Skip to main content
Global

1.6: Mali ya Hesabu halisi

  1. Last updated
  2. Save as PDF
  • Page ID
    176162

    • \( \newcommand{\vecs}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} } \) \( \newcommand{\vecd}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash {#1}}} \)\(\newcommand{\id}{\mathrm{id}}\) \( \newcommand{\Span}{\mathrm{span}}\) \( \newcommand{\kernel}{\mathrm{null}\,}\) \( \newcommand{\range}{\mathrm{range}\,}\) \( \newcommand{\RealPart}{\mathrm{Re}}\) \( \newcommand{\ImaginaryPart}{\mathrm{Im}}\) \( \newcommand{\Argument}{\mathrm{Arg}}\) \( \newcommand{\norm}[1]{\| #1 \|}\) \( \newcommand{\inner}[2]{\langle #1, #2 \rangle}\) \( \newcommand{\Span}{\mathrm{span}}\) \(\newcommand{\id}{\mathrm{id}}\) \( \newcommand{\Span}{\mathrm{span}}\) \( \newcommand{\kernel}{\mathrm{null}\,}\) \( \newcommand{\range}{\mathrm{range}\,}\) \( \newcommand{\RealPart}{\mathrm{Re}}\) \( \newcommand{\ImaginaryPart}{\mathrm{Im}}\) \( \newcommand{\Argument}{\mathrm{Arg}}\) \( \newcommand{\norm}[1]{\| #1 \|}\) \( \newcommand{\inner}[2]{\langle #1, #2 \rangle}\) \( \newcommand{\Span}{\mathrm{span}}\)\(\newcommand{\AA}{\unicode[.8,0]{x212B}}\)

    Malengo ya kujifunza

    Mwishoni mwa sehemu hii, utaweza:

    • Tumia mali za kubadilisha na za ushirika
    • Tumia mali ya utambulisho, inverse, na sifuri
    • Kurahisisha maneno kwa kutumia Mali ya Mgawanyo

    Tumia Mali za Comutative na Associative

    Ili sisi kuongeza namba mbili haiathiri matokeo. Kama sisi kuongeza\(8+9\) au\(9+8\), matokeo ni sawa-wote wawili sawa 17. Kwa hiyo,\(8+9=9+8\). Utaratibu ambao tunaongeza haijalishi!

    Vile vile, wakati wa kuzidisha namba mbili, utaratibu hauathiri matokeo. Kama sisi\(8·9\) kuzidisha\(9·8\) au matokeo ni sawa-wote wawili sawa 72. Kwa hiyo,\(9·8=8·9\). Utaratibu ambao tunazidisha haijalishi! Mifano hii inaonyesha Mali Comutative.

    MALI YA KUBADILISHA

    \[\begin{array}{lll} \textbf{of Addition} & \text{If }a \text{ and }b \text{are real numbers, then} & a+b=b+a. \\ \textbf{of Multiplication} & \text{If }a \text{ and }b \text{are real numbers, then} & a·b=b·a. \end{array} \]

    Wakati wa kuongeza au kuzidisha, kubadilisha utaratibu hutoa matokeo sawa.

    Mali Comutative ina nini na utaratibu. Sisi Ondoa\(9−8\) na\(8−9\), na kuona kwamba\(9−8\neq 8−9\). Kwa kuwa kubadilisha utaratibu wa uondoaji haitoi matokeo sawa, tunajua kwamba uondoaji sio kubadilisha.

    Idara si commutative ama. Tangu\(12÷3\neq 3÷12\), kubadilisha utaratibu wa mgawanyiko hakutoa matokeo sawa. Mali ya kubadilisha hutumika tu kwa kuongeza na kuzidisha!

    • Kuongezea na kuzidisha ni commutative.
    • Kutoa na mgawanyiko sio kubadilisha.

    Wakati wa kuongeza namba tatu, kubadilisha kikundi cha nambari hutoa matokeo sawa. Kwa mfano,\((7+8)+2=7+(8+2)\), tangu kila upande wa equation sawa 17.

    Hii ni kweli kwa kuzidisha, pia. Kwa mfano,\(\left(5·\frac{1}{3}\right)·3=5·\left(\frac{1}{3}·3\right)\), tangu kila upande wa equation sawa 5.

