-0.016 738 891 601 562 496 808 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.016 738 891 601 562 496 808(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
-0.016 738 891 601 562 496 808(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:

|-0.016 738 891 601 562 496 808| = 0.016 738 891 601 562 496 808


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: 0.016 738 891 601 562 496 808.

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) 0.016 738 891 601 562 496 808 × 2 = 0 + 0.033 477 783 203 124 993 616;
  • 2) 0.033 477 783 203 124 993 616 × 2 = 0 + 0.066 955 566 406 249 987 232;
  • 3) 0.066 955 566 406 249 987 232 × 2 = 0 + 0.133 911 132 812 499 974 464;
  • 4) 0.133 911 132 812 499 974 464 × 2 = 0 + 0.267 822 265 624 999 948 928;
  • 5) 0.267 822 265 624 999 948 928 × 2 = 0 + 0.535 644 531 249 999 897 856;
  • 6) 0.535 644 531 249 999 897 856 × 2 = 1 + 0.071 289 062 499 999 795 712;
  • 7) 0.071 289 062 499 999 795 712 × 2 = 0 + 0.142 578 124 999 999 591 424;
  • 8) 0.142 578 124 999 999 591 424 × 2 = 0 + 0.285 156 249 999 999 182 848;
  • 9) 0.285 156 249 999 999 182 848 × 2 = 0 + 0.570 312 499 999 998 365 696;
  • 10) 0.570 312 499 999 998 365 696 × 2 = 1 + 0.140 624 999 999 996 731 392;
  • 11) 0.140 624 999 999 996 731 392 × 2 = 0 + 0.281 249 999 999 993 462 784;
  • 12) 0.281 249 999 999 993 462 784 × 2 = 0 + 0.562 499 999 999 986 925 568;
  • 13) 0.562 499 999 999 986 925 568 × 2 = 1 + 0.124 999 999 999 973 851 136;
  • 14) 0.124 999 999 999 973 851 136 × 2 = 0 + 0.249 999 999 999 947 702 272;
  • 15) 0.249 999 999 999 947 702 272 × 2 = 0 + 0.499 999 999 999 895 404 544;
  • 16) 0.499 999 999 999 895 404 544 × 2 = 0 + 0.999 999 999 999 790 809 088;
  • 17) 0.999 999 999 999 790 809 088 × 2 = 1 + 0.999 999 999 999 581 618 176;
  • 18) 0.999 999 999 999 581 618 176 × 2 = 1 + 0.999 999 999 999 163 236 352;
  • 19) 0.999 999 999 999 163 236 352 × 2 = 1 + 0.999 999 999 998 326 472 704;
  • 20) 0.999 999 999 998 326 472 704 × 2 = 1 + 0.999 999 999 996 652 945 408;
  • 21) 0.999 999 999 996 652 945 408 × 2 = 1 + 0.999 999 999 993 305 890 816;
  • 22) 0.999 999 999 993 305 890 816 × 2 = 1 + 0.999 999 999 986 611 781 632;
  • 23) 0.999 999 999 986 611 781 632 × 2 = 1 + 0.999 999 999 973 223 563 264;
  • 24) 0.999 999 999 973 223 563 264 × 2 = 1 + 0.999 999 999 946 447 126 528;
  • 25) 0.999 999 999 946 447 126 528 × 2 = 1 + 0.999 999 999 892 894 253 056;
  • 26) 0.999 999 999 892 894 253 056 × 2 = 1 + 0.999 999 999 785 788 506 112;
  • 27) 0.999 999 999 785 788 506 112 × 2 = 1 + 0.999 999 999 571 577 012 224;
  • 28) 0.999 999 999 571 577 012 224 × 2 = 1 + 0.999 999 999 143 154 024 448;
  • 29) 0.999 999 999 143 154 024 448 × 2 = 1 + 0.999 999 998 286 308 048 896;
  • 30) 0.999 999 998 286 308 048 896 × 2 = 1 + 0.999 999 996 572 616 097 792;
  • 31) 0.999 999 996 572 616 097 792 × 2 = 1 + 0.999 999 993 145 232 195 584;
  • 32) 0.999 999 993 145 232 195 584 × 2 = 1 + 0.999 999 986 290 464 391 168;
  • 33) 0.999 999 986 290 464 391 168 × 2 = 1 + 0.999 999 972 580 928 782 336;
  • 34) 0.999 999 972 580 928 782 336 × 2 = 1 + 0.999 999 945 161 857 564 672;
  • 35) 0.999 999 945 161 857 564 672 × 2 = 1 + 0.999 999 890 323 715 129 344;
  • 36) 0.999 999 890 323 715 129 344 × 2 = 1 + 0.999 999 780 647 430 258 688;
  • 37) 0.999 999 780 647 430 258 688 × 2 = 1 + 0.999 999 561 294 860 517 376;
  • 38) 0.999 999 561 294 860 517 376 × 2 = 1 + 0.999 999 122 589 721 034 752;
  • 39) 0.999 999 122 589 721 034 752 × 2 = 1 + 0.999 998 245 179 442 069 504;
  • 40) 0.999 998 245 179 442 069 504 × 2 = 1 + 0.999 996 490 358 884 139 008;
  • 41) 0.999 996 490 358 884 139 008 × 2 = 1 + 0.999 992 980 717 768 278 016;
  • 42) 0.999 992 980 717 768 278 016 × 2 = 1 + 0.999 985 961 435 536 556 032;
  • 43) 0.999 985 961 435 536 556 032 × 2 = 1 + 0.999 971 922 871 073 112 064;
  • 44) 0.999 971 922 871 073 112 064 × 2 = 1 + 0.999 943 845 742 146 224 128;
  • 45) 0.999 943 845 742 146 224 128 × 2 = 1 + 0.999 887 691 484 292 448 256;
  • 46) 0.999 887 691 484 292 448 256 × 2 = 1 + 0.999 775 382 968 584 896 512;
  • 47) 0.999 775 382 968 584 896 512 × 2 = 1 + 0.999 550 765 937 169 793 024;
  • 48) 0.999 550 765 937 169 793 024 × 2 = 1 + 0.999 101 531 874 339 586 048;
  • 49) 0.999 101 531 874 339 586 048 × 2 = 1 + 0.998 203 063 748 679 172 096;
  • 50) 0.998 203 063 748 679 172 096 × 2 = 1 + 0.996 406 127 497 358 344 192;
  • 51) 0.996 406 127 497 358 344 192 × 2 = 1 + 0.992 812 254 994 716 688 384;
  • 52) 0.992 812 254 994 716 688 384 × 2 = 1 + 0.985 624 509 989 433 376 768;
  • 53) 0.985 624 509 989 433 376 768 × 2 = 1 + 0.971 249 019 978 866 753 536;
  • 54) 0.971 249 019 978 866 753 536 × 2 = 1 + 0.942 498 039 957 733 507 072;
  • 55) 0.942 498 039 957 733 507 072 × 2 = 1 + 0.884 996 079 915 467 014 144;
  • 56) 0.884 996 079 915 467 014 144 × 2 = 1 + 0.769 992 159 830 934 028 288;
  • 57) 0.769 992 159 830 934 028 288 × 2 = 1 + 0.539 984 319 661 868 056 576;
  • 58) 0.539 984 319 661 868 056 576 × 2 = 1 + 0.079 968 639 323 736 113 152;

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:


0.016 738 891 601 562 496 808(10) =

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

6. Positive number before normalization:

0.016 738 891 601 562 496 808(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:


0.016 738 891 601 562 496 808(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 -0.016 738 891 601 562 496 808 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