30}}}}{4} \end{aligned}$


  3. Write as a single fraction.

    (a) $\dfrac{1}{\sqrt{3}+1}+\dfrac{1}{\sqrt{3}-1}$


    $\begin{aligned} & \dfrac{1}{\sqrt{3}+1}+\dfrac{1}{\sqrt{3}-1} \\\\ =& \dfrac{1}{\sqrt{3}+1} \times \dfrac{\sqrt{3}-1}{\sqrt{3}-1}+\dfrac{1}{\sqrt{3}-1} \times \dfrac{\sqrt{3}+1}{\sqrt{3}+1} \\\\ =& \dfrac{\sqrt{3}-1}{(\sqrt{3})^{2}-1^{2}}+\dfrac{\sqrt{3}+1}{(\sqrt{3})^{2}-1^{2}} \\\\ =& \dfrac{\sqrt{3}-1}{3-1}+\dfrac{\sqrt{3}+1}{3-1} \\\\ =& \dfrac{\sqrt{3}-1+\sqrt{3}+1}{2} \\\\ =& \dfrac{2 \sqrt{3}}{2} \\\\ =& \sqrt{3} \end{aligned}$


    (b) $\dfrac{2}{\sqrt{7}+\sqrt{2}}+\dfrac{1}{\sqrt{7}-\sqrt{2}}$


    $\begin{aligned} & \dfrac{2}{\sqrt{7}+\sqrt{2}}+\dfrac{1}{\sqrt{7}-\sqrt{2}} \\\\ =& \dfrac{2}{\sqrt{7}+\sqrt{2}} \times \dfrac{\sqrt{7}-\sqrt{2}}{\sqrt{7}-\sqrt{2}}+\dfrac{1}{\sqrt{7}-\sqrt{2}} \times \dfrac{\sqrt{7}+\sqrt{2}}{\sqrt{7}+\sqrt{2}} \\\\ =& \dfrac{2(\sqrt{7}-\sqrt{2})}{(\sqrt{7})^{2}-(\sqrt{2})^{2}}+\dfrac{\sqrt{7}+\sqrt{2}}{(\sqrt{7})^{2}-(\sqrt{2})^{2}} \\\\ =& \dfrac{2(\sqrt{7}-\sqrt{2})}{7-2}+\dfrac{\sqrt{7}+\sqrt{2}}{7-2} \\\\ =& \dfrac{2 \sqrt{7}-2 \sqrt{2}+\sqrt{7}+\sqrt{2}}{5} \\\\ =& \dfrac{3 \sqrt{7}-\sqrt{2}}{5} \end{aligned}$


    (c) $\dfrac{1}{3+\sqrt{3}}+\dfrac{1}{\sqrt{3}-3}+\dfrac{1}{\sqrt{3}}$


    $\begin{aligned} & \dfrac{1}{3+\sqrt{3}}+\dfrac{1}{\sqrt{3}-3}+\dfrac{1}{\sqrt{3}} \\\\ =& \dfrac{3-3 \sqrt{3}}{\sqrt{3}(3+\sqrt{3})(\sqrt{3}-3)}+\dfrac{3 \sqrt{3}+3}{\sqrt{3}(3+\sqrt{3})(\sqrt{3}-3)}+\dfrac{-6}{\sqrt{3}(3+\sqrt{3})(\sqrt{3}-3)} \\\\ =& \dfrac{3-3 \sqrt{3}+3 \sqrt{3}+3-6}{\sqrt{3}(3+\sqrt{3})(\sqrt{3}-3)} \\\\ =& \dfrac{0}{\sqrt{3}(3+\sqrt{3})(\sqrt{3}-3)} \\\\ =& 0 \end{aligned}$


    (d) $\dfrac{7+\sqrt{5}}{7-\sqrt{5}}+\dfrac{\sqrt{11}-3}{\sqrt{11}+3}$


    $\begin{aligned} & \dfrac{7+\sqrt{5}}{7-\sqrt{5}}+\dfrac{\sqrt{11}-3}{\sqrt{11}+3} \\\\ =& \dfrac{(7+\sqrt{5})(7+\sqrt{5})}{(7-\sqrt{5})(7+\sqrt{5})}+\dfrac{(\sqrt{11}-3)(\sqrt{11}-3)}{(\sqrt{11}+3)(\sqrt{11}-3)} \\\\ =& \dfrac{54+14 \sqrt{5}}{49-5}+\dfrac{20-6 \sqrt{11}}{11-9} \\\\ =& \dfrac{54+14 \sqrt{5}}{44}+\dfrac{20-6 \sqrt{11}}{2} \\\\ =& \dfrac{27+7 \sqrt{5}}{22}+10-3 \sqrt{11} \\\\ =& \dfrac{27+7 \sqrt{5}+220-66 \sqrt{11}}{22} \\\\ =& \dfrac{247+7 \sqrt{5}-66 \sqrt{11}}{22} \end{aligned}$