    Mifano hii inaonyesha Mali Associative.

    MALI ASSOCIATIVE

    \[\begin{array}{lll} \textbf{of Addition} & \text{If }a,b, \text{ and }c \text{ are real numbers, then} & (a+b)+c=a+(b+c). \\ \textbf{of Multiplication} & \text{If }a,b,\text{ and }c \text{ are real numbers, then} & (a·b)·c=a·(b·c). \end{array} \]

    Wakati wa kuongeza au kuzidisha, kubadilisha kikundi hutoa matokeo sawa.

    Mali ya Associative inahusiana na kikundi. Ikiwa tunabadilisha jinsi namba zimeunganishwa, matokeo yatakuwa sawa. Taarifa ni sawa namba tatu katika utaratibu sawa-tofauti tu ni kambi.

    Tuliona kwamba uondoaji na mgawanyiko haukuwa wa kubadilisha. Wao si associative aidha.

    \[\begin{array}{cc} (10−3)−2\neq 10−(3−2) & (24÷4)÷2\neq 24÷(4÷2) \\ 7−2\neq 10−1 & 6÷2\neq 24÷2 \\ 5\neq 9 & 3\neq 12 \end{array}\]

    Wakati wa kurahisisha maneno, daima ni wazo nzuri kupanga mipango gani itakuwa. Ili kuchanganya maneno kama hayo katika mfano unaofuata, tutatumia Mali ya Kubadilisha ya kuongeza kuandika maneno kama hayo pamoja.

    Mfano\(\PageIndex{1}\)

    Kurahisisha:\(18p+6q+15p+5q\).

    Jibu

    \[\begin{array}{lc} \text{} & 18p+6q+15p+5q \\ \text{Use the Commutative Property of addition to} & 18p+15p+6q+5q \\ \text{reorder so that like terms are together.} & {} \\ \text{Add like terms.} & 33p+11q \end{array}\]

    Mfano\(\PageIndex{2}\)

    Kurahisisha:\(23r+14s+9r+15s\).

    Jibu

    \(32r+29s\)

    Mfano\(\PageIndex{3}\)

    Kurahisisha:\(37m+21n+4m−15n\).

    Jibu

    \(41m+6n\)

    Tunapopaswa kurahisisha maneno ya algebraic, mara nyingi tunaweza kufanya kazi iwe rahisi kwa kutumia Mali ya Commutative au Mali Associative kwanza.

    MFANO\(\PageIndex{4}\)

    Kurahisisha:\((\frac{5}{13}+\frac{3}{4})+\frac{1}{4}\).

    Jibu

    \( \begin{array}{lc} \text{} & (\frac{5}{13}+\frac{3}{4})+\frac{1}{4} \\ {\text{Notice that the last 2 terms have a common} \\ \text{denominator, so change the grouping.} } & \frac{5}{13}+(\frac{3}{4}+\frac{1}{4}) \\ \text{Add in parentheses first.} & \frac{5}{13}+(\frac{4}{4}) \\ \text{Simplify the fraction.} & \frac{5}{13}+1 \\ \text{Add.} & 1\frac{5}{13} \\ \text{Convert to an improper fraction.} & \frac{18}{13} \end{array}\)

    MFANO\(\PageIndex{5}\)

    Kurahisisha:\((\frac{7}{15}+\frac{5}{8})+\frac{3}{8}.\)

    Jibu

    \(1 \frac{7}{15}\)

    MFANO\(\PageIndex{6}\)

    Kurahisisha:\((\frac{2}{9}+\frac{7}{12})+\frac{5}{12}\).

    Jibu

    \(1\frac{2}{9}\)

    Tumia Mali ya Identity, Inverse, na Zero

    Nini kinatokea wakati sisi kuongeza 0 kwa idadi yoyote? Kuongeza 0 haina mabadiliko ya thamani. Kwa sababu hii, tunaita 0 utambulisho wa kuongezea. Identity Mali ya Aidha kwamba inasema kwamba kwa idadi yoyote halisi\(a,a+0=a\) na\(0+a=a.\)

    Nini kinatokea wakati sisi kuzidisha idadi yoyote kwa moja? Kuongezeka kwa 1 haina mabadiliko ya thamani. Hivyo tunaita 1 utambulisho multiplicative. Mali ya Utambulisho wa Kuzidisha ambayo inasema kwamba kwa idadi yoyote halisi\(a,a·1=a\) na\(1⋅a=a.\)

    Sisi muhtasari Mali Identity hapa.

