-0.016 738 891 601 562 496 530 293 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.016 738 891 601 562 496 530 293(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
-0.016 738 891 601 562 496 530 293(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:

|-0.016 738 891 601 562 496 530 293| = 0.016 738 891 601 562 496 530 293


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: 0.016 738 891 601 562 496 530 293.

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) 0.016 738 891 601 562 496 530 293 × 2 = 0 + 0.033 477 783 203 124 993 060 586;
  • 2) 0.033 477 783 203 124 993 060 586 × 2 = 0 + 0.066 955 566 406 249 986 121 172;
  • 3) 0.066 955 566 406 249 986 121 172 × 2 = 0 + 0.133 911 132 812 499 972 242 344;
  • 4) 0.133 911 132 812 499 972 242 344 × 2 = 0 + 0.267 822 265 624 999 944 484 688;
  • 5) 0.267 822 265 624 999 944 484 688 × 2 = 0 + 0.535 644 531 249 999 888 969 376;
  • 6) 0.535 644 531 249 999 888 969 376 × 2 = 1 + 0.071 289 062 499 999 777 938 752;
  • 7) 0.071 289 062 499 999 777 938 752 × 2 = 0 + 0.142 578 124 999 999 555 877 504;
  • 8) 0.142 578 124 999 999 555 877 504 × 2 = 0 + 0.285 156 249 999 999 111 755 008;
  • 9) 0.285 156 249 999 999 111 755 008 × 2 = 0 + 0.570 312 499 999 998 223 510 016;
  • 10) 0.570 312 499 999 998 223 510 016 × 2 = 1 + 0.140 624 999 999 996 447 020 032;
  • 11) 0.140 624 999 999 996 447 020 032 × 2 = 0 + 0.281 249 999 999 992 894 040 064;
  • 12) 0.281 249 999 999 992 894 040 064 × 2 = 0 + 0.562 499 999 999 985 788 080 128;
  • 13) 0.562 499 999 999 985 788 080 128 × 2 = 1 + 0.124 999 999 999 971 576 160 256;
  • 14) 0.124 999 999 999 971 576 160 256 × 2 = 0 + 0.249 999 999 999 943 152 320 512;
  • 15) 0.249 999 999 999 943 152 320 512 × 2 = 0 + 0.499 999 999 999 886 304 641 024;
  • 16) 0.499 999 999 999 886 304 641 024 × 2 = 0 + 0.999 999 999 999 772 609 282 048;
  • 17) 0.999 999 999 999 772 609 282 048 × 2 = 1 + 0.999 999 999 999 545 218 564 096;
  • 18) 0.999 999 999 999 545 218 564 096 × 2 = 1 + 0.999 999 999 999 090 437 128 192;
  • 19) 0.999 999 999 999 090 437 128 192 × 2 = 1 + 0.999 999 999 998 180 874 256 384;
  • 20) 0.999 999 999 998 180 874 256 384 × 2 = 1 + 0.999 999 999 996 361 748 512 768;
  • 21) 0.999 999 999 996 361 748 512 768 × 2 = 1 + 0.999 999 999 992 723 497 025 536;
  • 22) 0.999 999 999 992 723 497 025 536 × 2 = 1 + 0.999 999 999 985 446 994 051 072;
  • 23) 0.999 999 999 985 446 994 051 072 × 2 = 1 + 0.999 999 999 970 893 988 102 144;
  • 24) 0.999 999 999 970 893 988 102 144 × 2 = 1 + 0.999 999 999 941 787 976 204 288;
  • 25) 0.999 999 999 941 787 976 204 288 × 2 = 1 + 0.999 999 999 883 575 952 408 576;
  • 26) 0.999 999 999 883 575 952 408 576 × 2 = 1 + 0.999 999 999 767 151 904 817 152;
  • 27) 0.999 999 999 767 151 904 817 152 × 2 = 1 + 0.999 999 999 534 303 809 634 304;
  • 28) 0.999 999 999 534 303 809 634 304 × 2 = 1 + 0.999 999 999 068 607 619 268 608;
  • 29) 0.999 999 999 068 607 619 268 608 × 2 = 1 + 0.999 999 998 137 215 238 537 216;
