Intuitive explanation of solutions to a linear diophantine equation
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"Given a linear diophantine equation $ax+by=c$ with a particular solution $(x_0,y_0)$ the general solution is given by $$biggl(x_0-frac{b}{gcd(a,b)}t,y_0+frac{a}{gcd(a,b)}tbiggr)$$ for all $tin mathbb{Z}$"
I understand the proof of this theorem pretty well but would appreciate an intuitive explanation of why this general solution gives all the solutions to the equation...
diophantine-equations intuition linear-diophantine-equations
asked Jan 7 at 13:33
stochasticmrfoxstochasticmrfox
967
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add a comment |
$begingroup$
"Given a linear diophantine equation $ax+by=c$ with a particular solution $(x_0,y_0)$ the general solution is given by $$biggl(x_0-frac{b}{gcd(a,b)}t,y_0+frac{a}{gcd(a,b)}tbiggr)$$ for all $tin mathbb{Z}$"
I understand the proof of this theorem pretty well but would appreciate an intuitive explanation of why this general solution gives all the solutions to the equation...
diophantine-equations intuition linear-diophantine-equations
asked Jan 7 at 13:33
stochasticmrfoxstochasticmrfox
967
$endgroup$
add a comment |
$begingroup$
"Given a linear diophantine equation $ax+by=c$ with a particular solution $(x_0,y_0)$ the general solution is given by $$biggl(x_0-frac{b}{gcd(a,b)}t,y_0+frac{a}{gcd(a,b)}tbiggr)$$ for all $tin mathbb{Z}$"
I understand the proof of this theorem pretty well but would appreciate an intuitive explanation of why this general solution gives all the solutions to the equation...
diophantine-equations intuition linear-diophantine-equations
asked Jan 7 at 13:33
stochasticmrfoxstochasticmrfox
967
$endgroup$
"Given a linear diophantine equation $ax+by=c$ with a particular solution $(x_0,y_0)$ the general solution is given by $$biggl(x_0-frac{b}{gcd(a,b)}t,y_0+frac{a}{gcd(a,b)}tbiggr)$$ for all $tin mathbb{Z}$"
I understand the proof of this theorem pretty well but would appreciate an intuitive explanation of why this general solution gives all the solutions to the equation...
diophantine-equations intuition linear-diophantine-equations
diophantine-equations intuition linear-diophantine-equations
asked Jan 7 at 13:33
stochasticmrfoxstochasticmrfox
967
asked Jan 7 at 13:33
stochasticmrfoxstochasticmrfox
967
asked Jan 7 at 13:33
stochasticmrfoxstochasticmrfox
967
asked Jan 7 at 13:33
stochasticmrfoxstochasticmrfox
967
asked Jan 7 at 13:33
stochasticmrfoxstochasticmrfox
967
967
add a comment |
add a comment |
1 Answer
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Let's take, as an example, the case of money. We have banknotes whose values are $a$ and $b$. We want to exchange an amount of $c$.
Finding a solution $x_0, y_0$ then corresponds to finding some number of banknotes such that we exchange exactly $c$.
But, sort of as a dufus, I could insist on exchanging more bills.
So if I give you $x_0 + frac{b}{gdc(a,b)}t$ number of bills, valued $a$, and you give me $y_0 - frac{a}{gdc(a,b)}t$ number of bills valued $b$ back, we have exchanged exactly an additional $$frac{ab}{gcd(a,b)}t - frac{ba}{gcd(a,b)} t = 0$$
So we would still exchange exactly $c$
answered Jan 7 at 13:57
JuliusL33tJuliusL33t
1,36811017
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1 Answer
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1 Answer
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$begingroup$
Let's take, as an example, the case of money. We have banknotes whose values are $a$ and $b$. We want to exchange an amount of $c$.
Finding a solution $x_0, y_0$ then corresponds to finding some number of banknotes such that we exchange exactly $c$.
But, sort of as a dufus, I could insist on exchanging more bills.
So if I give you $x_0 + frac{b}{gdc(a,b)}t$ number of bills, valued $a$, and you give me $y_0 - frac{a}{gdc(a,b)}t$ number of bills valued $b$ back, we have exchanged exactly an additional $$frac{ab}{gcd(a,b)}t - frac{ba}{gcd(a,b)} t = 0$$
So we would still exchange exactly $c$
answered Jan 7 at 13:57
JuliusL33tJuliusL33t
1,36811017
$endgroup$
add a comment |
$begingroup$
Let's take, as an example, the case of money. We have banknotes whose values are $a$ and $b$. We want to exchange an amount of $c$.
Finding a solution $x_0, y_0$ then corresponds to finding some number of banknotes such that we exchange exactly $c$.
But, sort of as a dufus, I could insist on exchanging more bills.
So if I give you $x_0 + frac{b}{gdc(a,b)}t$ number of bills, valued $a$, and you give me $y_0 - frac{a}{gdc(a,b)}t$ number of bills valued $b$ back, we have exchanged exactly an additional $$frac{ab}{gcd(a,b)}t - frac{ba}{gcd(a,b)} t = 0$$
So we would still exchange exactly $c$
answered Jan 7 at 13:57
JuliusL33tJuliusL33t
1,36811017
$endgroup$
add a comment |
$begingroup$
Let's take, as an example, the case of money. We have banknotes whose values are $a$ and $b$. We want to exchange an amount of $c$.
Finding a solution $x_0, y_0$ then corresponds to finding some number of banknotes such that we exchange exactly $c$.
But, sort of as a dufus, I could insist on exchanging more bills.
So if I give you $x_0 + frac{b}{gdc(a,b)}t$ number of bills, valued $a$, and you give me $y_0 - frac{a}{gdc(a,b)}t$ number of bills valued $b$ back, we have exchanged exactly an additional $$frac{ab}{gcd(a,b)}t - frac{ba}{gcd(a,b)} t = 0$$
So we would still exchange exactly $c$
answered Jan 7 at 13:57
JuliusL33tJuliusL33t
1,36811017
$endgroup$
Let's take, as an example, the case of money. We have banknotes whose values are $a$ and $b$. We want to exchange an amount of $c$.
Finding a solution $x_0, y_0$ then corresponds to finding some number of banknotes such that we exchange exactly $c$.
But, sort of as a dufus, I could insist on exchanging more bills.
So if I give you $x_0 + frac{b}{gdc(a,b)}t$ number of bills, valued $a$, and you give me $y_0 - frac{a}{gdc(a,b)}t$ number of bills valued $b$ back, we have exchanged exactly an additional $$frac{ab}{gcd(a,b)}t - frac{ba}{gcd(a,b)} t = 0$$
So we would still exchange exactly $c$
answered Jan 7 at 13:57
JuliusL33tJuliusL33t
1,36811017
answered Jan 7 at 13:57
JuliusL33tJuliusL33t
1,36811017
answered Jan 7 at 13:57
JuliusL33tJuliusL33t
1,36811017
answered Jan 7 at 13:57
JuliusL33tJuliusL33t
1,36811017
1,36811017
add a comment |
add a comment |
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