6
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We have the following reasoning.

A < B < C < A

Can you make this statement solvable?
Find
-2 solutions by removing 2 straight lines and adding 2 straight lines.
-1 solution by adding 3 straight lines.
-1 solution by adding only 1 straight line.

NOTE
The solutions must be aesthetically acceptable.
The solutions must be simple. No fancy symbols accepted.

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  • 2
    $\begingroup$ That appears to be an inequality, not an equation. ;) $\endgroup$ – jpmc26 Nov 11 '16 at 2:53
9
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2 solutions by removing 2 straight lines and adding 2 straight lines.

$A\lt B\lt C+A, A\lt B\lt C\gt A$

1 solution by adding 3 straight lines.

$A\le B\le C\le A$

1 solution by adding only 1 straight line.

going with JA on this, $-A\lt B\lt C\lt A$

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  • $\begingroup$ I'll give you this one because you are the first one to give me 2 original solutions for the remove 2 add 2 part. Although what I had in mind was to change 1 A to "F" or "Z", in the end there was much more solutions that expected.(I had a feeling it would end up like that but I thought people would have fun trying to solve this puzzle so I made it anyway) $\endgroup$ – stack reader Nov 10 '16 at 6:50
  • $\begingroup$ A<B<C+A doesn't work always. Consider A=-100, B= 0, C=1. $\endgroup$ – CodeNewbie Nov 10 '16 at 7:22
  • 5
    $\begingroup$ @CodeNewbie 'Make the equation solvable' -> it's still solvable e.g. A, B, C = 1, 2, 3 $\endgroup$ – boboquack Nov 10 '16 at 8:05
  • $\begingroup$ Alternatively, for the adding one line, you can do a crossed out not less than $\endgroup$ – Conor O'Brien Nov 11 '16 at 4:11
8
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Seems like there may be a huge number of ways to achieve these, but here are some...

Removing 2 straight lines and adding 2 straight lines in two ways:

$H<B<C<A$
and
$A<B<C<H$

Adding 3 straight lines:

$L-A<B<C<A$

Adding 1 straight line:

$-A<B<C<A$

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  • $\begingroup$ The solution for 1 line is the one I had in mind. The solution with 3 lines is quite clever, I like it, mine was much simpler. and the 2 solutions for 2 lines I would consider them to be just 1, although that would be quite questionable. $\endgroup$ – stack reader Nov 10 '16 at 6:34
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2 solutions by removing 2 straight lines and adding 2 straight lines.

A > B < C < A
A < B < C > A
In both cases two straight lines removed are <, two straight lines added: >

1 solution by adding 3 straight lines.

A ≤ B ≤ C ≤ A
which resolves to A = B = C = A.
Three lines added are 'or-equal' lines in symbol made of <.

1 solution by adding only 1 straight line.

–A < B < C < A
A line added is the minus sign.

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  • $\begingroup$ +1 That is the solution I had in mind for the 3 lines, good job! $\endgroup$ – stack reader Nov 10 '16 at 6:46
5
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What do you think about this solution? It holds true without changing anything :)

rock < paper < scissors < rock

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  • 1
    $\begingroup$ Joined the community just to upvote this answer. In a similar vein, any set of objects works if we define a non-transitive $<$ order relation on them. Many such examples exist, for instance numbers in a modulus ring. $\endgroup$ – Konrad Rudolph Nov 10 '16 at 14:54
  • $\begingroup$ Thanks :) Rock, paper, scissors was just one example and it was more meant as a bonus answer since it technically doesn't fulfill the requirements of adding and removing lines. $\endgroup$ – Zibelas Nov 10 '16 at 14:59
4
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For the third question:

$A < B < C \nless A$

Of course this suggests two other possible solutions:

$A < B \nless C < A$

and

$A \nless B < C < A$

I don't think this counts as using "fancy symbols."

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4
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Some logical answers

2 solutions by removing 2 straight lines and adding 2 straight lines.

$A\lt B\lor C\lt A$, logical or
$A\lt B \land C\lt A$, logical and

1 solution by adding 3 straight lines.

$A\lt B\leftrightarrow C\lt A$, if and only if

1 solution by adding only 1 straight line.

$A\lt B\leftarrow C\lt A$, less common way of doing implication, latter implies former

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  • $\begingroup$ Nice. Do note, however, that these imply that the original is already "solvable" (evaluating to false) and that one of the first two* will evaluate to false. * Which one depending upon operator precedence, unless we define a rather strange intermingled ordering of precedence between logical and comparison operators. $\endgroup$ – Jonathan Allan Nov 10 '16 at 19:01
  • $\begingroup$ @JonathanAllan I would treat logical operations as the lowest precedence, so it is like (A<B)∨(C<A). All of my solutions can be evaluated as true which was my goal instead of just solvable, note 'or' is not 'xor'. $\endgroup$ – Angzuril Nov 10 '16 at 19:08
  • 1
    $\begingroup$ Ah yes, true, they can both evaluate as true, my bad. $\endgroup$ – Jonathan Allan Nov 10 '16 at 19:18
1
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2 solutions by removing 2 straight lines and adding 2 straight lines (slightly different from Jonathan’s):

A < B < C < VI
or
A < B < C < XI
(You can also do IV and IX, but the spacing is tight.)

1 solution by adding 3 straight lines (slightly different from Jonathan’s):

4A < B < C < A
(can be solved with A = −1, B = −3, and C = −2)

1 solution by adding only 1 straight line.

$A < B < \overline{C < A}$

The overline is a vinculum, which groups a subexpression.  So the above is equivalent to:

$A < B < (C < A)$

Using the standard (?) paradigm of TRUE ≡ 1 and FALSE ≡ 0, this can be solved with any A < B < 0 and C ≥ A,  or  C < A < B < 1.

Note that $\overline{A < B} < C < A$ would also work.

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  • $\begingroup$ P.S.   Yes, I also figured out the other three-line answer and the other one-line answers before I read the other posts. $\endgroup$ – Peregrine Rook Nov 11 '16 at 4:51

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