-0.016 738 891 601 561 39 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.016 738 891 601 561 39(10) to 64 bit double precision IEEE 754 binary floating point representation standard (1 bit for sign, 11 bits for exponent, 52 bits for mantissa)

What are the steps to convert decimal number
-0.016 738 891 601 561 39(10) to 64 bit double precision IEEE 754 binary floating point representation (1 bit for sign, 11 bits for exponent, 52 bits for mantissa)

1. Start with the positive version of the number:

|-0.016 738 891 601 561 39| = 0.016 738 891 601 561 39


2. First, convert to binary (in base 2) the integer part: 0.
Divide the number repeatedly by 2.

Keep track of each remainder.

We stop when we get a quotient that is equal to zero.


  • division = quotient + remainder;
  • 0 ÷ 2 = 0 + 0;

3. Construct the base 2 representation of the integer part of the number.

Take all the remainders starting from the bottom of the list constructed above.

0(10) =

0(2)


4. Convert to binary (base 2) the fractional part: 0.016 738 891 601 561 39.

Multiply it repeatedly by 2.

Keep track of each integer part of the results.

Stop when we get a fractional part that is equal to zero.


  • #) multiplying = integer + fractional part;
  • 1) 0.016 738 891 601 561 39 × 2 = 0 + 0.033 477 783 203 122 78;
  • 2) 0.033 477 783 203 122 78 × 2 = 0 + 0.066 955 566 406 245 56;
  • 3) 0.066 955 566 406 245 56 × 2 = 0 + 0.133 911 132 812 491 12;
  • 4) 0.133 911 132 812 491 12 × 2 = 0 + 0.267 822 265 624 982 24;
  • 5) 0.267 822 265 624 982 24 × 2 = 0 + 0.535 644 531 249 964 48;
  • 6) 0.535 644 531 249 964 48 × 2 = 1 + 0.071 289 062 499 928 96;
  • 7) 0.071 289 062 499 928 96 × 2 = 0 + 0.142 578 124 999 857 92;
  • 8) 0.142 578 124 999 857 92 × 2 = 0 + 0.285 156 249 999 715 84;
  • 9) 0.285 156 249 999 715 84 × 2 = 0 + 0.570 312 499 999 431 68;
  • 10) 0.570 312 499 999 431 68 × 2 = 1 + 0.140 624 999 998 863 36;
  • 11) 0.140 624 999 998 863 36 × 2 = 0 + 0.281 249 999 997 726 72;
  • 12) 0.281 249 999 997 726 72 × 2 = 0 + 0.562 499 999 995 453 44;
  • 13) 0.562 499 999 995 453 44 × 2 = 1 + 0.124 999 999 990 906 88;
  • 14) 0.124 999 999 990 906 88 × 2 = 0 + 0.249 999 999 981 813 76;
  • 15) 0.249 999 999 981 813 76 × 2 = 0 + 0.499 999 999 963 627 52;
  • 16) 0.499 999 999 963 627 52 × 2 = 0 + 0.999 999 999 927 255 04;
  • 17) 0.999 999 999 927 255 04 × 2 = 1 + 0.999 999 999 854 510 08;
  • 18) 0.999 999 999 854 510 08 × 2 = 1 + 0.999 999 999 709 020 16;
  • 19) 0.999 999 999 709 020 16 × 2 = 1 + 0.999 999 999 418 040 32;
  • 20) 0.999 999 999 418 040 32 × 2 = 1 + 0.999 999 998 836 080 64;
  • 21) 0.999 999 998 836 080 64 × 2 = 1 + 0.999 999 997 672 161 28;
  • 22) 0.999 999 997 672 161 28 × 2 = 1 + 0.999 999 995 344 322 56;