  • 30) 0.999 999 998 137 215 238 537 216 × 2 = 1 + 0.999 999 996 274 430 477 074 432;
  • 31) 0.999 999 996 274 430 477 074 432 × 2 = 1 + 0.999 999 992 548 860 954 148 864;
  • 32) 0.999 999 992 548 860 954 148 864 × 2 = 1 + 0.999 999 985 097 721 908 297 728;
  • 33) 0.999 999 985 097 721 908 297 728 × 2 = 1 + 0.999 999 970 195 443 816 595 456;
  • 34) 0.999 999 970 195 443 816 595 456 × 2 = 1 + 0.999 999 940 390 887 633 190 912;
  • 35) 0.999 999 940 390 887 633 190 912 × 2 = 1 + 0.999 999 880 781 775 266 381 824;
  • 36) 0.999 999 880 781 775 266 381 824 × 2 = 1 + 0.999 999 761 563 550 532 763 648;
  • 37) 0.999 999 761 563 550 532 763 648 × 2 = 1 + 0.999 999 523 127 101 065 527 296;
  • 38) 0.999 999 523 127 101 065 527 296 × 2 = 1 + 0.999 999 046 254 202 131 054 592;
  • 39) 0.999 999 046 254 202 131 054 592 × 2 = 1 + 0.999 998 092 508 404 262 109 184;
  • 40) 0.999 998 092 508 404 262 109 184 × 2 = 1 + 0.999 996 185 016 808 524 218 368;
  • 41) 0.999 996 185 016 808 524 218 368 × 2 = 1 + 0.999 992 370 033 617 048 436 736;
  • 42) 0.999 992 370 033 617 048 436 736 × 2 = 1 + 0.999 984 740 067 234 096 873 472;
  • 43) 0.999 984 740 067 234 096 873 472 × 2 = 1 + 0.999 969 480 134 468 193 746 944;
  • 44) 0.999 969 480 134 468 193 746 944 × 2 = 1 + 0.999 938 960 268 936 387 493 888;
  • 45) 0.999 938 960 268 936 387 493 888 × 2 = 1 + 0.999 877 920 537 872 774 987 776;
  • 46) 0.999 877 920 537 872 774 987 776 × 2 = 1 + 0.999 755 841 075 745 549 975 552;
  • 47) 0.999 755 841 075 745 549 975 552 × 2 = 1 + 0.999 511 682 151 491 099 951 104;
  • 48) 0.999 511 682 151 491 099 951 104 × 2 = 1 + 0.999 023 364 302 982 199 902 208;
  • 49) 0.999 023 364 302 982 199 902 208 × 2 = 1 + 0.998 046 728 605 964 399 804 416;
  • 50) 0.998 046 728 605 964 399 804 416 × 2 = 1 + 0.996 093 457 211 928 799 608 832;
  • 51) 0.996 093 457 211 928 799 608 832 × 2 = 1 + 0.992 186 914 423 857 599 217 664;
  • 52) 0.992 186 914 423 857 599 217 664 × 2 = 1 + 0.984 373 828 847 715 198 435 328;
  • 53) 0.984 373 828 847 715 198 435 328 × 2 = 1 + 0.968 747 657 695 430 396 870 656;
  • 54) 0.968 747 657 695 430 396 870 656 × 2 = 1 + 0.937 495 315 390 860 793 741 312;
  • 55) 0.937 495 315 390 860 793 741 312 × 2 = 1 + 0.874 990 630 781 721 587 482 624;
  • 56) 0.874 990 630 781 721 587 482 624 × 2 = 1 + 0.749 981 261 563 443 174 965 248;
  • 57) 0.749 981 261 563 443 174 965 248 × 2 = 1 + 0.499 962 523 126 886 349 930 496;
  • 58) 0.499 962 523 126 886 349 930 496 × 2 = 0 + 0.999 925 046 253 772 699 860 992;

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:


0.016 738 891 601 562 496 530 293(10) =

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

6. Positive number before normalization:

0.016 738 891 601 562 496 530 293(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:


0.016 738 891 601 562 496 530 293(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 -0.016 738 891 601 562 496 530 293 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