    (e) $\dfrac{3+2 \sqrt{2}}{(\sqrt{3}-1)^{2}}$


    $\begin{aligned} & \dfrac{3+2 \sqrt{2}}{(\sqrt{3}-1)^{2}} \\\\ =& \dfrac{3+2 \sqrt{2}}{(\sqrt{3})^{2}-2 \sqrt{3+1}} \\\\ =& \dfrac{3+2 \sqrt{2}}{4-2 \sqrt{3}} \times \dfrac{4+2 \sqrt{3}}{4+2 \sqrt{3}} \\\\ =& \dfrac{(3+2 \sqrt{2})(4+2 \sqrt{3})}{4^{2}-(2 \sqrt{3})^{2}} \\\\ =& \dfrac{12+6 \sqrt{3}+8 \sqrt{2}+4 \sqrt{6}}{16-12} \\\\ =& \dfrac{12+6 \sqrt{3}+8 \sqrt{2}+4 \sqrt{6}}{4} \\\\ =& \dfrac{6+3 \sqrt{3}+4 \sqrt{2}+2 \sqrt{6}}{2} \end{aligned}$


    (f) $\sqrt{\dfrac{x+1}{x-1}}+\sqrt{\dfrac{x-1}{x+1}}-\sqrt{\dfrac{1}{x^{2}-1}}$


    $\begin{aligned} & \sqrt{\dfrac{x+1}{x-1}}+\sqrt{\dfrac{x-1}{x+1}}-\sqrt{\dfrac{1}{x^{2}-1}} \\\\ =& \dfrac{\sqrt{x+1}}{\sqrt{x-1}}+\dfrac{\sqrt{x-1}}{\sqrt{x+1}}-\dfrac{1}{\sqrt{x^{2}-1}} \\\\ =&\left(\dfrac{\sqrt{x+1}}{\sqrt{x-1}} \times \dfrac{\sqrt{x-1}}{\sqrt{x-1}}\right)+\left(\dfrac{\sqrt{x-1}}{\sqrt{x+1}} \times \dfrac{\sqrt{x+1}}{\sqrt{x+1}}\right)-\left(\dfrac{1}{\sqrt{x^{2}-1}} \times \dfrac{\sqrt{x^{2}-1}}{\sqrt{x^{2}-1}}\right) \\\\ =& \dfrac{\sqrt{x^{2}-1}}{x-1}+\dfrac{\sqrt{x^{2}-1}}{x+1}-\dfrac{\sqrt{x^{2}-1}}{x^{2}-1} \\\\ =& \dfrac{(x+1) \sqrt{x^{2}-1}}{(x-1)(x+1)}+\dfrac{(x-1) \sqrt{x^{2}-1}}{(x+1)(x-1)}-\dfrac{\sqrt{x^{2}-1}}{x^{2}-1} \\\\ =&\left(\dfrac{x+1+x-1-1}{x^{2}-1}\right) \sqrt{x^{2}-1} \\\\ =& \dfrac{(2 x+1) \sqrt{x^{2}-1}}{x^{2}-1} \end{aligned}$


    (g) $\sqrt{\dfrac{\sqrt[5]{32}+\sqrt{4}}{2^{-2}-2^{-3}}}$


    $\begin{aligned} & \sqrt{\dfrac{\sqrt[5]{32}+\sqrt{4}}{2^{-2}-2^{-3}}} \\\\ =& \sqrt{\dfrac{\sqrt[5]{2^{5}}+\sqrt{4}}{\dfrac{1}{2^{2}}-\dfrac{1}{2^{3}}}} \\\\ =& \sqrt{\dfrac{2+2}{\dfrac{1}{4}-\dfrac{1}{8}}} \\\\ =& \sqrt{\dfrac{4}{1 / 8}} \\\\ =& \sqrt{4 \times 8} \\\\ =& \sqrt{32} \\\\ =& 4 \sqrt{2} \end{aligned}$


