-0.016 738 891 601 562 496 530 474 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.016 738 891 601 562 496 530 474(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
-0.016 738 891 601 562 496 530 474(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:

|-0.016 738 891 601 562 496 530 474| = 0.016 738 891 601 562 496 530 474


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: 0.016 738 891 601 562 496 530 474.

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) 0.016 738 891 601 562 496 530 474 × 2 = 0 + 0.033 477 783 203 124 993 060 948;
  • 2) 0.033 477 783 203 124 993 060 948 × 2 = 0 + 0.066 955 566 406 249 986 121 896;
  • 3) 0.066 955 566 406 249 986 121 896 × 2 = 0 + 0.133 911 132 812 499 972 243 792;
  • 4) 0.133 911 132 812 499 972 243 792 × 2 = 0 + 0.267 822 265 624 999 944 487 584;
  • 5) 0.267 822 265 624 999 944 487 584 × 2 = 0 + 0.535 644 531 249 999 888 975 168;
  • 6) 0.535 644 531 249 999 888 975 168 × 2 = 1 + 0.071 289 062 499 999 777 950 336;
  • 7) 0.071 289 062 499 999 777 950 336 × 2 = 0 + 0.142 578 124 999 999 555 900 672;
  • 8) 0.142 578 124 999 999 555 900 672 × 2 = 0 + 0.285 156 249 999 999 111 801 344;
  • 9) 0.285 156 249 999 999 111 801 344 × 2 = 0 + 0.570 312 499 999 998 223 602 688;
  • 10) 0.570 312 499 999 998 223 602 688 × 2 = 1 + 0.140 624 999 999 996 447 205 376;
  • 11) 0.140 624 999 999 996 447 205 376 × 2 = 0 + 0.281 249 999 999 992 894 410 752;
  • 12) 0.281 249 999 999 992 894 410 752 × 2 = 0 + 0.562 499 999 999 985 788 821 504;
  • 13) 0.562 499 999 999 985 788 821 504 × 2 = 1 + 0.124 999 999 999 971 577 643 008;
  • 14) 0.124 999 999 999 971 577 643 008 × 2 = 0 + 0.249 999 999 999 943 155 286 016;
  • 15) 0.249 999 999 999 943 155 286 016 × 2 = 0 + 0.499 999 999 999 886 310 572 032;
  • 16) 0.499 999 999 999 886 310 572 032 × 2 = 0 + 0.999 999 999 999 772 621 144 064;
  • 17) 0.999 999 999 999 772 621 144 064 × 2 = 1 + 0.999 999 999 999 545 242 288 128;
  • 18) 0.999 999 999 999 545 242 288 128 × 2 = 1 + 0.999 999 999 999 090 484 576 256;
  • 19) 0.999 999 999 999 090 484 576 256 × 2 = 1 + 0.999 999 999 998 180 969 152 512;
  • 20) 0.999 999 999 998 180 969 152 512 × 2 = 1 + 0.999 999 999 996 361 938 305 024;
  • 21) 0.999 999 999 996 361 938 305 024 × 2 = 1 + 0.999 999 999 992 723 876 610 048;
  • 22) 0.999 999 999 992 723 876 610 048 × 2 = 1 + 0.999 999 999 985 447 753 220 096;
  • 23) 0.999 999 999 985 447 753 220 096 × 2 = 1 + 0.999 999 999 970 895 506 440 192;
  • 24) 0.999 999 999 970 895 506 440 192 × 2 = 1 + 0.999 999 999 941 791 012 880 384;
  • 25) 0.999 999 999 941 791 012 880 384 × 2 = 1 + 0.999 999 999 883 582 025 760 768;
  • 26) 0.999 999 999 883 582 025 760 768 × 2 = 1 + 0.999 999 999 767 164 051 521 536;
  • 27) 0.999 999 999 767 164 051 521 536 × 2 = 1 + 0.999 999 999 534 328 103 043 072;
  • 28) 0.999 999 999 534 328 103 043 072 × 2 = 1 + 0.999 999 999 068 656 206 086 144;
  • 29) 0.999 999 999 068 656 206 086 144 × 2 = 1 + 0.999 999 998 137 312 412 172 288;
  • 30) 0.999 999 998 137 312 412 172 288 × 2 = 1 + 0.999 999 996 274 624 824 344 576;
  • 31) 0.999 999 996 274 624 824 344 576 × 2 = 1 + 0.999 999 992 549 249 648 689 152;
  • 32) 0.999 999 992 549 249 648 689 152 × 2 = 1 + 0.999 999 985 098 499 297 378 304;
  • 33) 0.999 999 985 098 499 297 378 304 × 2 = 1 + 0.999 999 970 196 998 594 756 608;
