-0.016 738 891 601 562 496 530 92 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.016 738 891 601 562 496 530 92(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
-0.016 738 891 601 562 496 530 92(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:

|-0.016 738 891 601 562 496 530 92| = 0.016 738 891 601 562 496 530 92


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: 0.016 738 891 601 562 496 530 92.

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) 0.016 738 891 601 562 496 530 92 × 2 = 0 + 0.033 477 783 203 124 993 061 84;
  • 2) 0.033 477 783 203 124 993 061 84 × 2 = 0 + 0.066 955 566 406 249 986 123 68;
  • 3) 0.066 955 566 406 249 986 123 68 × 2 = 0 + 0.133 911 132 812 499 972 247 36;
  • 4) 0.133 911 132 812 499 972 247 36 × 2 = 0 + 0.267 822 265 624 999 944 494 72;
  • 5) 0.267 822 265 624 999 944 494 72 × 2 = 0 + 0.535 644 531 249 999 888 989 44;
  • 6) 0.535 644 531 249 999 888 989 44 × 2 = 1 + 0.071 289 062 499 999 777 978 88;
  • 7) 0.071 289 062 499 999 777 978 88 × 2 = 0 + 0.142 578 124 999 999 555 957 76;
  • 8) 0.142 578 124 999 999 555 957 76 × 2 = 0 + 0.285 156 249 999 999 111 915 52;
  • 9) 0.285 156 249 999 999 111 915 52 × 2 = 0 + 0.570 312 499 999 998 223 831 04;
  • 10) 0.570 312 499 999 998 223 831 04 × 2 = 1 + 0.140 624 999 999 996 447 662 08;
  • 11) 0.140 624 999 999 996 447 662 08 × 2 = 0 + 0.281 249 999 999 992 895 324 16;
  • 12) 0.281 249 999 999 992 895 324 16 × 2 = 0 + 0.562 499 999 999 985 790 648 32;
  • 13) 0.562 499 999 999 985 790 648 32 × 2 = 1 + 0.124 999 999 999 971 581 296 64;
  • 14) 0.124 999 999 999 971 581 296 64 × 2 = 0 + 0.249 999 999 999 943 162 593 28;
  • 15) 0.249 999 999 999 943 162 593 28 × 2 = 0 + 0.499 999 999 999 886 325 186 56;
  • 16) 0.499 999 999 999 886 325 186 56 × 2 = 0 + 0.999 999 999 999 772 650 373 12;
  • 17) 0.999 999 999 999 772 650 373 12 × 2 = 1 + 0.999 999 999 999 545 300 746 24;
  • 18) 0.999 999 999 999 545 300 746 24 × 2 = 1 + 0.999 999 999 999 090 601 492 48;
  • 19) 0.999 999 999 999 090 601 492 48 × 2 = 1 + 0.999 999 999 998 181 202 984 96;
  • 20) 0.999 999 999 998 181 202 984 96 × 2 = 1 + 0.999 999 999 996 362 405 969 92;
  • 21) 0.999 999 999 996 362 405 969 92 × 2 = 1 + 0.999 999 999 992 724 811 939 84;
  • 22) 0.999 999 999 992 724 811 939 84 × 2 = 1 + 0.999 999 999 985 449 623 879 68;
  • 23) 0.999 999 999 985 449 623 879 68 × 2 = 1 + 0.999 999 999 970 899 247 759 36;
  • 24) 0.999 999 999 970 899 247 759 36 × 2 = 1 + 0.999 999 999 941 798 495 518 72;
  • 25) 0.999 999 999 941 798 495 518 72 × 2 = 1 + 0.999 999 999 883 596 991 037 44;
  • 26) 0.999 999 999 883 596 991 037 44 × 2 = 1 + 0.999 999 999 767 193 982 074 88;
  • 27) 0.999 999 999 767 193 982 074 88 × 2 = 1 + 0.999 999 999 534 387 964 149 76;
  • 28) 0.999 999 999 534 387 964 149 76 × 2 = 1 + 0.999 999 999 068 775 928 299 52;
  • 29) 0.999 999 999 068 775 928 299 52 × 2 = 1 + 0.999 999 998 137 551 856 599 04;
  • 30) 0.999 999 998 137 551 856 599 04 × 2 = 1 + 0.999 999 996 275 103 713 198 08;
  • 31) 0.999 999 996 275 103 713 198 08 × 2 = 1 + 0.999 999 992 550 207 426 396 16;
  • 32) 0.999 999 992 550 207 426 396 16 × 2 = 1 + 0.999 999 985 100 414 852 792 32;
  • 33) 0.999 999 985 100 414 852 792 32 × 2 = 1 + 0.999 999 970 200 829 705 584 64;