Principle of Mathematical Induction

Mathematical Induction แ€€ို domino effect แ€–ြแ€„့် แ€แ€„်แ€…ားแ€–ော်แ€•ြแ€œေ့แ€›ှိแ€žแ€Š်။แ€กောแ€€်แ€•ါ แ€•ုံแ€€ိုแ€œေ့แ€œာแ€€ြแ€Š့်แ€•ါ။



แ€•แ€‘แ€™ domino แ€œဲแ€™แ€€ျแ€œျှแ€„် ၎แ€„်းแ€”ောแ€€်แ€™ှ แ€…ီแ€‘ားแ€žော domino แ€™ျားแ€œဲแ€€ျแ€™แ€Š်แ€™แ€Ÿုแ€်แ€•ါ။ แ€•แ€‘แ€™ domino แ€œဲแ€€ျแ€žแ€Š်။ แ€‘ို့แ€”ောแ€€် แ€…ီแ€‘ားแ€žော domino แ€‘ဲแ€™ှ แ€€ြိုแ€€်แ€›ာแ€แ€…်แ€ုแ€œဲแ€€ျแ€žแ€Š်แ€Ÿု แ€šူแ€†แ€•ြီး၊ ၎แ€„်းแ€”ောแ€€်แ€›ှိ แ€€แ€•်แ€œျแ€€်แ€›ှိ‌แ€žော domino แ€œဲแ€€ျแ€€ြောแ€„်း แ€žแ€€်แ€žေแ€•ြแ€”ိုแ€„်แ€œျှแ€„် แ€™แ€Š်แ€žแ€Š့် domino แ€™แ€†ို แ€œဲแ€€ျแ€žแ€Š်แ€Ÿု แ€žแ€€်แ€žေแ€•ြแ€”ိုแ€„်แ€•ါแ€žแ€Š်။ แ€คแ€žแ€Š်แ€™ှာ Mathematical Induction แ€†ိုแ€„်แ€›ာ แ€žแ€€်แ€žေแ€•ြแ€™ှုแ€ွแ€„် แ€กแ€žုံးแ€•ြုแ€žော แ€กแ€šူแ€กแ€†แ€–ြแ€…်แ€žแ€Š်။

Mathematical Induction แ€žแ€Š် แ€กแ€•ေါแ€„်းแ€€ိแ€”်းแ€•ြแ€Š့် แ€†ိုแ€„်แ€›ာ แ€กแ€†ိုแ€™ျား แ€™ှแ€”်แ€€แ€”်แ€ျแ€€်แ€™ျားแ€€ို แ€žแ€€်แ€žေแ€•ြแ€›ာแ€ွแ€„် แ€™ျားแ€…ွာ แ€กแ€žုံးแ€แ€„်แ€žแ€Š်။

Mathematical Induction แ€€ို แ€กแ€žုံးแ€•ြု၍ แ€žแ€€်แ€žေแ€•ြแ€›ာแ€ွแ€„် แ€กแ€“ိแ€€ แ€กแ€†แ€„့်แ€”ှแ€…်แ€†แ€„့်แ€–ြแ€„့် แ€žแ€€်แ€žေแ€•ြแ€›แ€™แ€Š်။

  • Step (1) - แ€•ေးแ€‘ားแ€žော แ€กแ€†ိုแ€•ြုแ€ျแ€€်แ€แ€…်แ€ုแ€กแ€ွแ€€် แ€•แ€‘แ€™แ€†ုံးแ€กแ€€ြိแ€™် [แ€šျေแ€˜ုแ€šျแ€กားแ€–ြแ€„့် (n=1)] แ€กแ€ွแ€€် แ€™ှแ€”်แ€€แ€”်แ€€ြောแ€„်း แ€žแ€€်แ€žေแ€•ြแ€›แ€™แ€Š်။
  • Step (2) - แ€‘ိုแ€กแ€†ိုแ€žแ€Š် แ€กแ€ြား แ€™แ€Š့်แ€žแ€Š့် $k$ แ€กแ€€ြိแ€™်แ€กแ€ွแ€€်แ€™แ€†ို แ€™ှแ€”်แ€€แ€”်แ€žแ€Š်แ€Ÿု แ€žแ€်แ€™ှแ€်แ€œိုแ€€်แ€•ြီး ၎แ€„်းแ€”ောแ€€်แ€›ှိ แ€€แ€•်แ€œျแ€€်แ€›ှိ‌แ€žော $k+1$ แ€กแ€€ြိแ€™်แ€กแ€ွแ€€် แ€™ှแ€”်แ€€ြောแ€„်း แ€žแ€€်แ€žေแ€•ြแ€”ိုแ€„်แ€œျှแ€„် แ€•ေးแ€‘ားแ€žော แ€กแ€†ိုแ€žแ€Š် แ€กแ€™ြဲแ€™ှแ€”်แ€€แ€”်แ€žแ€Š်။