    UTAMBULISHO MALI

    \[\begin{array}{ll} \textbf{of Addition} \text{ For any real number }a:a+0=a & 0+a=a \\ \\ \\ \textbf{0} \text{ is the } \textbf{additive identity} \\ \textbf{of Multiplication} \text{ For any real number } a:a·1=a & 1·a=a \\ \\ \\ \textbf{1} \text{ is the } \textbf{multiplicative identity} \end{array}\]

    Nini idadi aliongeza kwa 5 anatoa livsmedelstillsats utambulisho, 0? Tunajua

    alt

    Nambari ya kukosa ilikuwa kinyume cha idadi!

    Sisi wito\(−a\) inverse livsmedelstillsats ya\(a\). Kinyume cha nambari ni inverse yake ya kuongezea. Nambari na kinyume chake huongeza sifuri, ambayo ni utambulisho wa kuongezea. Hii inasababisha Mali Inverse ya Aidha kwamba inasema kwa idadi yoyote halisi\(a,a+(−a)=0.\)

    Nini idadi tele na\(\frac{2}{3}\) anatoa utambulisho multiplicative, 1? Kwa maneno mengine,\(\frac{2}{3}\) mara nini matokeo katika 1? Tunajua

    alt

    Nambari ya kukosa ilikuwa ya kawaida ya idadi!

    Tunatoa\(\frac{1}{a}\) wito inverse multiplicative ya. Utoaji wa nambari ni inverse yake ya kuzidisha. Hii inasababisha Mali Inverse ya Kuzidisha ambayo inasema kwamba kwa idadi yoyote halisi\(a,a\neq 0,a·\frac{1}{a}=1.\)

    Tutaweza rasmi hali mali inverse hapa.

    MALI YA INVERSE

    \[\begin{array}{lc} \textbf{of addition} \text{For any real number }a, & a+(−a)=0 \\ \;\;\;\; −a \text{ is the } \textbf{additive inverse }\text{ of }a & {} \\ \;\;\;\; \text{A number and its } \textit{opposite } \text{add to zero.} \\ \\ \\ \textbf{of multiplication } \text{For any real number }a,a\neq 0 & a·\dfrac{1}{a}=1 \\ \;\;\;\;\;\dfrac{1}{a} \text{ is the } \textbf{multiplicative inverse} \text{ of }a \\ \;\;\;\; \text{A number and its } \textit{reciprocal} \text{ multiply to one.} \end{array}\]

    Identity Mali ya kuongeza anasema kwamba wakati sisi kuongeza 0 kwa idadi yoyote, matokeo ni kwamba idadi sawa. Nini kinatokea wakati sisi kuzidisha idadi kwa 0? Kuongezeka kwa 0 hufanya bidhaa sawa na sifuri.

    Nini kuhusu mgawanyiko kuwashirikisha sifuri? Ni nini\(0÷3\)? Fikiria juu ya mfano halisi: Ikiwa hakuna cookies katika jar ya kuki na watu 3 watawashirikisha, ni vidakuzi ngapi ambavyo kila mtu hupata? Hakuna vidakuzi vya kushiriki, hivyo kila mtu anapata cookies 0. Hivyo,\(0÷3=0.\)

    Tunaweza kuangalia mgawanyiko na kuhusiana kuzidisha ukweli. Hivyo tunajua\(0÷3=0\) kwa sababu\(0·3=0\).

    Sasa fikiria juu ya kugawa na sifuri. Matokeo ya kugawa 4 na 0 ni nini? Fikiria juu ya ukweli unaohusiana na kuzidisha:

    alt

    Je, kuna idadi kwamba tele kwa 0 anatoa 4? Kwa kuwa idadi yoyote halisi tele kwa 0 anatoa 0, hakuna idadi halisi ambayo inaweza kuzidishwa na 0 kupata 4. Tunahitimisha kuwa hakuna jibu kwa\(4÷0\) na hivyo tunasema kuwa mgawanyiko na 0 haujafafanuliwa.

    Sisi muhtasari mali ya sifuri hapa.

