- WELCOME TO A LESSON ON HOW TO WRITE BASE 10 NUMBERS
IN DIFFERENT BASES.
THE BASE SYSTEM IS THE STRUCTURE WITHIN WHICH WE COUNT.
THE EASIEST WAY TO DESCRIBE A BASE SYSTEM
IS TO THINK ABOUT OUR OWN BASE-10 SYSTEM.
THE BASE-10 SYSTEM, WHICH WE CALL THE DECIMAL SYSTEM,
REQUIRES A TOTAL OF 10 DIFFERENT SYMBOLS OR DIGITS
TO WRITE ANY NUMBER.
THEY ARE OF COURSE ZERO THROUGH NINE, LISTED HERE.
THE DECIMAL SYSTEM IS ALSO AN EXAMPLE
OF A POSITIONAL-BASE SYSTEM,
WHICH SIMPLY MEANS THAT THE POSITION OF A DIGIT
GIVES ITS PLACE VALUE.
NOT ALL CIVILIZATIONS HAD A POSITIONAL SYSTEM,
EVEN THOUGH THEY DID HAVE A BASE WITH WHICH THEY WORKED.
IN OUR BASE-10 SYSTEM, A NUMBER LIKE 5,783,216 HAS MEANING TO US
BECAUSE WE ARE FAMILIAR WITH THE SYSTEM AND ITS PLACES.
WE CAN TAKE THIS NUMBER
AND ASSIGN EACH OF ITS DIGITS A VALUE.
ONE WAY TO DO THIS IS GIVEN IN THE TABLE.
NOTICE HOW THE FIVE REPRESENTS FIVE ONE MILLIONS,
WHICH WE CAN ALSO WRITE AS 5 x 10 TO THE 6th.
THE SEVEN REPRESENTS SEVEN 100,000S,
WHICH WE CAN ALSO WRITE AS 7 x 10 TO THE 5th, AND SO ON.
NOTICE HOW TO DETERMINE THE VALUE OF EACH POSITION
OR PLACE VALUE,
WE COULD TAKE OUR BASE 10, STARTING WITH 10 TO THE 0,
WHICH WOULD GIVE US 1,
AND THEN INCREASE THE EXPONENT BY ONE
TO DETERMINE PLACE VALUES.
10 TO THE 1ST GIVES US THE TENS PLACE VALUE,
10 TO THE 2nd GIVES US THE HUNDREDS,
10 TO THE 3rd GIVES US THE THOUSANDS, AND SO ON.
HOWEVER, BASE 10 IS NOT THE ONLY OPTION WE HAVE.
PRACTICALLY ANY POSITIVE INTEGER GREATER THAN OR = 2
CAN BE USED AS A BASE FOR A NUMBER SYSTEM.
SUCH SYSTEMS CAN WORK JUST LIKE THE DECIMAL SYSTEM,
EXCEPT THE NUMBER OF SYMBOLS WILL BE DIFFERENT,
AND EACH POSITION WILL DEPEND ON THE BASE ITSELF.
FOR EXAMPLE, LET'S SUPPOSE WE ADOPT A BASE-5 SYSTEM.
THE ONLY MODERN DIGITS WE WOULD NEED FOR THIS SYSTEM
ARE ZERO, ONE, TWO, THREE, AND FOUR.
SO TO DETERMINE THE PLACE VALUES IN SUCH A SYSTEM,
WE WOULD TAKE THE BASE-5, START WITH 5 TO THE 0,
WHICH WOULD GIVE US THE ONES PLACE VALUE,
AND THEN START INCREASING THE EXPONENT BY ONE.
5 TO THE 1st WOULD GIVE US THE FIVES PLACE VALUE,
5 TO THE 2nd WOULD GIVE US THE 25S PLACE VALUE,
5 TO THE 3rd WOULD GIVE US THE 125S PLACE VALUE, AND SO ON.
SO BY KNOWING THE BASE
WE CAN DETERMINE THE VALUE OF EACH PLACE VALUE.