  • 23) 0.999 999 995 344 322 56 × 2 = 1 + 0.999 999 990 688 645 12;
  • 24) 0.999 999 990 688 645 12 × 2 = 1 + 0.999 999 981 377 290 24;
  • 25) 0.999 999 981 377 290 24 × 2 = 1 + 0.999 999 962 754 580 48;
  • 26) 0.999 999 962 754 580 48 × 2 = 1 + 0.999 999 925 509 160 96;
  • 27) 0.999 999 925 509 160 96 × 2 = 1 + 0.999 999 851 018 321 92;
  • 28) 0.999 999 851 018 321 92 × 2 = 1 + 0.999 999 702 036 643 84;
  • 29) 0.999 999 702 036 643 84 × 2 = 1 + 0.999 999 404 073 287 68;
  • 30) 0.999 999 404 073 287 68 × 2 = 1 + 0.999 998 808 146 575 36;
  • 31) 0.999 998 808 146 575 36 × 2 = 1 + 0.999 997 616 293 150 72;
  • 32) 0.999 997 616 293 150 72 × 2 = 1 + 0.999 995 232 586 301 44;
  • 33) 0.999 995 232 586 301 44 × 2 = 1 + 0.999 990 465 172 602 88;
  • 34) 0.999 990 465 172 602 88 × 2 = 1 + 0.999 980 930 345 205 76;
  • 35) 0.999 980 930 345 205 76 × 2 = 1 + 0.999 961 860 690 411 52;
  • 36) 0.999 961 860 690 411 52 × 2 = 1 + 0.999 923 721 380 823 04;
  • 37) 0.999 923 721 380 823 04 × 2 = 1 + 0.999 847 442 761 646 08;
  • 38) 0.999 847 442 761 646 08 × 2 = 1 + 0.999 694 885 523 292 16;
  • 39) 0.999 694 885 523 292 16 × 2 = 1 + 0.999 389 771 046 584 32;
  • 40) 0.999 389 771 046 584 32 × 2 = 1 + 0.998 779 542 093 168 64;
  • 41) 0.998 779 542 093 168 64 × 2 = 1 + 0.997 559 084 186 337 28;
  • 42) 0.997 559 084 186 337 28 × 2 = 1 + 0.995 118 168 372 674 56;
  • 43) 0.995 118 168 372 674 56 × 2 = 1 + 0.990 236 336 745 349 12;
  • 44) 0.990 236 336 745 349 12 × 2 = 1 + 0.980 472 673 490 698 24;
  • 45) 0.980 472 673 490 698 24 × 2 = 1 + 0.960 945 346 981 396 48;
  • 46) 0.960 945 346 981 396 48 × 2 = 1 + 0.921 890 693 962 792 96;
  • 47) 0.921 890 693 962 792 96 × 2 = 1 + 0.843 781 387 925 585 92;
  • 48) 0.843 781 387 925 585 92 × 2 = 1 + 0.687 562 775 851 171 84;
  • 49) 0.687 562 775 851 171 84 × 2 = 1 + 0.375 125 551 702 343 68;
  • 50) 0.375 125 551 702 343 68 × 2 = 0 + 0.750 251 103 404 687 36;
  • 51) 0.750 251 103 404 687 36 × 2 = 1 + 0.500 502 206 809 374 72;
  • 52) 0.500 502 206 809 374 72 × 2 = 1 + 0.001 004 413 618 749 44;
  • 53) 0.001 004 413 618 749 44 × 2 = 0 + 0.002 008 827 237 498 88;
  • 54) 0.002 008 827 237 498 88 × 2 = 0 + 0.004 017 654 474 997 76;
  • 55) 0.004 017 654 474 997 76 × 2 = 0 + 0.008 035 308 949 995 52;
  • 56) 0.008 035 308 949 995 52 × 2 = 0 + 0.016 070 617 899 991 04;
  • 57) 0.016 070 617 899 991 04 × 2 = 0 + 0.032 141 235 799 982 08;
  • 58) 0.032 141 235 799 982 08 × 2 = 0 + 0.064 282 471 599 964 16;

We didn't get any fractional part that was equal to zero. But we had enough iterations (over Mantissa limit) and at least one integer that was different from zero => FULL STOP (Losing precision - the converted number we get in the end will be just a very good approximation of the initial one).


