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- Thread starter Mr Davis 97
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An example of an implication step which is not an equivalence would be squaring the equation ##x = 2## to obtain ##x^2 = 4##. You then introduce a false root ##x = -2##.

Edit: That being said, it is of course

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This still tells us nothing unless you specify the problem that was to be solved and the logical steps taken in doing so.

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fresh_42

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This is correct. But as you performed equivalence deductions, you already have proven the backward direction. In general, however, this might not be the case, so you will have to plug in the deduced statement to see, if it also satisfies the original equation, because a deduction only provides a necessary condition.I read that in solving an equation, the result is only a possibility for a solution, and not necessarily a solution. For example, in solving x+3 = 5, x=2 is not a solution until you explicitly plug it back in to show that it actually. Why is the statement x=2 not enough to show that 2 is actually a solution, from a strictly logical point of view? This could apply to any equation-solving process.

$$ x+3=5 \Longleftrightarrow x=2$$

On the other hand, you could start with ##x^3+y^3=z^3## and find some necessary conditions on ##x,y,z \in \mathbb{Z}##. But whatever you will find, it cannot satisfy the original equation.

Edit: Or more obvious: ##2x= 4 \Longrightarrow 4x^2=16 \Longrightarrow x \in \{-2,2\}## but ##x=-2## isn't a solution. (Oops, already been said.)

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fresh_42

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Another famous pit is if you multiply an equation or greater-relation by a term, which isn't a known number at the time, e.g. something like ##xy+z^2## or similar, and you don't rule out the possibility that it might be zero or negative, then you get wrong results without knowing.

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FactChecker

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Strictly speaking, you have proven in one direction but not in the reverse direction.

You have proven "If 5+x=2, then x=-3."

But to prove that "-3 is a solution of 5+x=2", you want to prove "If x=-3, then 5+x=2."

In this example, it seems like a silly distinction but in complicated examples it can be an important distinction.

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Stephen Tashi

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Algebraic manipulations abbreviate a complicated process of reasoning. Solutions can be deduced by using "logical equivalences". Doing "reversible operations" in the sense of "doing the same operation to both sides" does not necessarily produce an equation that is logically equivalent to the original equation. Try using reversible operations to solve these equations:

1) ## x + \frac{1}{x-2} = 2 + \frac{1}{x-2}##

2) ## x \ln(x-2) = \ln(x-2)##

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pwsnafu

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Consider ##\sqrt{x^2+x+3} = x##.

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We've been through this before with you!

https://www.physicsforums.com/threads/show-that-an-integer-is-unique.895228/#post-5631947

It's all about the reverse implication of the steps you take.

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