AS AN EXAMPLE, 4,322 WE KNOW IN BASE 10,
WHICH WE COULD INDICATE WITH A SMALL 10 HERE.
THE 4 MEANS 4 ONE THOUSANDS, THE 3 MEANS 3 ONE HUNDREDS,
THE 2 HERE MEANS 2 TENS, AND THIS 2 MEANS 2 ONES.
BUT IF WE CHANGE THIS TO BASE 5,
NOW THE 4 REPRESENTS 4 5 TO THE 3rds, OR 4 125S,
THE 3 INDICATES 3 5 TO THE 2nds, OR 25S,
THIS 2 HERE MEANS 2 5 TO THE 1sts, OR 2 5S,
AND THIS 2 MEANS 2 1S.
SO IF WE WANTED TO DETERMINE THE VALUE OF THIS NUMBER IN BASE 10,
WE COULD DO SO BY MULTIPLYING THE DIGIT BY ITS PLACE VALUE
AS WE SEE HERE.
SO IF WE SIMPLIFY ON THE RIGHT,
NOTICE HOW WE GET 587 IN BASE 10,
SO WE CAN SAY THAT 4,322 IN BASE 5 = 587 IN BASE 10.
BUT IN THIS LESSON WE'LL FOCUS ON HOW TO CONVERT A NUMBER
IN BASE 10 TO A DIFFERENT BASE.
NOW WE'RE GOING TO USE WHAT'S CALLED THE REMAINDER METHOD.
IN ANOTHER VIDEO WE'LL TAKE A LOOK AT HOW WE CAN USE DECIMALS
TO DO THESE CONVERSIONS USING A CALCULATOR.
SO FOR THE REMAINDER METHOD, TO CONVERT BASE 10 TO BASE B,
NUMBER ONE, WE'LL FIND THE HIGHEST POWER OF THE BASE
THAT WILL DIVIDE INTO THE GIVEN NUMBER AT LEAST ONCE,
AND THEN DIVIDE.
STEP TWO, WE'LL WRITE DOWN THE WHOLE NUMBER PART,
THEN USE THE REMAINDER FROM THE DIVISION IN THE NEXT STEP.
STEP THREE, WE'LL REPEAT STEP TWO,
DIVIDING BY THE NEXT HIGHEST POWER OF THE BASE B,
WRITING DOWN THE WHOLE NUMBER PART INCLUDING ZERO,
AND USING THE REMAINDER IN THE NEXT STEP.
STEP FOUR, CONTINUE
UNTIL THE REMAINDER IS SMALLER THAN THE BASE.
THIS LAST REMAINDER WILL BE THE ONES PLACE.
STEP FIVE, COLLECT ALL YOUR WHOLE NUMBER PARTS
TO GET THE NUMBER IN BASE-B NOTATION.
SO LET'S TAKE A LOOK AT TWO EXAMPLES.
WE WANT TO CONVERT THE BASE-10 NUMBER 83 TO BASE 2.
TO BEGIN, I'D LIKE TO DIVIDE OUT THE PLACE VALUES IN BASE 2.
SO WE'LL START WITH TWO TO THE ZERO POWER, WHICH = 1,
AND THEN WE'LL INCREASE THE EXPONENT BY ONE
TO FIND SEVERAL MORE PLACE VALUES.
SO 2 TO THE 1st = 2, 2 TO THE 2nd = 4,
2 TO THE 3rd = 8, 2 TO THE 4th = 16, AND SO ON.
THIS INDICATES THE PLACE VALUES FOR BASE 2.
AND NOW WE'RE LOOKING FOR THE HIGHEST POWER OF 2
THAT WILL DIVIDE INTO THE GIVE NUMBER, 83, AT LEAST ONCE.
NOTICE THAT THAT WOULD BE 2 TO THE 6th, OR 64,
BECAUSE 2 TO THE 7th WOULD BE TOO LARGE.
SO BECAUSE WE KNOW THAT TWO TO THE 6th
IS THE HIGHEST POWER OF 2 THAT DIVIDES AT LEAST ONCE INTO 83,
WE KNOW OUR NUMBER IN BASE 2 IS GOING TO HAVE ONE, TWO,
THREE, FOUR, FIVE, SIX, SEVEN DIGITS.