    MALI YA SIFURI

    Kuzidisha na Zero: Kwa yoyote ya kweli idadi a,

    \[a⋅0=0 \; \; \; 0⋅a=0 \; \; \; \; \text{The product of any number and 0 is 0.}\]

    Idara na Zero: Kwa yoyote ya kweli idadi a,\(a\neq 0\)

    \[\begin{array}{cl} \dfrac{0}{a}=0 & \text{Zero divided by any real number, except itself, is zero.} \\ \dfrac{a}{0} \text{ is undefined} & \text{Division by zero is undefined.} \end{array}\]

    Sasa tutatumia kutumia mali ya utambulisho, inverses, na sifuri ili kurahisisha maneno.

    MFANO\(\PageIndex{7}\)

    Kurahisisha:\(−84n+(−73n)+84n.\)

    Jibu

    \(\begin{array}{lc} \text{} & −84n+(−73n)+84n \\ \text{Notice that the first and third terms are} \\ \text{opposites; use the Commutative Property of} & −84n+84n+(−73n) \\ \text{addition to re-order the terms.} \\ \text{Add left to right.} & 0+(−73n) \\ \text{Add.} & −73n \end{array}\)

    MFANO\(\PageIndex{8}\)

    Kurahisisha:\(−27a+(−48a)+27a\).

    Jibu

    \(−48a\)

    MFANO\(\PageIndex{9}\)

    Kurahisisha:\(39x+(−92x)+(−39x)\).

    Jibu

    \(−92x\)

    Sasa tutaona jinsi kutambua usawa ni muhimu. Kabla ya kuzidisha kushoto kwenda kulia, angalia kurudisha-bidhaa zao ni 1.

    MFANO\(\PageIndex{10}\)

    Kurahisisha:\(\frac{7}{15}⋅\frac{8}{23}⋅\frac{15}{7}\).

    Jibu

    \(\begin{array}{lc} \text{} & \frac{7}{15}⋅\frac{8}{23}⋅\frac{15}{7} \\ \text{Notice the first and third terms} \\ {\text{are reciprocals, so use the Commutative} \\ \text{Property of multiplication to re-order the} \\ \text{factors.}} & \frac{7}{15}·\frac{15}{7}·\frac{8}{23} \\ \text{Multiply left to right.} & 1·\frac{8}{23} \\ \text{Multiply.} & \frac{8}{23} \end{array}\)

    MFANO\(\PageIndex{11}\)

    Kurahisisha:\(\frac{9}{16}⋅\frac{5}{49}⋅\frac{16}{9}\).

    Jibu

    \(\frac{5}{49}\)

    Kurahisisha:\(\frac{6}{17}⋅\frac{11}{25}⋅\frac{17}{6}\).

    Jibu

    \(\frac{11}{25}\)

    Mfano unaofuata unatufanya tufahamu wa tofauti kati ya kugawa 0 kwa idadi fulani au idadi fulani ikigawanywa na 0.

    Kurahisisha: a.\(\frac{0}{n+5}\), ambapo\(n\neq −5\) b.\(\frac{10−3p}{0}\) wapi\(10−3p\neq 0.\)

    Jibu

    a.

    \(\begin{array}{lc} {} & \dfrac{0}{n+5} \\ \text{Zero divided by any real number except itself is 0.} & 0 \end{array}\)

    b.

    \(\begin{array}{lc} {} & \dfrac{10−3p}{0} \\ \text{Division by 0 is undefined.} & \text{undefined} \end{array}\)

    MFANO\(\PageIndex{14}\)

    Kurahisisha: a.\(\frac{0}{m+7}\), ambapo\(m\neq −7\) b.\(\frac{18−6c}{0}\), wapi\(18−6c\neq 0\).

    Jibu

    a. 0 b.
    haijulikani

    MFANO\(\PageIndex{15}\)

    Kurahisisha: a.\(\frac{0}{d−4}\), ambapo\(d\neq 4\) b.\(\frac{15−4q}{0}\), wapi\(15−4q\neq 0\).

    Jibu

    a. 0 b.
    haijulikani

    Kurahisisha Maneno Kutumia Mali ya Usambazaji

    Tuseme kwamba marafiki watatu wanaenda kwenye sinema. Kila mmoja anahitaji $9.25 - hiyo ni dola 9 na robo-1 kulipa tiketi zao. Ni kiasi gani cha fedha wanahitaji wote pamoja?

