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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, 7 months ago

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

Eg:

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. · 2 years, 6 months ago

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. · 2 years, 6 months ago

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. · 2 years, 6 months ago

I think that the second step is invalid because you squared without accounting for the second root that is produced. :P · 2 years, 5 months ago

what if we take, 1/∞=0, then, 1=0×∞ since (z×0=0) Therefore , 1=0 Sir, please tell me if i am wrong · 1 year, 12 months ago

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

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

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 · 2 years ago

a^0=1 how give proof · 2 years, 6 months ago

simple proof

1= $$a^n/a^n$$

1=$$a^{n-n}$$

hence 1=$$a^0$$ · 2 years, 5 months ago

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 $$1={a}^{m}:{a}^{m}={a}^{0}$$. · 2 years, 6 months ago