Translation examples
noun
• yhteen-ja vähennyslaskua, kertolasku, jakolasku.
• Addition, Subtraction, Multiplication, Division.
Suorita arvoille vähennyslasku Excelissä miinusmerkin (-) avulla.
Subtract values in Excel using the minus sign (-).
Probleemat 21–23 ovat täydennysprobleemoja, jota nykyisin ratkaistaisiin yksinkertaisesti murtolukujen vähennyslaskulla.
Problems 21–23 are problems in completion, which in modern notation are simply subtraction problems.
Lukusuoraa käytetään apuvälineenä yhteen- ja vähennyslaskun opetuksessa, etenkin negatiivisten lukujen osalta.
It is often used as an aid in teaching simple addition and subtraction, especially involving negative numbers.
Helpoimmillaan esimerkiksi 2×3-tapauksessa (vaikeusaste 0) ne sopivat esimerkiksi koululaisille yhteen- ja vähennyslaskun harjoitustehtäviksi.
In their simplest form, as in the following 2 × 3 case (degree of difficulty 0) Survo puzzles are suitable exercises in addition and subtraction.
Niinpä riittää todeta, että jos e i G 1 ⋅ R = 1 {\displaystyle e^{i\mathbf {G} _{1}\cdot \mathbf {R} }=1} ja e i G 2 ⋅ R = 1 {\displaystyle e^{i\mathbf {G} _{2}\cdot \mathbf {R} }=1} summa ja erotus G 1 ± G 2 {\displaystyle \mathbf {G} _{1}\pm \mathbf {G} _{2}} toteuttavat saman yhtälön. e i ( G 1 + G 2 ) R = ( e i G 1 ⋅ R ) ( e i G 2 ⋅ R ) = ( 1 ) ( 1 ) = 1 e i ( G 1 − G 2 ) ⋅ R = e i G 1 ⋅ R e i G 2 ⋅ R = 1 {\displaystyle {\begin{aligned}e^{i\left(\mathbf {G} _{1}+\mathbf {G} _{2}\right)\mathbf {R} }&=\left(e^{i\mathbf {G} _{1}\cdot \mathbf {R} }\right)\left(e^{i\mathbf {G} _{2}\cdot \mathbf {R} }\right)=\left(1\right)\left(1\right)=1\\e^{i\left(\mathbf {G} _{1}-\mathbf {G} _{2}\right)\cdot \mathbf {R} }&={\frac {e^{i\mathbf {G} _{1}\cdot \mathbf {R} }}{e^{i\mathbf {G} _{2}\cdot \mathbf {R} }}}=1\end{aligned}}} Tämä osoittaa, että käänteishilakin on sulkettu vektorien yhteen- ja vähennyslaskun suhteen, toisin sanoen kahden käänteishilaan kuuluvan vektorin summa ja erotus kuuluvat siihen myös.
Thus it is sufficient to say that if we have e i G 1 ⋅ R = 1 {\displaystyle e^{i\mathbf {G} _{1}\cdot \mathbf {R} }=1} and e i G 2 ⋅ R = 1 {\displaystyle e^{i\mathbf {G} _{2}\cdot \mathbf {R} }=1} then the sum and difference G 1 ± G 2 {\displaystyle \mathbf {G} _{1}\pm \mathbf {G} _{2}} satisfy the same. e i ( G 1 + G 2 ) R = ( e i G 1 ⋅ R ) ( e i G 2 ⋅ R ) = ( 1 ) ( 1 ) = 1 e i ( G 1 − G 2 ) ⋅ R = e i G 1 ⋅ R e i G 2 ⋅ R = 1 {\displaystyle {\begin{aligned}e^{i\left(\mathbf {G} _{1}+\mathbf {G} _{2}\right)\mathbf {R} }&=\left(e^{i\mathbf {G} _{1}\cdot \mathbf {R} }\right)\left(e^{i\mathbf {G} _{2}\cdot \mathbf {R} }\right)=\left(1\right)\left(1\right)=1\\e^{i\left(\mathbf {G} _{1}-\mathbf {G} _{2}\right)\cdot \mathbf {R} }&={\frac {e^{i\mathbf {G} _{1}\cdot \mathbf {R} }}{e^{i\mathbf {G} _{2}\cdot \mathbf {R} }}}=1\end{aligned}}} Thus we have shown the reciprocal lattice is closed under vector addition and subtraction.