    Unaweza kufikiri juu ya dola tofauti na robo. Wanahitaji mara 3 $9 hivyo $27 na 3 mara 1 robo, hivyo senti 75. Kwa jumla, wanahitaji $27.75. Ikiwa unafikiri juu ya kufanya hesabu kwa njia hii, unatumia Mali ya Usambazaji.

    MALI YA KUSAMBAZA

    \(\begin{array}{lc} \text{If }a,b \text{,and }c \text{are real numbers, then} \; \; \; \; \; & a(b+c)=ab+ac \\ {} & (b+c)a=ba+ca \\ {} & a(b−c)=ab−ac \\{} & (b−c)a=ba−ca \end{array}\)

    Katika algebra, tunatumia Mali Distributive kuondoa mabano kama sisi kurahisisha maneno.

    MFANO\(\PageIndex{16}\)

    Kurahisisha:\(3(x+4)\).

    Jibu

    \(\begin{array} {} & 3(x+4) \\ \text{Distribute.} \; \; \; \; \; \; \; \; & 3·x+3·4 \\ \text{Multiply.} & 3x+12 \end{array}\)

    Kurahisisha:\(4(x+2)\).

    Jibu

    \(4x8\)

    MFANO\(\PageIndex{18}\)

    Kurahisisha:\(6(x+7)\).

    Jibu

    \(6x42\)

    Wanafunzi wengine wanaona ni muhimu kuteka mishale kuwakumbusha jinsi ya kutumia Mali ya Usambazaji. Kisha hatua ya kwanza katika Mfano ingeonekana kama hii:

    alt

    MFANO\(\PageIndex{19}\)

    Kurahisisha:\(8(\frac{3}{8}x+\frac{1}{4})\).

    Jibu
      alt
    Kusambaza. alt
    Kuzidisha. alt
    MFANO\(\PageIndex{20}\)

    Kurahisisha:\(6(\frac{5}{6}y+\frac{1}{2})\).

    Jibu

    \(5y+3\)

    MFANO\(\PageIndex{21}\)

    Kurahisisha:\(12(\frac{1}{3}n+\frac{3}{4})\)

    Jibu

    \(4n+9\)

    Kutumia Mali ya Usambazaji kama inavyoonekana katika mfano unaofuata itakuwa muhimu sana wakati tunatatua maombi ya fedha katika sura za baadaye.

    MFANO\(\PageIndex{22}\)

    Kurahisisha:\(100(0.3+0.25q)\).

    Jibu
      alt
    Kusambaza. alt
    Kuzidisha. alt
    MFANO\(\PageIndex{23}\)

    Kurahisisha:\(100(0.7+0.15p).\)

    Jibu

    \(70+15p\)

    MFANO\(\PageIndex{24}\)

    Kurahisisha:\(100(0.04+0.35d)\).

    Jibu

    \(4+35d\)

    Tunaposambaza namba hasi, tunahitaji kuwa makini zaidi ili kupata ishara sahihi!

    MFANO\(\PageIndex{25}\)

    Kurahisisha:\(−11(4−3a).\)

    Jibu

    \(\begin{array}{lc} {} & −11(4−3a) \\ \text{Distribute. } \; \; \; \; \; \; \; \; \; \;& −11·4−(−11)·3a \\ \text{Multiply.} & −44−(−33a) \\ \text{Simplify.} & −44+33a \end{array}\)

    Kumbuka kwamba unaweza pia kuandika matokeo kama\(33a−44.\) Unajua kwa nini?

    Kurahisisha:\(−5(2−3a)\).

    Jibu

    \(−10+15a\)

    MFANO\(\PageIndex{27}\)

    Kurahisisha:\(−7(8−15y).\)

    Jibu

    \(−56+105y\)

    Katika mfano unaofuata, tutaonyesha jinsi ya kutumia Mali ya Mgawanyo ili kupata kinyume cha maneno.

    Kurahisisha:\(−(y+5)\).

