Here is an easy one. Enjoy!

Can you find a way to make the following expression false by only changing the order of the symbols?



closed as too broad by Peregrine Rook, JMP, IAmInPLS, Ankoganit, Gamow Nov 12 '16 at 15:50

Please edit the question to limit it to a specific problem with enough detail to identify an adequate answer. Avoid asking multiple distinct questions at once. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.

  • $\begingroup$ Are the ones symbols or only the operators? $\endgroup$ – Jonathan Allan Nov 11 '16 at 1:53
  • $\begingroup$ Furthermore, how about the equals sign? May we move that as well? $\endgroup$ – greenturtle3141 Nov 11 '16 at 1:54
  • $\begingroup$ It is meant to be an easy puzzle with only a small twist. Answering these questions would pretty much give the answer so lets just say that symbols are symbols and numbers and numbers. $\endgroup$ – stack reader Nov 11 '16 at 1:58
  • $\begingroup$ @stackreader: Now that you've accepted an answer, would you care to explain what the "small twist" is? $\endgroup$ – Peregrine Rook Nov 11 '16 at 17:32
  • $\begingroup$ small as in very small. It's just that some people would naturally try to change the symbols to achieve the result of 1, but in fact the only way to do it is to move the equal symbol to change desired equality. $\endgroup$ – stack reader Nov 11 '16 at 17:40

Well here is a way

$1\times 1+1=1 \div 1 - 1 \qquad\rightarrow\qquad 2=0$

or only moving 3 symbols (is less possible?):
$1-1=1\div1\times 1+1 \qquad\rightarrow\qquad 0=2$

  • $\begingroup$ Too easy was it? It was meant to be easy but solved within 4 views is a little bit sad haha.... Well as long as people are having fun it's fine by me. Unless it's too easy to the point of not even being fun. $\endgroup$ – stack reader Nov 11 '16 at 2:04
  • $\begingroup$ I found it quite easy, but easy does not imply bad and hard does not imply good; plus it's good to have diversity. Others do not have to look at the solution I gave - maybe someone will give a solution with less changes...? $\endgroup$ – Jonathan Allan Nov 11 '16 at 4:13
  • $\begingroup$ 2 moves is possible e.g. 1+1=1/1*1-1 which is the same as 2=0. Should I post as separate solution? $\endgroup$ – boboquack Nov 11 '16 at 5:51
  • $\begingroup$ @boboquack Nice. I don't see why not?! $\endgroup$ – Jonathan Allan Nov 11 '16 at 6:00

My comment on @JohnathonAllan's answer that moves two symbols that are not numbers:



You can do it by just moving one symbol:



A trivial variation on Jonathan’s answer:

 $~1+1~~~~~~~\div1\times1=1-1 \qquad\implies\qquad~~~~ 2=0$

And, since you refused to nail down the rules,


Each answer moves only two symbols.

  • $\begingroup$ I suggest you stop being so literal. Symbols are symbols and numbers are numbers as the OP said should be enough for you to work out which is which. $\endgroup$ – boboquack Nov 11 '16 at 7:17
  • $\begingroup$ I seem to have two options: (1) Take the OP’s words literally, and (2) read the OP’s mind.  “Guess what I’m thinking”-style questions are not welcome at Stack Exchange; not even at Puzzling.  … … … … … … …  P.S. Nick Dewitt seems to have interpreted the OP’s ambiguity the same way I did — are you going to go negative on him, too? $\endgroup$ – Peregrine Rook Nov 11 '16 at 17:46
  • $\begingroup$ To be fair to stack reader the puzzle is titled "Symbols and ones", my question was really just for double clarification. My guess is that the answer given was meant to be read how I automatically read it, "...so lets just say that symbols are symbols and numbers [are] numbers", and that was just a typo (if not it's pretty confusing). $\endgroup$ – Jonathan Allan Nov 11 '16 at 18:46

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