Principle of Mathematical Induction

Mathematical Induction is a mathematical technique which is used to prove a statement, a formula or a theorem is true for every natural number. The technique involves two steps to prove a statement, as stated below −

Step 1 (Base step) − It proves that a statement is true for the initial value.

Step 2 (Inductive step) − It proves that if the statement is true for the $k^{\text{th}}$ iteration (or number $k$), then it is also true for $(k+1)^{\text{th}}$ iteration ( or number $k+1$).


Terminology of Statements
  • Definition: an explanation of the mathematical meaning of a word.
  • แ€…แ€€ားแ€œုံးแ€แ€…်แ€œုံး၏ แ€žแ€„်္แ€ျာแ€†ိုแ€„်แ€›ာ แ€กแ€“ိแ€•္แ€•ာแ€š်แ€–ွแ€„့်แ€†ိုแ€ျแ€€်แ€€ို Definition แ€Ÿုแ€ေါ်แ€žแ€Š်။

  • Axiom: a basic assumption about a mathematical situation (model) which requires no proof.
  • แ€žแ€€်แ€žေแ€•ြแ€›แ€”်แ€™แ€œိုแ€กแ€•်แ€žော แ€žแ€„်္แ€ျာแ€†ိုแ€„်แ€›ာ แ€šူแ€†แ€ျแ€€်แ€€ို Axiom แ€Ÿုแ€ေါ်แ€žแ€Š်။

  • Theorem: a very important true statement that is provable in terms of definitions and axioms.
  • Definition , Axiom แ€ို့แ€€ို แ€กแ€žုံးแ€•ြု၍ แ€กแ€™ြဲแ€™ှแ€”်แ€€แ€”်แ€€ြောแ€„်း แ€žแ€€်แ€žေแ€•ြแ€”ိုแ€„်แ€žော แ€กแ€›ေးแ€•ါแ€žแ€Š့် แ€žแ€„်္แ€ျာแ€†ိုแ€„်แ€›ာ แ€กแ€†ိုแ€•ြုแ€ျแ€€် แ€–ြแ€…်แ€žแ€Š်။

  • Proposition: a statement of fact that is true and interesting in a given context.
  • แ€•ေးแ€‘ားแ€žော แ€กแ€€ြောแ€„်းแ€กแ€›ာแ€žแ€Š် แ€กแ€™ြဲแ€™ှแ€”်แ€€แ€”်แ€žแ€Š်แ€Ÿု แ€†ိုแ€”ိုแ€„်แ€žော แ€žแ€„်္แ€ျာแ€†ိုแ€„်แ€›ာ แ€กแ€†ိုแ€•ြုแ€ျแ€€် แ€–ြแ€…်แ€žแ€Š်။

  • Lemma: a true statement used in proving other true statements.
  • แ€กแ€ြားแ€กแ€†ိုแ€•ြုแ€ျแ€€်แ€แ€…်แ€ု แ€™ှแ€”်แ€€แ€”်แ€€ြောแ€„်း แ€žแ€€်แ€žေแ€•ြแ€›แ€”် แ€€ိုးแ€€ားแ€กแ€žုံးแ€•ြုแ€žแ€Š့် แ€กแ€™ြဲแ€™ှแ€”်แ€žော แ€กแ€€ြောแ€„်းแ€กแ€›ာ แ€–ြแ€…်แ€žแ€Š်။