    Jibu

    \(\begin{array}{lc} {} & −(y+5) \\ \text{Multiplying by }−1 \text{ results in the opposite.}& −1(y+5) \\ \text{Distribute.} & −1·y+(−1)·5 \\ \text{Simplify.} & −y+(−5) \\ \text{Simplify.} & −y−5 \end{array} \)

    MFANO\(\PageIndex{29}\)

    Kurahisisha:\(−(z−11)\).

    Jibu

    \(−z+11\)

    MFANO\(\PageIndex{30}\)

    Kurahisisha:\(−(x−4)\).

    Jibu

    \(−x+4\)

    Kutakuwa na nyakati ambapo tutahitaji kutumia Mali Distributive kama sehemu ya utaratibu wa shughuli. Anza kwa kuangalia mabano. Ikiwa maneno ndani ya mabano hayawezi kurahisishwa, hatua inayofuata itakuwa kuzidisha kwa kutumia Mali ya Mgawanyo, ambayo huondoa mabano. Mifano miwili ijayo itaonyesha hili.

    MFANO\(\PageIndex{31}\)

    Kurahisisha:\(8−2(x+3)\)

    Jibu

    Tunafuata utaratibu wa shughuli. Kuzidisha huja kabla ya kuondoa, kwa hiyo tutasambaza 2 kwanza na kisha tuondoe.

    \(\begin{array}{lc} {} & \text{8−2(x+3)} \\ \text{Distribute.} & 8−2·x−2·3 \\ \text{Multiply.} & 8−2x−6 \\ \text{Combine like terms.} &−2x+2 \end{array}\)

    MFANO\(\PageIndex{32}\)

    Kurahisisha:\(9−3(x+2)\).

    Jibu

    \(3−3x\)

    MFANO\(\PageIndex{33}\)

    Kurahisisha:\(7x−5(x+4)\).

    Jibu

    \(2x−20\)

    MFANO\(\PageIndex{34}\)

    Kurahisisha:\(4(x−8)−(x+3)\).

    Jibu

    \(\begin{array}{lc} {} & 4(x−8)−(x+3) \\ \text{Distribute.} & 4x−32−x−3 \\ \text{Combine like terms.} & 3x−35 \end{array}\)

    MFANO\(\PageIndex{35}\)

    Kurahisisha:\(6(x−9)−(x+12)\).

    Jibu

    \(5x−66\)

    MFANO\(\PageIndex{36}\)

    Kurahisisha:\(8(x−1)−(x+5)\).

    Jibu

    \(7x−13\)

    Mali yote ya namba halisi tumetumia katika sura hii ni muhtasari hapa.

    Comutative Mali

    Wakati wa kuongeza au kuzidisha, kubadilisha utaratibu hutoa matokeo sawa

    \[\begin{array}{lll} \textbf{of Addition} & \text{If }a \text{ and }b \text{are real numbers, then} & a+b=b+a. \\ \textbf{of Multiplication} & \text{If }a \text{ and }b \text{are real numbers, then} & a·b=b·a. \end{array} \]
    Associative Mali

    Wakati wa kuongeza au kuzidisha, kubadilisha kikundi hutoa matokeo sawa.

    \[\begin{array}{lll} \textbf{of Addition} & \text{If }a,b, \text{ and }c \text{ are real numbers, then} & (a+b)+c=a+(b+c). \\ \textbf{of Multiplication} & \text{If }a,b,\text{ and }c \text{ are real numbers, then} & (a·b)·c=a·(b·c). \end{array} \]
    Mali ya Kusambaza

    \[\begin{array}{lc} \text{If }a,b \text{,and }c \text{are real numbers, then} \; \; \; \; \; & a(b+c)=ab+ac \\ {} & (b+c)a=ba+ca \\ {} & a(b−c)=ab−ac \\{} & (b−c)a=ba−ca \end{array}\]
    Mali ya Identity
    \[\begin{array}{ll} \textbf{of Addition} \text{ For any real number }a:a+0=a & 0+a=a \\ \;\;\;\; \textbf{0} \text{ is the } \textbf{additive identity} \\ \textbf{of Multiplication} \text{ For any real number } a:a·1=a & 1·a=a \\ \;\;\;\; \textbf{1} \text{ is the } \textbf{multiplicative identity} \end{array}\]
    Inverse Mali