  • 34) 0.999 999 970 200 829 705 584 64 × 2 = 1 + 0.999 999 940 401 659 411 169 28;
  • 35) 0.999 999 940 401 659 411 169 28 × 2 = 1 + 0.999 999 880 803 318 822 338 56;
  • 36) 0.999 999 880 803 318 822 338 56 × 2 = 1 + 0.999 999 761 606 637 644 677 12;
  • 37) 0.999 999 761 606 637 644 677 12 × 2 = 1 + 0.999 999 523 213 275 289 354 24;
  • 38) 0.999 999 523 213 275 289 354 24 × 2 = 1 + 0.999 999 046 426 550 578 708 48;
  • 39) 0.999 999 046 426 550 578 708 48 × 2 = 1 + 0.999 998 092 853 101 157 416 96;
  • 40) 0.999 998 092 853 101 157 416 96 × 2 = 1 + 0.999 996 185 706 202 314 833 92;
  • 41) 0.999 996 185 706 202 314 833 92 × 2 = 1 + 0.999 992 371 412 404 629 667 84;
  • 42) 0.999 992 371 412 404 629 667 84 × 2 = 1 + 0.999 984 742 824 809 259 335 68;
  • 43) 0.999 984 742 824 809 259 335 68 × 2 = 1 + 0.999 969 485 649 618 518 671 36;
  • 44) 0.999 969 485 649 618 518 671 36 × 2 = 1 + 0.999 938 971 299 237 037 342 72;
  • 45) 0.999 938 971 299 237 037 342 72 × 2 = 1 + 0.999 877 942 598 474 074 685 44;
  • 46) 0.999 877 942 598 474 074 685 44 × 2 = 1 + 0.999 755 885 196 948 149 370 88;
  • 47) 0.999 755 885 196 948 149 370 88 × 2 = 1 + 0.999 511 770 393 896 298 741 76;
  • 48) 0.999 511 770 393 896 298 741 76 × 2 = 1 + 0.999 023 540 787 792 597 483 52;
  • 49) 0.999 023 540 787 792 597 483 52 × 2 = 1 + 0.998 047 081 575 585 194 967 04;
  • 50) 0.998 047 081 575 585 194 967 04 × 2 = 1 + 0.996 094 163 151 170 389 934 08;
  • 51) 0.996 094 163 151 170 389 934 08 × 2 = 1 + 0.992 188 326 302 340 779 868 16;
  • 52) 0.992 188 326 302 340 779 868 16 × 2 = 1 + 0.984 376 652 604 681 559 736 32;
  • 53) 0.984 376 652 604 681 559 736 32 × 2 = 1 + 0.968 753 305 209 363 119 472 64;
  • 54) 0.968 753 305 209 363 119 472 64 × 2 = 1 + 0.937 506 610 418 726 238 945 28;
  • 55) 0.937 506 610 418 726 238 945 28 × 2 = 1 + 0.875 013 220 837 452 477 890 56;
  • 56) 0.875 013 220 837 452 477 890 56 × 2 = 1 + 0.750 026 441 674 904 955 781 12;
  • 57) 0.750 026 441 674 904 955 781 12 × 2 = 1 + 0.500 052 883 349 809 911 562 24;
  • 58) 0.500 052 883 349 809 911 562 24 × 2 = 1 + 0.000 105 766 699 619 823 124 48;

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:


0.016 738 891 601 562 496 530 92(10) =

0.0000 0100 0100 1000 1111 1111 1111 1111 1111 1111 1111 1111 1111 1111 11(2)

6. Positive number before normalization:

0.016 738 891 601 562 496 530 92(10) =

0.0000 0100 0100 1000 1111 1111 1111 1111 1111 1111 1111 1111 1111 1111 11(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:


0.016 738 891 601 562 496 530 92(10) =

0.0000 0100 0100 1000 1111 1111 1111 1111 1111 1111 1111 1111 1111 1111 11(2) =

0.0000 0100 0100 1000 1111 1111 1111 1111 1111 1111 1111 1111 1111 1111 11(2) × 20 =

1.0001 0010 0011 1111 1111 1111 1111 1111 1111 1111 1111 1111 1111(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 1111 1111 1111


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 1111 1111 1111 =

0001 0010 0011 1111 1111 1111 1111 1111 1111 1111 1111 1111 1111


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 1111 1111 1111


Decimal number -0.016 738 891 601 562 496 530 92 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 1111 1111 1111

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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