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# 9/14/2017 - An Abstract Way To Equalize

• #41
dsf001 wrote:
Gil-Gandel wrote:
aHorseWithNoName wrote:

Come to my intelligent forum

2+2=5 can it be true?

Let a + b = c
Note that 5a - 4a = a
Note that 5a - (2 + 2)a = a

Then 5a - (2 + 2)a + 5b - (2 + 2)b = 5c - (2 + 2)c
Rearranging,

5a + 5b - 5c = (2 + 2)a + (2 + 2)b - (2 + 2)c

Factorising,

5(a + b - c) = (2 + 2)(a + b - c)
Cancelling the common factor (a + b - c), and removing the brackets as they are now redundant,

5 = 2 + 2

Hope this helps.

The fallacy is that if a+b=c, then a+b-c=0, so when you write

"Factorising,

5(a + b - c) = (2 + 2)(a + b - c)"

what you are really saying is that 5 x 0 = (2+2) x 0, which is true, but you can't cancel the 0s to leave 5 = 2+2 because 5x0/0 does not equal 5 and (2+2)x0/0 does not equal (2+2).

This took about the same amount of time for me to solve as the chess puzzle.

instead of trying to bring new math rules that don't exist, just come to my forum and learn

• #42

I'm late today

• #43

@dsf001  I see now Kg5 is big blunder.   Many Thx.

• #44

Hello and good morning. That was practice puzzle par excellence [pardon my bad French)!

• #45

but this puzzle will end up being a draw. so white cant win

• #46
aHorseWithNoName wrote:
dsf001 wrote:
Gil-Gandel wrote:
aHorseWithNoName wrote:

Come to my intelligent forum

2+2=5 can it be true?

Let a + b = c
Note that 5a - 4a = a
Note that 5a - (2 + 2)a = a

Then 5a - (2 + 2)a + 5b - (2 + 2)b = 5c - (2 + 2)c
Rearranging,

5a + 5b - 5c = (2 + 2)a + (2 + 2)b - (2 + 2)c

Factorising,

5(a + b - c) = (2 + 2)(a + b - c)
Cancelling the common factor (a + b - c), and removing the brackets as they are now redundant,

5 = 2 + 2

Hope this helps.

The fallacy is that if a+b=c, then a+b-c=0, so when you write

"Factorising,

5(a + b - c) = (2 + 2)(a + b - c)"

what you are really saying is that 5 x 0 = (2+2) x 0, which is true, but you can't cancel the 0s to leave 5 = 2+2 because 5x0/0 does not equal 5 and (2+2)x0/0 does not equal (2+2).

This took about the same amount of time for me to solve as the chess puzzle.

instead of trying to bring new math rules that don't exist, just come to my forum and learn

That is an abstract way to equalise.

• #47
vagga2 wrote:
aHorseWithNoName wrote:
dsf001 wrote:
Gil-Gandel wrote:
aHorseWithNoName wrote:

Come to my intelligent forum

2+2=5 can it be true?

Let a + b = c
Note that 5a - 4a = a
Note that 5a - (2 + 2)a = a

Then 5a - (2 + 2)a + 5b - (2 + 2)b = 5c - (2 + 2)c
Rearranging,

5a + 5b - 5c = (2 + 2)a + (2 + 2)b - (2 + 2)c

Factorising,

5(a + b - c) = (2 + 2)(a + b - c)
Cancelling the common factor (a + b - c), and removing the brackets as they are now redundant,

5 = 2 + 2

Hope this helps.

The fallacy is that if a+b=c, then a+b-c=0, so when you write

"Factorising,

5(a + b - c) = (2 + 2)(a + b - c)"

what you are really saying is that 5 x 0 = (2+2) x 0, which is true, but you can't cancel the 0s to leave 5 = 2+2 because 5x0/0 does not equal 5 and (2+2)x0/0 does not equal (2+2).

This took about the same amount of time for me to solve as the chess puzzle.

instead of trying to bring new math rules that don't exist, just come to my forum and learn

That is an abstract way to equalise.

ur rating is 719. that is way abstract

• #48

easy!

• #49

Bit tricky once you get the first move the rest follows up easily

• #50

another queen sac

• #51

Done.

• #52

A Queen sacrifice, putting the Black King in check and threatening the Black Rook. Wunderbar!

• #53

That was cool. Not that obvious either.

• #54
nice
• #55
easy
• #56

Fortunately for Black  the White Q sacrifice is accepted. Both first moves lead to the best outcome (draw) that either player can expect. Fun Puzzle, and it only took two tries to solve.

• #57
• #58

hi.

• #59

thankyou

• #60

Had me stumped - Not used to playing for stalemate.  Must look for alternatives for black.

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