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Proof that 0=1

Here is a proof that \(0=1\):

\[ \begin{aligned} &\mbox{Let} & & 1=x \quad &(1) \\ &\mbox{We know that} & & 1=x^2 \quad &(2) \\ &\mbox{Taking the difference, we get} && 0 = x^2 - x = x(x-1) \quad & (3)\\ &\mbox{Hence, } && x=0, 1 & (4)\\ &\mbox{From (4), we have} & & x=0 & \\ &\mbox{From (1), we have} & & x=1 & \\ &\mbox{Hence, we have} && 0=x=1. & \\ \end{aligned} \]

At which step did we go wrong?

We did not divide by 0. We may have introduced an extra solution \( x=-1\), but that didn't affect anything.

Note: The common objection is that I can't claim that the roots of a quadratic are equal. I agree with that, and that they in fact are (often) not equal. However, that wasn't how the proof was obtained. We obtained \( x=0\) from (4) and have the original condition that \( x=1\) from (1).

This post arose from a discussion that we had on a problem. The original problem that led to this discussion is:

[Nov 11 Algebra and Number Theory] The point \( (x,y)\) lies on both conics \( x^2+xy+x=81\) and \( y^2+xy+y=51\). Given that \( x+y\) is positive, determine the value of \( x+y\).

Note by Calvin Lin
2 years, 9 months ago

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At step 4. We get two solutions. One of them is our number.


a.Let's say \(x={x}_{1}\) is root for a quadratic equation. Does that mean that \({x}_{1}={x}_{2}\)?

b. Let \(f=0\) a polynomial. We have \(x=0\) a root but also \(x=12\) is a root. Does that mean that 0=12?

That other problem is easy. If you add the equations and use S and P for the sum and product you get: \({S}^{2}-2P+2P+S=132\) and S will be 11. Adrian Neacșu · 2 years, 9 months ago

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@Adrian Neacșu I agree. We can toss out the "x=0" solution we obtain in step 4, after we substitute the value of "0" for x in steps 1 and 2. We see that it creates an equation that is obviously incorrect ("0=1"), so we know "x=0" is an extraneous solution. Bhagirath Mehta · 2 years, 9 months ago

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@Bhagirath Mehta In step (4) you get \({x} _ {1} = 0, {x} _ {2} = 1\). And yes 1 is the root of this equation. Some numbers have some common properties that doesn't mean that the numbers are equal. Adrian Neacșu · 2 years, 9 months ago

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I think that the second step is invalid because you squared without accounting for the second root that is produced. :P Finn Hulse · 2 years, 8 months ago

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what if we take, 1/∞=0, then, 1=0×∞ since (z×0=0) Therefore , 1=0 Sir, please tell me if i am wrong San Bha · 2 years, 2 months ago

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It should be either x=0 or x=1, the two values of x can't be equal at the same time.. Asama Zaldy Jr. · 2 years, 3 months ago

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Squaring gives us an extraneous root, after the final solution we check that the solutions satify the original equation or not. Krishna Sharma · 2 years, 3 months ago

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this happens because whenever we square any equation after arriving at the answer we must substitute and into our original condition as squaring always stands and gives possibilities of +/- value of roots Chirag Shetty · 2 years, 3 months ago

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a^0=1 how give proof Hash Jain · 2 years, 9 months ago

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@Hash Jain simple proof

1= \( a^n/a^n\)


hence 1=\(a^0\) Prajwal Kavad · 2 years, 7 months ago

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@Hash Jain It's easy to prove that \({a}^{0}=1\) for any \(a>0\).

We know that \({a}^{m}:{a}^{n}={a}^{m-n}\).

For \(m=n\) we get \(1={a}^{m}:{a}^{m}={a}^{0}\). Adrian Neacșu · 2 years, 9 months ago

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