5. Construct the base 2 representation of the fractional part of the number.

Take all the integer parts of the multiplying operations, starting from the top of the constructed list above:


0.016 738 891 601 561 39(10) =

0.0000 0100 0100 1000 1111 1111 1111 1111 1111 1111 1111 1111 1011 0000 00(2)

6. Positive number before normalization:

0.016 738 891 601 561 39(10) =

0.0000 0100 0100 1000 1111 1111 1111 1111 1111 1111 1111 1111 1011 0000 00(2)

7. Normalize the binary representation of the number.

Shift the decimal mark 6 positions to the right, so that only one non zero digit remains to the left of it:


0.016 738 891 601 561 39(10) =

0.0000 0100 0100 1000 1111 1111 1111 1111 1111 1111 1111 1111 1011 0000 00(2) =

0.0000 0100 0100 1000 1111 1111 1111 1111 1111 1111 1111 1111 1011 0000 00(2) × 20 =

1.0001 0010 0011 1111 1111 1111 1111 1111 1111 1111 1110 1100 0000(2) × 2-6


8. Up to this moment, there are the following elements that would feed into the 64 bit double precision IEEE 754 binary floating point representation:

Sign 1 (a negative number)

Exponent (unadjusted): -6

Mantissa (not normalized):
1.0001 0010 0011 1111 1111 1111 1111 1111 1111 1111 1110 1100 0000


9. Adjust the exponent.

Use the 11 bit excess/bias notation:


Exponent (adjusted) =

Exponent (unadjusted) + 2(11-1) - 1 =

-6 + 2(11-1) - 1 =

(-6 + 1 023)(10) =

1 017(10)


10. Convert the adjusted exponent from the decimal (base 10) to 11 bit binary.

Use the same technique of repeatedly dividing by 2:


  • division = quotient + remainder;
  • 1 017 ÷ 2 = 508 + 1;
  • 508 ÷ 2 = 254 + 0;
  • 254 ÷ 2 = 127 + 0;
  • 127 ÷ 2 = 63 + 1;
  • 63 ÷ 2 = 31 + 1;
  • 31 ÷ 2 = 15 + 1;
  • 15 ÷ 2 = 7 + 1;
  • 7 ÷ 2 = 3 + 1;
  • 3 ÷ 2 = 1 + 1;
  • 1 ÷ 2 = 0 + 1;

11. Construct the base 2 representation of the adjusted exponent.

Take all the remainders starting from the bottom of the list constructed above.


Exponent (adjusted) =

1017(10) =

011 1111 1001(2)


12. Normalize the mantissa.

a) Remove the leading (the leftmost) bit, since it's allways 1, and the decimal point, if the case.

b) Adjust its length to 52 bits, only if necessary (not the case here).


Mantissa (normalized) =

1. 0001 0010 0011 1111 1111 1111 1111 1111 1111 1111 1110 1100 0000 =

0001 0010 0011 1111 1111 1111 1111 1111 1111 1111 1110 1100 0000


13. The three elements that make up the number's 64 bit double precision IEEE 754 binary floating point representation:

Sign (1 bit) =
1 (a negative number)

Exponent (11 bits) =
011 1111 1001

Mantissa (52 bits) =
0001 0010 0011 1111 1111 1111 1111 1111 1111 1111 1110 1100 0000


Decimal number -0.016 738 891 601 561 39 converted to 64 bit double precision IEEE 754 binary floating point representation:

1 - 011 1111 1001 - 0001 0010 0011 1111 1111 1111 1111 1111 1111 1111 1110 1100 0000

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How to convert numbers from the decimal system (base ten) to 64 bit double precision IEEE 754 binary floating point standard

Follow the steps below to convert a base 10 decimal number to 64 bit double precision IEEE 754 binary floating point:

  • 1. If the number to be converted is negative, start with its the positive version.
  • 2. First convert the integer part. Divide repeatedly by 2 the positive representation of the integer number that is to be converted to binary, until we get a quotient that is equal to zero, keeping track of each remainder.
  • 3. Construct the base 2 representation of the positive integer part of the number, by taking all the remainders from the previous operations, starting from the bottom of the list constructed above. Thus, the last remainder of the divisions becomes the first symbol (the leftmost) of the base two number, while the first remainder becomes the last symbol (the rightmost).
  • 4. Then convert the fractional part. Multiply the number repeatedly by 2, until we get a fractional part that is equal to zero, keeping track of each integer part of the results.
  • 5. Construct the base 2 representation of the fractional part of the number, by taking all the integer parts of the multiplying operations, starting from the top of the list constructed above (they should appear in the binary representation, from left to right, in the order they have been calculated).
  • 6. Normalize the binary representation of the number, shifting the decimal mark (the decimal point) "n" positions either to the left, or to the right, so that only one non zero digit remains to the left of the decimal mark.
  • 7. Adjust the exponent in 11 bit excess/bias notation and then convert it from decimal (base 10) to 11 bit binary, by using the same technique of repeatedly dividing by 2, as shown above:
    Exponent (adjusted) = Exponent (unadjusted) + 2(11-1) - 1
  • 8. Normalize mantissa, remove the leading (leftmost) bit, since it's allways '1' (and the decimal mark, if the case) and adjust its length to 52 bits, either by removing the excess bits from the right (losing precision...) or by adding extra bits set on '0' to the right.
  • 9. Sign (it takes 1 bit) is either 1 for a negative or 0 for a positive number.

