# Two Variables One Equation

Hi everyone!

Here is an interesting solution to a two-variable one-equation system.

We want to solve $$x+y=1$$.

Subtracting $$x$$ from both sides, we see that $$y=1-x$$.

Subtracting $$y$$ from both sides of the original equation, we get that $$x=1-y$$.

Now we have a system of equations: $\left\{\begin{array}{l}x=1-y\\y=1-x\end{array}\right.$

Plugging the second equation into the first, we get that $$x=1-(1-x)$$.

Now, plugging this equation into itself recursively, we see that $$x=1-(1-(1-\cdots))$$

To solve $$x$$, we observe the following sequence: $1,1-(1),1-(1-(1)),1-(1-(1-(1))),\ldots$

Agree that the limit of this sequence as it goes on to its infinite term is $$x$$.

Also, note that this sequence is equivalent to the following sequence: $1,0,1,0,1,\ldots$

As we all know from various proofs which I will not outline here (this sequence is pretty famous), the limit of this sequence is $$\dfrac{1}{2}$$.

However, this limit is equivalent to $$x$$, as we know that $$x=1-(1-(1-\cdots))$$.

Therefore, our solution to our original system of equation is $$\boxed{(x,y)=\left(\dfrac{1}{2},\dfrac{1}{2}\right)}$$, and we are done. $$\Box$$

Note by Daniel Liu
4 years ago

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Here is a proof that $$1-1+1-1+\cdots=\dfrac{1}{2}$$.

Let $$S=1-1+1-1+\cdots$$.

Note that $$2S=1-1+1-1+\cdots +1-1+1-1+\cdots$$.

Pairing up the $$n$$th term of the first sequence with the $$n+1$$th term of the second sequence, we see that $$2S=1+(-1+1)+(-1+1)+\cdots$$.

Therefore, $$2S=1$$, and so $$\boxed{S=\dfrac{1}{2}}$$.

- 4 years ago

Daniel, what is the finite value of this series

$\sum_{n=0}^\infty 10^n=\,...$

- 4 years ago

Let the summation be $$S$$

We have $$S = 10^0 + 10^1 + 10^2 + 10^3 + 10^4 + \dotsb$$

Multiplying both sides by 9, we get

$$9S = 9(10^0 + 10^1 + 10^2 + 10^3 + 10^4 + \dotsb )$$

We can write $$9$$ as $$10 - 1$$

Hence, $$9S = (10 - 1)(10^0 + 10^1 + 10^2 + 10^3 + 10^4 + \dotsb$$

Expanding the $$RHS$$

$$\Rightarrow 9S = [10^1 + 10^2 + 10^3 + 10^4 + 10^5 + \dotsb ] - [10^0 + 10^1 + 10^2 + 10^3 + 10^4 + \dotsb ]$$

$$\Rightarrow 9S = - 10^0$$

subsequent terms get cancelled, only $$-1$$ remains

We get $$S = - \frac {1}{9}$$

- 4 years ago

\begin{align} S&=10^0+10^1+10^2+10^3+\cdots\\ S&=10^0+10(10^0+10^1+10^2+10^3+\cdots)\\ S&=1+10S\\ -9S&=1\\ S&=-\frac{1}{9}. \end{align}

- 4 years ago

Clearly,the expression is divergent as it is composed of only positive nos. So,there can't be a finite value.Also,there seems no flaw in the proof given by Deepanshu to me at least.But surely there's a fallacy somewhere.Can someone point it out?

- 4 years ago

You might wanna check out this

- 4 years ago

Yes... It is a famous Grandi series in which somehow I believe or I don't believe which made the sum of natural numbers be -1/12. But (1/2, 1/2) is not just one solution we are looking for. Say we have infinitely many solutions (integral, rational, etc.) that lies on this locus....

- 4 years ago