# 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
5 years, 7 months ago

This discussion board is a place to discuss our Daily Challenges and the math and science related to those challenges. Explanations are more than just a solution — they should explain the steps and thinking strategies that you used to obtain the solution. Comments should further the discussion of math and science.

When posting on Brilliant:

• Use the emojis to react to an explanation, whether you're congratulating a job well done , or just really confused .
• Ask specific questions about the challenge or the steps in somebody's explanation. Well-posed questions can add a lot to the discussion, but posting "I don't understand!" doesn't help anyone.
• Try to contribute something new to the discussion, whether it is an extension, generalization or other idea related to the challenge.

MarkdownAppears as
*italics* or _italics_ italics
**bold** or __bold__ bold
- bulleted- list
• bulleted
• list
1. numbered2. list
1. numbered
2. list
Note: you must add a full line of space before and after lists for them to show up correctly
paragraph 1paragraph 2

paragraph 1

paragraph 2

[example link](https://brilliant.org)example link
> This is a quote
This is a quote
    # I indented these lines
# 4 spaces, and now they show
# up as a code block.

print "hello world"
# I indented these lines
# 4 spaces, and now they show
# up as a code block.

print "hello world"
MathAppears as
Remember to wrap math in $$ ... $$ or $ ... $ to ensure proper formatting.
2 \times 3 $2 \times 3$
2^{34} $2^{34}$
a_{i-1} $a_{i-1}$
\frac{2}{3} $\frac{2}{3}$
\sqrt{2} $\sqrt{2}$
\sum_{i=1}^3 $\sum_{i=1}^3$
\sin \theta $\sin \theta$
\boxed{123} $\boxed{123}$

Sort by:

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 $\boxed{S=\dfrac{1}{2}}$.

- 5 years, 7 months ago

Daniel, what is the finite value of this series

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

- 5 years, 7 months 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 $S = - \frac {1}{9}$

- 5 years, 7 months ago

\begin{aligned} 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{aligned}

- 5 years, 7 months 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?

- 5 years, 7 months ago

You might wanna check out this

- 5 years, 7 months 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....

- 5 years, 7 months ago