  • Corollary: a true statement that is a simple deduction from a theorem or proposition.
  • Theorem แ€”ှแ€„့် Proporsition แ€ို့แ€™ှ แ€†แ€„်းแ€žแ€€်แ€œာแ€žแ€Š့် แ€†แ€„့်แ€•ွားแ€™ှแ€”်แ€€แ€”်แ€ျแ€€် แ€–ြแ€…်แ€žแ€Š်။

  • Proof: the explanation of why a statement is true.
  • แ€กแ€€ြောแ€„်းแ€กแ€›ာแ€แ€…်แ€ု แ€™ှแ€”်แ€€แ€”်แ€€ြောแ€„်း แ€›ှแ€„်းแ€•ြแ€ျแ€€်แ€€ို แ€žแ€€်แ€žေแ€•ြแ€ျแ€€်แ€Ÿု แ€ေါ်แ€žแ€Š်။

  • Conjecture: a statement believed to be true, but for which we have no proof.
  • แ€žแ€€်แ€žေแ€•ြแ€ြแ€„်း แ€™แ€›ှိแ€•ဲ แ€™ှแ€”်แ€€แ€”်แ€žแ€Š်แ€Ÿု แ€šုံแ€€ြแ€Š်แ€‘ားแ€žော แ€กแ€†ိုแ€แ€…်แ€ုแ€€ို Conjecture แ€Ÿုแ€ေါ်แ€žแ€Š်။


Evaluation Steps (Steps of Proof)
  • Step 1: Let $P(n)$ be a result or statement formulated in terms of $n$ (given question).

  • Step 2: Prove that $P(1)$ or $P(\text{initial value})$ is true.

  • Step 3: Assume that $P(k)$ is true.

  • Step 4: Using Step 3 , prove that $P(k+1)$ is true.

  • Step 5: Thus $P(1)$ is true and $P(k+1)$ is true whenever $P(k)$ is true.

  • Conclusion: Hence, by the Principle of Mathematical Induction, $P(n)$ is true for all natural numbers $n$.


Example (1) Use mathematical induction to prove that $1 + 3 + 5 + \ldots + (2n - 1) = n^2$ is true for every positive integer $n$.

Solution

Let $P(n) : 1 + 3 + 5 + \ldots + (2n - 1) = n^2$.

When $n=1$,

LHS $= 1$

RHS $= 1^2 = 1$

$\therefore\ P(1)$ is true.

Assume that $P(k)$ is true for $n=k$.

$\therefore\ P(k) : 1 + 3 + 5 + \ldots + (2k - 1) = k^2 $

When $n=k+1$,

$ \quad 1 + 3 + 5 + \ldots + (2k - 1)+ \left[2(k+1) - 1\right] $

$ = k^2 + (2k + 2 - 1) $

$ = k^2 + 2k + 1 $

$ = (k+1)^2 $

$\therefore\ P(k + 1 )$ is true.

Hence, by the Principle of Mathematical Induction, $P(n)$ is true for all natural numbers $n$.

Example (2) Use mathematical induction to prove that $1 + 2 + 3 + \ldots + n = \dfrac{n(n+1)}{2}$ is true for every $n\in\mathbb{N}$.

Solution

Let $P(n) : 1 + 2 + 3 + \ldots + n = \dfrac{n(n+1)}{2}$

When $n=1$,

LHS $= 1$

RHS $= \dfrac{1(1+1)}{2} = 1$

$\therefore\ P(1)$ is true.

Assume that $P(k)$ is true for $n=k$.

$\therefore\ P(k) : 1 + 2 + 3 + \ldots + k = \dfrac{k(k+1)}{2}$

When $n=k+1$,

$\quad 1 + 2 + 3 + \ldots + k+ (k+1) $

$ = \dfrac{k(k+1)}{2} + (k+1) $

$ = \dfrac{k^2+3k+2}{2} $

$ = \dfrac{k^2+2k+1+ k + 1}{2} $

$ = \dfrac{(k+1)^2+(k + 1)}{2} $

$ = \dfrac{(k + 1) \left[(k+1) + 1\right]}{2} $

$ \therefore\ P(k + 1 )$ is true.

Hence, by the Principle of Mathematical Induction, $P(n)$ is true for all positive integers $n$.

Example (3) Prove that for any positive integer number $n$ , $n^3 + 2 n$ is divisible by $3$ by using mathematical induction.

Solution

Let $P(n) :n^3 + 2 n$ is divisible by $3$

When $n=1$,

$1^3+2(1) = 3$

Since $3$ is divisible by $3$, $P(1)$ is true.