  • 34) 0.999 999 970 196 998 594 756 608 × 2 = 1 + 0.999 999 940 393 997 189 513 216;
  • 35) 0.999 999 940 393 997 189 513 216 × 2 = 1 + 0.999 999 880 787 994 379 026 432;
  • 36) 0.999 999 880 787 994 379 026 432 × 2 = 1 + 0.999 999 761 575 988 758 052 864;
  • 37) 0.999 999 761 575 988 758 052 864 × 2 = 1 + 0.999 999 523 151 977 516 105 728;
  • 38) 0.999 999 523 151 977 516 105 728 × 2 = 1 + 0.999 999 046 303 955 032 211 456;
  • 39) 0.999 999 046 303 955 032 211 456 × 2 = 1 + 0.999 998 092 607 910 064 422 912;
  • 40) 0.999 998 092 607 910 064 422 912 × 2 = 1 + 0.999 996 185 215 820 128 845 824;
  • 41) 0.999 996 185 215 820 128 845 824 × 2 = 1 + 0.999 992 370 431 640 257 691 648;
  • 42) 0.999 992 370 431 640 257 691 648 × 2 = 1 + 0.999 984 740 863 280 515 383 296;
  • 43) 0.999 984 740 863 280 515 383 296 × 2 = 1 + 0.999 969 481 726 561 030 766 592;
  • 44) 0.999 969 481 726 561 030 766 592 × 2 = 1 + 0.999 938 963 453 122 061 533 184;
  • 45) 0.999 938 963 453 122 061 533 184 × 2 = 1 + 0.999 877 926 906 244 123 066 368;
  • 46) 0.999 877 926 906 244 123 066 368 × 2 = 1 + 0.999 755 853 812 488 246 132 736;
  • 47) 0.999 755 853 812 488 246 132 736 × 2 = 1 + 0.999 511 707 624 976 492 265 472;
  • 48) 0.999 511 707 624 976 492 265 472 × 2 = 1 + 0.999 023 415 249 952 984 530 944;
  • 49) 0.999 023 415 249 952 984 530 944 × 2 = 1 + 0.998 046 830 499 905 969 061 888;
  • 50) 0.998 046 830 499 905 969 061 888 × 2 = 1 + 0.996 093 660 999 811 938 123 776;
  • 51) 0.996 093 660 999 811 938 123 776 × 2 = 1 + 0.992 187 321 999 623 876 247 552;
  • 52) 0.992 187 321 999 623 876 247 552 × 2 = 1 + 0.984 374 643 999 247 752 495 104;
  • 53) 0.984 374 643 999 247 752 495 104 × 2 = 1 + 0.968 749 287 998 495 504 990 208;
  • 54) 0.968 749 287 998 495 504 990 208 × 2 = 1 + 0.937 498 575 996 991 009 980 416;
  • 55) 0.937 498 575 996 991 009 980 416 × 2 = 1 + 0.874 997 151 993 982 019 960 832;
  • 56) 0.874 997 151 993 982 019 960 832 × 2 = 1 + 0.749 994 303 987 964 039 921 664;
  • 57) 0.749 994 303 987 964 039 921 664 × 2 = 1 + 0.499 988 607 975 928 079 843 328;
  • 58) 0.499 988 607 975 928 079 843 328 × 2 = 0 + 0.999 977 215 951 856 159 686 656;

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:


0.016 738 891 601 562 496 530 474(10) =

0.0000 0100 0100 1000 1111 1111 1111 1111 1111 1111 1111 1111 1111 1111 10(2)

6. Positive number before normalization:

0.016 738 891 601 562 496 530 474(10) =

0.0000 0100 0100 1000 1111 1111 1111 1111 1111 1111 1111 1111 1111 1111 10(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:


0.016 738 891 601 562 496 530 474(10) =

0.0000 0100 0100 1000 1111 1111 1111 1111 1111 1111 1111 1111 1111 1111 10(2) =

0.0000 0100 0100 1000 1111 1111 1111 1111 1111 1111 1111 1111 1111 1111 10(2) × 20 =

1.0001 0010 0011 1111 1111 1111 1111 1111 1111 1111 1111 1111 1110(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 1110


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 1110 =

0001 0010 0011 1111 1111 1111 1111 1111 1111 1111 1111 1111 1110


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 1110


Decimal number -0.016 738 891 601 562 496 530 474 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 1110

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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