SO LET'S GO AHEAD AND LEAVE ROOM FOR ALL SEVEN DIGITS.
AND NOW WE'LL START DIVIDING.
WE'LL DIVIDE 83 BY 64.
64 DIVIDES ONCE INTO 83.
1 x 64 = 64.
NOTICE HERE WE HAVE A REMAINDER OF 19.
THE WHOLE NUMBER PART OF OUR QUOTIENT,
IN THIS CASE ONE, GIVES US THE DIGIT IN THE 2
TO THE 6th PLACE VALUE, WHICH IS HERE.
NOW WE REPEAT THE PROCESS WITH 19.
WE'RE LOOKING FOR THE HIGHEST POWER OF 2
THAT DIVIDES AT LEAST ONCE INTO 19.
NOTICE 2 TO THE 5th IS TOO LARGE,
WHICH MEANS WE'LL HAVE THE 0 IN THE 2 TO THE 5th PLACE VALUE.
AND THEN WE GO TO 2 TO THE 4th, WHICH IS 16.
16 WILL DIVIDE INTO 19 AT LEAST ONCE.
SO NOW WE'LL TAKE 19 AND DIVIDE BY 16.
IT GOES IN ONE TIME, AND NOW WE HAVE A REMAINDER OF 3.
SO THE WHOLE NUMBER 1 GIVES US THE DIGIT IN THE 2
TO THE 4th PLACE VALUE, WHICH IS HERE.
AND WE REPEAT THE PROCESS WITH 3.
WE'RE GOING TO FIND THE HIGHEST POWER OF 2
THAT WILL DIVIDE AT LEAST ONCE INTO 3.
WELL, 2 TO THE 3rd, OR 8, IS TOO LARGE,
WHICH MEANS WE'LL HAVE A 0 IN THE 2 TO THE 3rd PLACE VALUE,
WHICH IS HERE.
2 TO THE 2nd, OR 4, IS ALSO TOO LARGE,
SO WE'LL HAVE ANOTHER 0 IN THIS PLACE VALUE,
WHICH BRINGS US TO 2 TO THE 1st, OR 2.
SO 3 DIVIDED BY 2 = 1, 1 x 2 OF COURSE = 2,
WE SUBTRACT THE REMAINDER [IS] 1.
THIS 1 GOES INTO THE 2 TO THE 1st, OR THE 2 PLACE VALUE,
WHICH IS HERE.
THIS REMAINDER IS LESS THAN THE BASE,
WHICH GIVES US THE DIGIT IN THE ONES PLACE VALUE,
WHICH WOULD BE HERE.
SO WHAT WE FOUND IS 83 IN BASE 10 = 1010011 IN BASE 2.
OF COURSE IF WE WANT TO CHECK THIS
WE CAN GO IN THE REVERSE ORDER,
WHICH I'VE ALREADY DONE TO SAVE SOME TIME.
AND SO THIS ONE IS IN 2 TO THE 6th PLACE VALUE,
THIS ONE HERE IS IN THE 2 TO THE 4th PLACE VALUE,
THIS ONE IS IN THE 2 TO THE 1st PLACE VALUE,
AND FINALLY THIS ONE IS IN THE 2 TO THE 0 OR ONES PLACE VALUE.
[IF] WE FIND THE SUM OF THESE PRODUCTS,
IT DOES GIVE US 83 IN BASE 10.
LET'S TAKE A LOOK AT ONE MORE EXAMPLE.
WE WANT TO CONVERT THE BASE-10 NUMBER 2,973 TO BASE 6.
SO AGAIN, WE FIRST WANT TO FIND THE PLACE VALUES
BY STARTING WITH 6 TO THE 0, INCREASING THE EXPONENT BY 1,
TO FIND SEVERAL PLACE VALUES TO THE LEFT, WHICH WE SEE HERE--
6 TO THE 1st, 6 TO THE 2nd, 6 TO THE 3rd, 6 TO THE 4th,
AND SO ON.