When $n=k,\ P(k) :k^3 + 2 k$ is divisible by $3$.

Assume that $P(k)$ is true.

Let $k^3 + 2 k = 3p$ where $p$ is a positive integer.

When $n=k+1$,

$\quad (k + 1)^3 + 2 (k + 1) $

$ = k^3+3k^2+3k+1+2k+2 $

$ = (k^3+2k)+3k^2+3k+3$

$ = 3p+3(k^2+k+1)$

$ = 3(p+k^2+k+1)$

Since $p$ and $k$ are positive integers, $3(p+k^2+k+1)$ is an integer which is a multiple of $3$.

Therefore, $3(p+k^2+k+1)$ is divisible by $3$.

$\therefore\ P(k + 1 )$ is true.

Hence, by the Principle of Mathematical Induction, $P(n)$ is true for all positive integers $n$.

Example (4) Prove that $n ! > 2 n$ where n is a positive integer and $n\ge 4$.

Solution

Let $P(n) :n ! > 2 n$

When $n=4$,

$4! = 4\times3\times2\times1=24$

$2(4) = 8$

$\therefore\ 24 > 8\Rightarrow 4!>2(4)$ $\therefore\ P(4)$ is true.

Assume that $P(k)$ is true for $n=k$ where $k>4$.

$\therefore\ P(k) :k! > 2k$.

When $n=k+1$,

$(k+1)!= (k+1)k! $

Since $k>4,\ k!> 2$.

$ \therefore\ (k+1)k! > 2(k+1) $

$ \therefore\ (k+1)! > 2(k+1) $

$\therefore\ P(k + 1 )$ is true.

Hence, by the Principle of Mathematical Induction, $P(n)$ is true for all positive integers $n$.

Example (5) For every positive integer $n$, prove that $7^n – 3^n$ is divisible by $4$.

Solution

Let $P(n) :7^n – 3^n$ is divisible by $4$.

When $n=1$,

$7^1 – 3^1=7-3 = 4$

Since $4$ is divisible by $4$, $P(1)$ is true.

When $n=k,\ P(k) :7^k – 3^k$ is divisible by $4$.

Assume that $P(k)$ is true.

Let $7^k – 3^k = 4p$ where $p$ is a positive integer. When $n=k+1$,

$\quad 7^{k+1} – 3^{k+1} $

$ = 7\times7^k – 3\times3^k $

$ = 7\times7^k -(7\times3^k) +(7\times3^k) – (3\times3^k) $

$ = 7(7^k -3^k) +3^k(7 – 3) $

$ = 7(4p) +4\times3^k$

$ = 4 (7p +3^k)$

Since $p$ and $k$ are positive integers, $4 (7p +3^k)$ is an integer which is a multiple of $7$.

Therefore, $4 (7p +3^k)$ is divisible by $7$.

$\therefore\ P(k + 1 )$ is true.

Hence, by the Principle of Mathematical Induction, $P(n)$ is true for all positive integers $n$.

Example (6) Using mathematical induction, show that $9^{n}-8 n-1$ where $n \in\mathbb{N}$ is divisible by $64$.

Solution

Let $P(n) :9^{n}-8 n-1$ where $n \in\mathbb{N}$ is divisible by $64$.

When $n=1$,

$9^{1}-8 (1)-1 = 0$

Since $0$ is divisible by $64$, $P(1)$ is true.

When $n=k,\ P(k) :9^{k}-8 k-1$ is divisible by $64$.

Assume that $P(k)$ is true.

Let $9^{k}-8 k-1 = 64p$ where $p$ is a positive integer.

When $n=k+1$,

$\quad 9^{k+1}-8 (k+1)-1$

$ = 9\times 9^{k}-8k-9 $

$ = 9\times 9^{k}-(9\times 8k)+ (9\times 8k) -8k-9 $

$ = 9\times 9^{k}-(9\times 8k)-9 + (9\times 8k) -8k$

$ = 9(9^{k}-8k-1) + 64k$

$ = 9(64p) + 64k$

$ = 64 (9p +k)$

Since $p$ and $k$ are positive integers, $64 (9p +k)$ is an integer which is a multiple of $64$.

Therefore, $64 (9p +k)$ is divisible by $64$.

$\therefore\ P(k + 1 )$ is true.

