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Mathematics 18 Online
OpenStudy (anonymous):

Explain this hexadecimal subtraction

OpenStudy (anonymous):

FFE,A7809- 13ABC,00000

OpenStudy (anonymous):

Can't you write it more nicely? :P

OpenStudy (anonymous):

FFE,A7809- 13ABC,00000 ------------ ?? 542,A7809

OpenStudy (anonymous):

The two numbers are not equal in lengths?

OpenStudy (anonymous):

@amistre64 @anwaarullah @eliassaab @tkhunny @undeadknight26 @Preetha @satellite73 @freckles @glamourglory @hartnn @Kamille @Loser66 @camerondoherty And the answer is NO @waterineyes

OpenStudy (anonymous):

\(FFEA7809\) -\(13ABC00000\) ----------------

OpenStudy (anonymous):

Last \(5\) digits will remain same..

OpenStudy (anonymous):

@waterineyes you skipped the comma those are not integer hex values

OpenStudy (anonymous):

Then from the right, it is \(2\).

OpenStudy (anonymous):

What does that "comma" signify?

ganeshie8 (ganeshie8):

hexadecimal point i suppose..

OpenStudy (anonymous):

okay let's skip the comma I'm interested only on this: FFE- 13ABC ------

OpenStudy (anonymous):

@ganeshie8 you're right

ganeshie8 (ganeshie8):

you want to work the subtraction in hexadecimal system without converting is it ?

OpenStudy (anonymous):

yes

OpenStudy (anonymous):

If we can find the 16's Compliment of 13ABC, then may be we can add them just simply.. :)

OpenStudy (anonymous):

@waterineyes solve this exercise without using any Compliment :D

OpenStudy (anonymous):

Dear, I am here only, so still you feel need of tagging?? :P

OpenStudy (anonymous):

Can't we switch the numbers by taking \(-\) sign out?

OpenStudy (anonymous):

\(13ABC,00000\) -\(FFE,A7809\) ------------- -...................

ganeshie8 (ganeshie8):

13ABC FFE -------- 12ABE

OpenStudy (anonymous):

D??

OpenStudy (anonymous):

\(12ABD\)

OpenStudy (anonymous):

actually the answer is: FFFFFFFFFFFED542

ganeshie8 (ganeshie8):

step1 : 1 13ABC FFE -------- E 1C - E = E

OpenStudy (anonymous):

guys you're not helping me

ganeshie8 (ganeshie8):

you borrow a 16 from B and add it to units place C then subtract..

OpenStudy (anonymous):

ffe-13abc not reverse

ganeshie8 (ganeshie8):

i am trying to help you, how about you ? are u trying ?

OpenStudy (anonymous):

No, firstly we will help ourselves then you... :P

ganeshie8 (ganeshie8):

how do you work 456 - 121314 in decimal ?

ganeshie8 (ganeshie8):

a-b = -(b-a) right ?

OpenStudy (anonymous):

You think we are pearls here knowing everything.. :P Have patience..!!

ganeshie8 (ganeshie8):

since you don't want to use complements, there is no way to represent a negative number without using the minus sign

OpenStudy (anonymous):

Yeah, ganesh I agree..!!

OpenStudy (anonymous):

I'm here to figure out why: 00- 13 FFFFFFFFFFFED

ganeshie8 (ganeshie8):

I thought we're working on ``` okay let's skip the comma I'm interested only on this: FFE- 13ABC ``` ?

OpenStudy (amistre64):

0 0F FE + ------- 1 3A BC

OpenStudy (amistre64):

thats my idea

OpenStudy (anonymous):

@ganeshie8 put 2 of 0 in front of ffe

OpenStudy (anonymous):

ffe-abc=542 00-13=????????

ganeshie8 (ganeshie8):

filling 0's on left side is for machines, they are not needed for carrying out subtraction manually

ganeshie8 (ganeshie8):

00-13 = -13

OpenStudy (anonymous):

http://www.csgnetwork.com/hexaddsubcalc.html

OpenStudy (anonymous):

@ganeshie8 you mean this subtraction can't be done without using the compliment or the reverse code?

OpenStudy (anonymous):

we should calculator now.. :P

ganeshie8 (ganeshie8):

how do you represent -13 on a machine ?

OpenStudy (anonymous):

with the bit of sign

ganeshie8 (ganeshie8):

thats one way, and we call it signed magnitude system

ganeshie8 (ganeshie8):

however the one you're trying here is called 2's complement representation : 00- 13 FFFFFFFFFFFED

OpenStudy (anonymous):

2's?? or 16's ??

ganeshie8 (ganeshie8):

...FFFFFFFFFFFED = 16's complement of 13 thnks for correcting water :)

ganeshie8 (ganeshie8):

you're using it to represent negative numbers without wasting a bit for sign

ganeshie8 (ganeshie8):

`00` - `13` = `00` + `(16's complement of 13)` = `00` + `...FFFFED` = `...FFFFED`

OpenStudy (anonymous):

okay so if I'm right to find the 16's complement of -13 you do this: FFFFFFFFFFFFF- 13 FFFFFFFFFFFED

ganeshie8 (ganeshie8):

yes subtract each digit from F and in the end add 1

OpenStudy (anonymous):

YEAH THAT'S WHY F-3+1=D

OpenStudy (anonymous):

thank you all for the patience

OpenStudy (anonymous):

@Alexander95 this is cheating, you used Complements here.. :P

OpenStudy (anonymous):

@waterineyes I can use complements I said that hoping I can find another way

OpenStudy (anonymous):

But you limited my approach to "just subtracting" only.. :P It is fine.. :)

ganeshie8 (ganeshie8):

maybe take an example in base10 to see why the complement stuff works

ganeshie8 (ganeshie8):

base 10 : `20` - `30` = `20` + `(10's complement of 30`) = `20` + `70` = `90`

hartnn (hartnn):

you mean 100's compliment ? :P

ganeshie8 (ganeshie8):

Note that 10's complement of `90` is `10` as you expect the result to be... the left most bit of 1 tells you that the stored number is negative

ganeshie8 (ganeshie8):

lol techically 100's complement is possible too as the base can be any real number

OpenStudy (anonymous):

@ganeshie8 how many complements are:1's;2's;...;16's;...

ganeshie8 (ganeshie8):

there are only two useful complements in base N system : 1) N's complement 2) N-1 complement

hartnn (hartnn):

and base can be anything, 11's compliment of 5 is 11-5

ganeshie8 (ganeshie8):

if you're in base 10, you call them 10's complement and 9's complement

OpenStudy (anonymous):

so for base 2 we got:1,2 10 we got:9,10 .......... 16 we got:16,15 .................

ganeshie8 (ganeshie8):

looks good, subtracting each digit by 9 or F may not reveal you much about whats going on.. there is a nice way to make sense of this complement business

ganeshie8 (ganeshie8):

\(10\)'s complement of a \(\large k \) digit number, \(\large x\), in base \(\large 10\) = \(\large 10^k-x\)

ganeshie8 (ganeshie8):

If you want to work the subtraction : \(\large w-x\) : \[\large \begin{align} w - x &= w + (-x)\\&=w + (10^k-x)\\\end{align}\] next, take the 10's complement of result, you get : \(\large 10^k- \left[ w+(10^k-x)\right] = x-w \) the computer adds a negative sign because it knows that the left most digit of the subtraction (w-x) is 1.

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