Niitä voidaan laskea yhteen tavalliseen tapaan: A + B = ( A 0 , A 1 , A 2 , A 3 ) + ( B 0 , B 1 , B 2 , B 3 ) = ( A 0 + B 0 , A 1 + B 1 , A 2 + B 2 , A 3 + B 3 ) {\displaystyle \mathbf {A} +\mathbf {B} =(A^{0},A^{1},A^{2},A^{3})+(B^{0},B^{1},B^{2},B^{3})=(A^{0}+B^{0},A^{1}+B^{1},A^{2}+B^{2},A^{3}+B^{3})} ja ne voidaan kertoa skalaarilla λ komponenteittain: λ A = λ ( A 0 , A 1 , A 2 , A 3 ) = ( λ A 0 , λ A 1 , λ A 2 , λ A 3 ) {\displaystyle \lambda \mathbf {A} =\lambda (A^{0},A^{1},A^{2},A^{3})=(\lambda A^{0},\lambda A^{1},\lambda A^{2},\lambda A^{3})} Samoin vähennyslasku on nelivektoreillakin yhteenlaskun käänteistoimitus, joka määritellään komponenteittain: A + ( − 1 ) B = ( A 0 , A 1 , A 2 , A 3 ) + ( − 1 ) ( B 0 , B 1 , B 2 , B 3 ) = ( A 0 − B 0 , A 1 − B 1 , A 2 − B 2 , A 3 − B 3 ) {\displaystyle \mathbf {A} +(-1)\mathbf {B} =(A^{0},A^{1},A^{2},A^{3})+(-1)(B^{0},B^{1},B^{2},B^{3})=(A^{0}-B^{0},A^{1}-B^{1},A^{2}-B^{2},A^{3}-B^{3})} Katso myös: Intervalli (fysiikka) Kahden nelivektorin A ja B sisätulo eli skalaaritulo määritellään Einsteinin notaatiota käyttäen seuraavasti: A ⋅ B = A μ η μ ν B ν {\displaystyle \mathbf {A} \cdot \mathbf {B} =A^{\mu }\eta _{\mu \nu }B^{\nu }} missä η {\displaystyle \eta } on Minkowskin metriikka.
They can be added in the usual entrywise way: A + B = ( A 0 , A 1 , A 2 , A 3 ) + ( B 0 , B 1 , B 2 , B 3 ) = ( A 0 + B 0 , A 1 + B 1 , A 2 + B 2 , A 3 + B 3 ) {\displaystyle \mathbf {A} +\mathbf {B} =(A^{0},A^{1},A^{2},A^{3})+(B^{0},B^{1},B^{2},B^{3})=(A^{0}+B^{0},A^{1}+B^{1},A^{2}+B^{2},A^{3}+B^{3})} and similarly scalar multiplication by a scalar λ is defined entrywise by: λ A = λ ( A 0 , A 1 , A 2 , A 3 ) = ( λ A 0 , λ A 1 , λ A 2 , λ A 3 ) {\displaystyle \lambda \mathbf {A} =\lambda (A^{0},A^{1},A^{2},A^{3})=(\lambda A^{0},\lambda A^{1},\lambda A^{2},\lambda A^{3})} Then subtraction is the inverse operation of addition, defined entrywise by: A + ( − 1 ) B = ( A 0 , A 1 , A 2 , A 3 ) + ( − 1 ) ( B 0 , B 1 , B 2 , B 3 ) = ( A 0 − B 0 , A 1 − B 1 , A 2 − B 2 , A 3 − B 3 ) {\displaystyle \mathbf {A} +(-1)\mathbf {B} =(A^{0},A^{1},A^{2},A^{3})+(-1)(B^{0},B^{1},B^{2},B^{3})=(A^{0}-B^{0},A^{1}-B^{1},A^{2}-B^{2},A^{3}-B^{3})} Applying the Minkowski tensor ημν to two four-vectors A and B, writing the result in dot product notation, we have, using Einstein notation: A ⋅ B = A μ η μ ν B ν {\displaystyle \mathbf {A} \cdot \mathbf {B} =A^{\mu }\eta _{\mu \nu }B^{\nu }} It is convenient to rewrite the definition in matrix form: A ⋅ B = ( A 0 A 1 A 2 A 3 ) ( η 00 η 01 η 02 η 03 η 10 η 11 η 12 η 13 η 20 η 21 η 22 η 23 η 30 η 31 η 32 η 33 ) ( B 0 B 1 B 2 B 3 ) {\displaystyle \mathbf {A\cdot B} ={\begin{pmatrix}A^{0}&A^{1}&A^{2}&A^{3}\end{pmatrix}}{\begin{pmatrix}\eta _{00}&\eta _{01}&\eta _{02}&\eta _{03}\\\eta _{10}&\eta _{11}&\eta _{12}&\eta _{13}\\\eta _{20}&\eta _{21}&\eta _{22}&\eta _{23}\\\eta _{30}&\eta _{31}&\eta _{32}&\eta _{33}\end{pmatrix}}{\begin{pmatrix}B^{0}\\B^{1}\\B^{2}\\B^{3}\end{pmatrix}}} in which case ημν above is the entry in row μ and column ν of the Minkowski metric as a square matrix.
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