2+2=5

Originally posted by wolfgang59
You call me 'mr dumb' and tell me to get a book on binary! .. I call that offensive.

And why message me? I really don't want your insight into a branch of maths I covered when I was 12.

Please do not message me again.

Add him to your Ignore List [found in the "My Home" menu]. Problem solved.

Is this debate really still active?

Of course 10+10=100 (binary) as 2+2=4 (decimal).
Why is this even debated? Simple arithmetics.
Don't confuse binary arithmetics with boolean algebra.

Originally posted by eatmybishop
decimal has nothing to do with it.... it is positive, on, true... not 4

You're missing what binary is all about.
For the record -
01100111 + 00011011 = 01111111
or in decimal : 103 + 27 = 127 ???
is wrong
The correct answer is - 0110111 + 00011011 = 10000010
or in decimal: 103 + 27 = 130

Hypothetical -
I have FOURTEEN apples - looks like this
oooooooooooooo
(I know, small apples - but they're tasty)

In decimal I write this as
14

(that's '1' for ten [in the "tens" column] and '4' for four [in the "units" column] .. so ... ten + four = fourteen). It looks like this -
oooooooooooooo

In binary I write this as
1110

(that's '1' for eight [in the "eights" column], '1' for four [in the "fours" column], '1' for two [in the "twos" column] and '0' [in the "units" column] ... so ... eight + four + two = fourteen. It STILL looks like this -
oooooooooooooo


My point - Take any ordinal number and it can be represented in any number base. Fourteen apples are fourteen apples whether we write that as 1110 (in binary) or 14 (in decimal) or 'E' in hexadecimal or whatever base we want - there are still fourteen apples.


Anyone want an apple?

Originally posted by orangutan
Anyone want an apple?

Sure. Can I have one of the binary ones?

Originally posted by Nordlys
Sure. Can I have one of the binary ones?

Here -

_

(oops, I dropped it 😛), how about a hexadecimal one?

1
Any good?

Originally posted by orangutan
Here -

_

(oops, I dropped it 😛), how about a hexadecimal one?

1
Any good?

Delicious, thank you! My first hexadecimapple!

edit:

Step 1: -1/1 = 1/-1

Step 2: Taking the square root of both sides: SQRT -1/1 = SQRT 1/-1

Step 3: Simplifying: SQRT -1/ SQRT 1 = SQRT 1/ SQRT -1

Step 4: In other words, i/1 = 1/i.

Step 5: Therefore, i / 2 = 1 / (2i),

Step 6: i/2 + 3/(2i) = 1/(2i) + 3/(2i),

Step 7: i (i/2 + 3/(2i) ) = i ( 1/(2i) + 3/(2i) ),

Step 8: (i^2)/2 + (3i)/2i = i/2i + (3i)/(2i)

Step 9: (-1)/2 + 3/2 = 1/2 + 3/2,

Step 10: and this shows that 1=2.

Originally posted by EinsteinMind
edit:

Step 1: -1/1 = 1/-1

Step 2: Taking the square root of both sides: SQRT -1/1 = SQRT 1/-1

Step 3: Simplifying: SQRT -1/ SQRT 1 = SQRT 1/ SQRT -1

Step 4: In other words, i/1 = 1/i.

Something is alread wrong at this point, because:

i(i/1) = i(1/i)
i^2/1 = 1i/i
-1/1 = 1
-1 =1

I suspect it has to do with the fact that SQRT(1) can be either -1 or 1, and SQRT(-1) can either be i or -i.

Originally posted by EinsteinMind
edit:

Step 1: -1/1 = 1/-1

Step 2: Taking the square root of both sides: SQRT -1/1 = SQRT 1/-1

Step 3: Simplifying: SQRT -1/ SQRT 1 = SQRT 1/ SQRT -1

Step 4: In other words, i/1 = 1/i.

Step 5: Therefore, i / 2 = 1 / (2i),

Step 6: i/2 + 3/(2i) = 1/(2i) + 3/(2i),

Step 7: i (i/2 + 3/(2i) ) = i ( 1/(2i) + 3/(2i) ),

Step 8: (i^2 ...[text shortened]... 2i = i/2i + (3i)/(2i)

Step 9: (-1)/2 + 3/2 = 1/2 + 3/2,

Step 10: and this shows that 1=2.

Tough to find a flaw! Using square roots generally results in loss of sign information. This must be the answer somehow as has been noted before me.

Originally posted by BobbyG
Something is alread wrong at this point, because:

i(i/1) = i(1/i)
i^2/1 = 1i/i
-1/1 = 1
-1 =1

I suspect it has to do with the fact that SQRT(1) can be either -1 or 1, and SQRT(-1) can either be i or -i.

This step is not the source of the fallacy.

However! you have found a mistake!

The square root symbol only has an unambiguous meaning when applied to positive numbers.

when x = any positive integer, x has two square roots, one positive and one negative.

This convention will not work when x is any negative integer. for instance, the two square roots of -1 are indeed i and -i. These cannot be distinguished on the basis of "positive" and "negative"; so how do we know which one is being meant by SQRT -1?

Therefore this step of the proof may seem unclear.

HOwever, this can be easily remedied.

Just say that
"when x is a negative integer, we are using the notation SQRT x to stand for the square root which is a positive multiple of i, rather than the other one which is a negative multiple of i."

It simply rationalizes away by defining which SQRT you are after.

does anyone no the .99=1?? i do🙂

Originally posted by AThousandYoung
Tough to find a flaw! Using square roots generally results in loss of sign information. This must be the answer somehow as has been noted before me.

Wrong. It's a mistake, but an asusmption or a definition of which SQRT you are after, and the problem continues.

Originally posted by tournymangr
does anyone no the .99=1?? i do🙂

You mean .9 recurring = 1 yes?

let X= 0.999999 .....

then 10X = 9.99999999 .....
X = 0.99999999 ....
subtract

therefore 10X - X = 9

therefore 9X = 9

therefore X = 1 QED

Originally posted by wolfgang59
You mean .9 recurring = 1 yes?

let X= 0.999999 .....

then 10X = 9.99999999 .....
X = 0.99999999 ....
subtract

therefore 10X - X = 9

therefore 9X = 9

therefore X = 1 QED

the .99=1 has nothing to do with that formula, it is not based on the .9 recurring = 1 principles... you couldnt be more wrong

Originally posted by eatmybishop
the .99=1 has nothing to do with that formula, it is not based on the .9 recurring = 1 principles... you couldnt be more wrong

I may have misunderstood the question (which is why my first line ends with a '?' ). I gave the correct answer to another question 😀

However I will be intrigued as to how any proof of 0.99=1 can be correct!!

Perhaps you can show me?