Example: convert the negative number -31.640 215 from the decimal system (base ten) to 64 bit double precision IEEE 754 binary floating point:

  • 1. Start with the positive version of the number:

    |-31.640 215| = 31.640 215

  • 2. First convert the integer part, 31. Divide it repeatedly by 2, keeping track of each remainder, until we get a quotient that is equal to zero:
    • division = quotient + remainder;
    • 31 ÷ 2 = 15 + 1;
    • 15 ÷ 2 = 7 + 1;
    • 7 ÷ 2 = 3 + 1;
    • 3 ÷ 2 = 1 + 1;
    • 1 ÷ 2 = 0 + 1;
    • We have encountered a quotient that is ZERO => FULL STOP
  • 3. Construct the base 2 representation of the integer part of the number by taking all the remainders of the previous dividing operations, starting from the bottom of the list constructed above:

    31(10) = 1 1111(2)

  • 4. Then, convert the fractional part, 0.640 215. Multiply repeatedly by 2, keeping track of each integer part of the results, until we get a fractional part that is equal to zero:
    • #) multiplying = integer + fractional part;
    • 1) 0.640 215 × 2 = 1 + 0.280 43;
    • 2) 0.280 43 × 2 = 0 + 0.560 86;
    • 3) 0.560 86 × 2 = 1 + 0.121 72;
    • 4) 0.121 72 × 2 = 0 + 0.243 44;
    • 5) 0.243 44 × 2 = 0 + 0.486 88;
    • 6) 0.486 88 × 2 = 0 + 0.973 76;
    • 7) 0.973 76 × 2 = 1 + 0.947 52;
    • 8) 0.947 52 × 2 = 1 + 0.895 04;
    • 9) 0.895 04 × 2 = 1 + 0.790 08;
    • 10) 0.790 08 × 2 = 1 + 0.580 16;
    • 11) 0.580 16 × 2 = 1 + 0.160 32;
    • 12) 0.160 32 × 2 = 0 + 0.320 64;
    • 13) 0.320 64 × 2 = 0 + 0.641 28;
    • 14) 0.641 28 × 2 = 1 + 0.282 56;
    • 15) 0.282 56 × 2 = 0 + 0.565 12;
    • 16) 0.565 12 × 2 = 1 + 0.130 24;
    • 17) 0.130 24 × 2 = 0 + 0.260 48;
    • 18) 0.260 48 × 2 = 0 + 0.520 96;
    • 19) 0.520 96 × 2 = 1 + 0.041 92;
    • 20) 0.041 92 × 2 = 0 + 0.083 84;
    • 21) 0.083 84 × 2 = 0 + 0.167 68;
    • 22) 0.167 68 × 2 = 0 + 0.335 36;
    • 23) 0.335 36 × 2 = 0 + 0.670 72;
    • 24) 0.670 72 × 2 = 1 + 0.341 44;
    • 25) 0.341 44 × 2 = 0 + 0.682 88;
    • 26) 0.682 88 × 2 = 1 + 0.365 76;
    • 27) 0.365 76 × 2 = 0 + 0.731 52;
    • 28) 0.731 52 × 2 = 1 + 0.463 04;
    • 29) 0.463 04 × 2 = 0 + 0.926 08;
    • 30) 0.926 08 × 2 = 1 + 0.852 16;
    • 31) 0.852 16 × 2 = 1 + 0.704 32;
    • 32) 0.704 32 × 2 = 1 + 0.408 64;
    • 33) 0.408 64 × 2 = 0 + 0.817 28;
    • 34) 0.817 28 × 2 = 1 + 0.634 56;
    • 35) 0.634 56 × 2 = 1 + 0.269 12;
    • 36) 0.269 12 × 2 = 0 + 0.538 24;
    • 37) 0.538 24 × 2 = 1 + 0.076 48;
    • 38) 0.076 48 × 2 = 0 + 0.152 96;
    • 39) 0.152 96 × 2 = 0 + 0.305 92;
    • 40) 0.305 92 × 2 = 0 + 0.611 84;
    • 41) 0.611 84 × 2 = 1 + 0.223 68;
    • 42) 0.223 68 × 2 = 0 + 0.447 36;
    • 43) 0.447 36 × 2 = 0 + 0.894 72;
    • 44) 0.894 72 × 2 = 1 + 0.789 44;
    • 45) 0.789 44 × 2 = 1 + 0.578 88;
    • 46) 0.578 88 × 2 = 1 + 0.157 76;
    • 47) 0.157 76 × 2 = 0 + 0.315 52;
    • 48) 0.315 52 × 2 = 0 + 0.631 04;
    • 49) 0.631 04 × 2 = 1 + 0.262 08;
    • 50) 0.262 08 × 2 = 0 + 0.524 16;
    • 51) 0.524 16 × 2 = 1 + 0.048 32;
    • 52) 0.048 32 × 2 = 0 + 0.096 64;
    • 53) 0.096 64 × 2 = 0 + 0.193 28;
    • We didn't get any fractional part that was equal to zero. But we had enough iterations (over Mantissa limit = 52) and at least one integer part that was different from zero => FULL STOP (losing precision...).
  • 5. Construct the base 2 representation of the fractional part of the number, by taking all the integer parts of the previous multiplying operations, starting from the top of the constructed list above:

    0.640 215(10) = 0.1010 0011 1110 0101 0010 0001 0101 0111 0110 1000 1001 1100 1010 0(2)

  • 6. Summarizing - the positive number before normalization:

    31.640 215(10) = 1 1111.1010 0011 1110 0101 0010 0001 0101 0111 0110 1000 1001 1100 1010 0(2)

  • 7. Normalize the binary representation of the number, shifting the decimal mark 4 positions to the left so that only one non-zero digit stays to the left of the decimal mark:

    31.640 215(10) =
    1 1111.1010 0011 1110 0101 0010 0001 0101 0111 0110 1000 1001 1100 1010 0(2) =
    1 1111.1010 0011 1110 0101 0010 0001 0101 0111 0110 1000 1001 1100 1010 0(2) × 20 =
    1.1111 1010 0011 1110 0101 0010 0001 0101 0111 0110 1000 1001 1100 1010 0(2) × 24

  • 8. Up to this moment, there are the following elements that would feed into the 64 bit double precision IEEE 754 binary floating point representation:

    Sign: 1 (a negative number)

    Exponent (unadjusted): 4

    Mantissa (not-normalized): 1.1111 1010 0011 1110 0101 0010 0001 0101 0111 0110 1000 1001 1100 1010 0

  • 9. Adjust the exponent in 11 bit excess/bias notation and then convert it from decimal (base 10) to 11 bit binary (base 2), by using the same technique of repeatedly dividing it by 2, as shown above:

    Exponent (adjusted) = Exponent (unadjusted) + 2(11-1) - 1 = (4 + 1023)(10) = 1027(10) =
    100 0000 0011(2)

  • 10. Normalize mantissa, remove the leading (leftmost) bit, since it's allways '1' (and the decimal sign) and adjust its length to 52 bits, by removing the excess bits, from the right (losing precision...):

    Mantissa (not-normalized): 1.1111 1010 0011 1110 0101 0010 0001 0101 0111 0110 1000 1001 1100 1010 0

    Mantissa (normalized): 1111 1010 0011 1110 0101 0010 0001 0101 0111 0110 1000 1001 1100

  • Conclusion:

    Sign (1 bit) = 1 (a negative number)

    Exponent (8 bits) = 100 0000 0011

    Mantissa (52 bits) = 1111 1010 0011 1110 0101 0010 0001 0101 0111 0110 1000 1001 1100

  • Number -31.640 215, converted from decimal system (base 10) to 64 bit double precision IEEE 754 binary floating point =
    1 - 100 0000 0011 - 1111 1010 0011 1110 0101 0010 0001 0101 0111 0110 1000 1001 1100