I JUST NOTICED THIS SHOULD SAY,
"WE ONLY USE 0 THROUGH 5 FOR BASE 6, NOT BASE 2.
AND NOW WE BEGIN BY DETERMINING THE HIGHEST POWER OF 6
THAT WILL DIVIDE AT LEAST ONCE INTO 2,973,
WHICH WOULD BE 6 TO THE 4th,
BECAUSE 6 TO THE 5th WOULD BE TO LARGE,
WHICH MEANS OUR NUMBER IN BASE 6
IS GOING TO HAVE ONE, TWO, THREE, FOUR, FIVE DIGITS--
SO ONE, TWO, THREE, FOUR, FIVE.
NOW WE BEGIN BY DIVIDING 2,973 BY 6 TO THE 4th, OR 1,296.
THIS DIVIDES IN TWO TIMES.
2 x OUR DIVISOR = 2,592,
AND THIS DIFFERENCE WITH THE REMAINDER IS 381.
SO THE WHOLE NUMBER OF 2 GIVES US THE DIGIT IN THE 6
TO THE 4th PLACE VALUE, WHICH IS HERE.
NOW WE REPEAT THE PROCESS USING THE REMAINDER OF 381.
WE WANT TO FIND THE HIGHEST POWER OF 6
THAT'LL DIVIDE AT LEAST ONCE INTO 381.
AND NOTICE HOW 216 WOULD DIVIDE AT LEAST ONCE INTO 381.
SO NOW WE'LL DIVIDE BY 216 TO FIND OUR NEXT DIGIT.
216 GOES ONCE INTO 381, WE SUBTRACT--THE REMAINDER IS 165.
SO 1 IS OUR NEXT DIGIT.
NOW WE REPEAT AGAIN.
WE WANT TO DETERMINE THE HIGHEST POWER OF 6
THAT WOULD DIVIDE INTO 165 AT LEAST ONCE.
AND NOTICE HOW 6 TO THE 2nd, OR 36,
WILL DIVIDE AT LEAST ONCE INTO 165.
SO WE TAKE 165 AND DIVIDE BY 36.
36 GOES INTO 165 FOUR TIMES.
4 x 36 = 144, WITH A REMAINDER OF 21.
SO OUR NEXT DIGIT IS 4,
WHICH WILL BE IN THE 6 TO THE 2nd PLACE VALUE, OR HERE.
REPEAT USING 21.
6 DIVIDES AT LEAST ONCE INTO 21,
SO WE'LL TAKE 21 AND DIVIDE BY 6.
6 DIVIDES INTO 21 THREE TIMES.
6 X 3 = 18.
WE SUBTRACT--NOTICE HERE OUR REMAINDER IS 3,
LESS THAN OUR BASE OF 6, SO WE'RE ACTUALLY DONE.
THE NEXT DIGIT WOULD BE THIS 3,
THE WHOLE NUMBER OF OUR QUOTIENT,
AND THE LAST DIGIT IN THE ONES PLACE VALUE
IS OUR REMAINDER THAT'S LESS THAN THE BASE OF 3.
SO WE FOUND THAT 2,973 BASE 10 = 21,433 IN BASE 6.
AND AGAIN, TO VERIFY THIS WE CAN GO IN THE OPPOSITE DIRECTION.
WE CAN TAKE OUR NUMBER IN BASE 6 AND WRITE OUT EACH PLACE VALUE.
SO THIS 2 REPRESENTS 2 x 6 TO THE 4th,
THIS 1 REPRESENTS 1 x 6 TO THE 3rd,
THIS 4 REPRESENTS 4 x 6 TO THE 2nd,
THIS 3 REPRESENTS 3 x 6 TO THE 1st,
AND FINALLY THIS 3 REPRESENTS 3 x 6 TO THE 0.
SIMPLIFYING, WE CAN VERIFY THIS = 2,973 IN BASE 10.
I HOPE YOU FOUND THIS HELPFUL