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. $\Box$

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2 \times 3

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2^{34}

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Math	Appears 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}`	$\boxed{123}$

Comments

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

Daniel Liu - 7 years, 2 months ago

Daniel, what is the finite value of this series

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

Tunk-Fey Ariawan - 7 years, 2 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 $S = - \frac {1}{9}$

Deepansh Mathur - 7 years, 2 months ago

@Deepansh Mathur – $\begin{aligned} S&=10^0+10^1+10^2+10^3+\cdots\\ S&=10^0+10(10^0+10^1+10^2+10^3+\cdots)\\ S&=1+10S\\ -9S&=1\\ S&=-\frac{1}{9}. \end{aligned}$

@Deepansh Mathur – 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?

Bhargav Das - 7 years, 2 months ago

@Bhargav Das – You might wanna check out this

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....

John Ashley Capellan - 7 years, 2 months ago

Math	Appears 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}`	$\boxed{123}$