    \[\begin{array}{lc} \textbf{of addition } \text{For any real number }a, & a+(−a)=0 \\ \;\;\;\; −a \text{ is the } \textbf{additive inverse }\text{ of }a & {} \\ \;\;\;\; \text{A number and its } \textit{opposite } \text{add to zero.} \\ \\ \\ \textbf{of multiplication } \text{For any real number }a,a\neq 0 & a·\dfrac{1}{a}=1 \\ \;\;\;\;\;\dfrac{1}{a} \text{ is the } \textbf{multiplicative inverse} \text{ of }a \\ \;\;\;\; \text{A number and its } \textit{reciprocal} \text{ multiply to one.} \end{array}\]
    Mali ya Zero
    \[\begin{array}{lc} \text{For any real number }a, & a·0=0 \\ {} & 0·a=0 \\ \text{For any real number }a,a\neq 0, & \dfrac{0}{a}=0 \\ \text{For any real number }a, & \dfrac{a}{0} \text{ is undefined} \end{array}\]

    Dhana muhimu

    Mali ya kubadilisha
    Wakati wa kuongeza au kuzidisha, kubadilisha utaratibu hutoa matokeo sawa

    \[\begin{array}{lll} \textbf{of Addition} & \text{If }a \text{ and }b \text{are real numbers, then} & a+b=b+a. \\ \textbf{of Multiplication} & \text{If }a \text{ and }b \text{are real numbers, then} & a·b=b·a. \end{array} \]
    Mali ya Associative Wakati wa kuongeza au kuzidisha, kubadilisha kikundi hutoa matokeo sawa. \[\begin{array}{lll} \textbf{of Addition} & \text{If }a,b, \text{ and }c \text{ are real numbers, then} & (a+b)+c=a+(b+c). \\ \textbf{of Multiplication} & \text{If }a,b,\text{ and }c \text{ are real numbers, then} & (a·b)·c=a·(b·c). \end{array} \]
    Mali ya Kusambaza

    \[\begin{array}{lc} \text{If }a,b \text{,and }c \text{are real numbers, then} \; \; \; \; \; & a(b+c)=ab+ac \\ {} & (b+c)a=ba+ca \\ {} & a(b−c)=ab−ac \\{} & (b−c)a=ba−ca \end{array}\]
    Mali ya Identity

    \[\begin{array}{ll} \textbf{of Addition} \text{ For any real number }a:a+0=a & 0+a=a \\ \;\;\;\; \textbf{0} \text{ is the } \textbf{additive identity} \\ \textbf{of Multiplication} \text{ For any real number } a:a·1=a & 1·a=a \\ \;\;\;\; \textbf{1} \text{ is the } \textbf{multiplicative identity} \end{array}\]
    Inverse Mali

    \[\begin{array}{lc} \textbf{of addition} \text{For any real number }a, & a+(−a)=0 \\ \;\;\;\; −a \text{ is the } \textbf{additive inverse }\text{ of }a & {} \\ \;\;\;\; \text{A number and its } \textit{opposite } \text{add to zero.} \\ \\ \\ \textbf{of multiplication } \text{For any real number }a,a\neq 0 & a·\dfrac{1}{a}=1 \\ \;\;\;\;\;\dfrac{1}{a} \text{ is the } \textbf{multiplicative inverse} \text{ of }a \\ \;\;\;\; \text{A number and its } \textit{reciprocal} \text{ multiply to one.} \end{array}\]
    Mali ya Zero

    \[\begin{array}{lc} \text{For any real number }a, & a·0=0 \\ {} & 0·a=0 \\ \text{For any real number }a,a\neq 0, & \dfrac{0}{a}=0 \\ \text{For any real number }a, & \dfrac{a}{0} \text{ is undefined} \end{array}\]

    faharasa

    utambulisho wa nyongeza
    Nambari 0 ni utambulisho wa kuongezea kwa sababu kuongeza 0 kwa nambari yoyote haibadili thamani yake.
    nyongeza inverse
    Kinyume cha nambari ni inverse yake ya kuongezea.
    utambulisho wa kuzidisha
    Nambari ya 1 ni utambulisho wa kuzidisha kwa sababu kuzidisha 1 kwa namba yoyote hakubadilisha thamani yake.
    inverse ya kuzidisha
    Utoaji wa nambari ni inverse yake ya kuzidisha.