Hence, by the Principle of Mathematical Induction, $P(n)$ is true for all positive integers $n$.

Exercises Prove the following by using the principle of mathematical induction for all $n\in\mathbb{N}$:

  1. $1+3+3^{2}+\ldots+3^{n-1}=\dfrac{\left(3^{n}-1\right)}{2}$.

  2. $1^{3}+2^{3}+3^{3}+\ldots+n^{3}=\left(\dfrac{n(n+1)}{2}\right)^{2}$.

  3. $1+\dfrac{1}{(1+2)}+\dfrac{1}{(1+2+3)}+\ldots+\dfrac{1}{(1+2+3+\ldots n)}=\dfrac{2n}{n+1}$

  4. $1\cdot2 \cdot3+2\cdot3 \cdot4+\ldots+n(n+1)(n+2)=\dfrac{n(n+1)(n+2)(n+3)}{4}$.

  5. $1\cdot3+2\cdot3^{2}+3\cdot3^{3}+\ldots+n \cdot3^{n}=\dfrac{(2 n-1) 3^{n+1}+3}{4}$.

  6. $1\cdot2+2\cdot3+3\cdot4+\ldots+n \cdot(n+1)=\left[\dfrac{n(n+1)(n+2)}{3}\right]$.

  7. $1\cdot3+3\cdot5+5\cdot7+\ldots+(2 n-1)(2 n+1)=\dfrac{n\left(4 n^{2}+6 n-1\right)}{3}$.

  8. $1\cdot2+2\cdot2^{2}+3\cdot2^{3}+\ldots+n \cdot 2^{n}=(n-1) 2^{n+1}+2$.

  9. $\dfrac{1}{2}+\dfrac{1}{4}+\dfrac{1}{8}+\ldots+\dfrac{1}{2^{n}}=1-\dfrac{1}{2^{n}}$.

  10. $\dfrac{1}{2\cdot5}+\dfrac{1}{5\cdot8}+\dfrac{1}{8\cdot11}+\ldots+\dfrac{1}{(3 n-1)(3 n+2)}=\dfrac{n}{(6 n+4)}$

  11. $\dfrac{1}{1\cdot2 \cdot3}+\dfrac{1}{2\cdot3 \cdot4}+\dfrac{1}{3\cdot4 \cdot5}+\ldots+\dfrac{1}{n(n+1)(n+2)}=\dfrac{n(n+3)}{4(n+1)(n+2)}$.

  12. $a+a r+a r^{2}+\ldots+a r^{n-1}=\dfrac{a\left(r^{n}-1\right)}{r-1}$.

  13. $\left(1+\dfrac{3}{1}\right)\left(1+\dfrac{5}{4}\right)\left(1+\dfrac{7}{9}\right) \ldots\left(1+\dfrac{(2 n+1)}{n^{2}}\right)=(n+1)^{2}$.

  14. $\left(1+\dfrac{1}{1}\right)\left(1+\dfrac{1}{2}\right)\left(1+\dfrac{1}{3}\right) \ldots\left(1+\dfrac{1}{n}\right)=(n+1)$.

  15. $1^{2}+3^{2}+5^{2}+\ldots+(2 n-1)^{2}=\dfrac{n(2 n-1)(2 n+1)}{3}$.

  16. $\dfrac{1}{1\cdot4}+\dfrac{1}{4\cdot7}+\dfrac{1}{7\cdot10}+\ldots+\dfrac{1}{(3 n-2)(3 n+1)}=\dfrac{n}{(3 n+1)}$.

  17. $\dfrac{1}{3\cdot5}+\dfrac{1}{5\cdot7}+\dfrac{1}{7\cdot9}+\ldots+\dfrac{1}{(2 n+1)(2 n+3)}=\dfrac{n}{3(2 n+3)}$.

  18. $1+2+3+\ldots+n<\dfrac{1}{8}(2 n+1)^{2}$.

  19. $n(n+1)(n+5)$ is a multiple of $3$ .

  20. $10^{2 n-1}+1$ is divisible by $11$ .

  21. $x^{2 n}-y^{2 n}$ is divisible by $x+y$.

  22. $3^{2 n+2}-8 n-9$ is divisible by $8$ .

  23. $41^{n}-14^{n}$ is a multiple of $27$ .

  24. $(2 n+7)< (n+3)^{2}$.