-0.016 738 891 601 562 496 452 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.016 738 891 601 562 496 452(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
-0.016 738 891 601 562 496 452(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:

|-0.016 738 891 601 562 496 452| = 0.016 738 891 601 562 496 452


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: 0.016 738 891 601 562 496 452.

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) 0.016 738 891 601 562 496 452 × 2 = 0 + 0.033 477 783 203 124 992 904;
  • 2) 0.033 477 783 203 124 992 904 × 2 = 0 + 0.066 955 566 406 249 985 808;
  • 3) 0.066 955 566 406 249 985 808 × 2 = 0 + 0.133 911 132 812 499 971 616;
  • 4) 0.133 911 132 812 499 971 616 × 2 = 0 + 0.267 822 265 624 999 943 232;
  • 5) 0.267 822 265 624 999 943 232 × 2 = 0 + 0.535 644 531 249 999 886 464;
  • 6) 0.535 644 531 249 999 886 464 × 2 = 1 + 0.071 289 062 499 999 772 928;
  • 7) 0.071 289 062 499 999 772 928 × 2 = 0 + 0.142 578 124 999 999 545 856;
  • 8) 0.142 578 124 999 999 545 856 × 2 = 0 + 0.285 156 249 999 999 091 712;
  • 9) 0.285 156 249 999 999 091 712 × 2 = 0 + 0.570 312 499 999 998 183 424;
  • 10) 0.570 312 499 999 998 183 424 × 2 = 1 + 0.140 624 999 999 996 366 848;
  • 11) 0.140 624 999 999 996 366 848 × 2 = 0 + 0.281 249 999 999 992 733 696;
  • 12) 0.281 249 999 999 992 733 696 × 2 = 0 + 0.562 499 999 999 985 467 392;
  • 13) 0.562 499 999 999 985 467 392 × 2 = 1 + 0.124 999 999 999 970 934 784;
  • 14) 0.124 999 999 999 970 934 784 × 2 = 0 + 0.249 999 999 999 941 869 568;
  • 15) 0.249 999 999 999 941 869 568 × 2 = 0 + 0.499 999 999 999 883 739 136;
  • 16) 0.499 999 999 999 883 739 136 × 2 = 0 + 0.999 999 999 999 767 478 272;
  • 17) 0.999 999 999 999 767 478 272 × 2 = 1 + 0.999 999 999 999 534 956 544;
  • 18) 0.999 999 999 999 534 956 544 × 2 = 1 + 0.999 999 999 999 069 913 088;
  • 19) 0.999 999 999 999 069 913 088 × 2 = 1 + 0.999 999 999 998 139 826 176;
  • 20) 0.999 999 999 998 139 826 176 × 2 = 1 + 0.999 999 999 996 279 652 352;
  • 21) 0.999 999 999 996 279 652 352 × 2 = 1 + 0.999 999 999 992 559 304 704;
  • 22) 0.999 999 999 992 559 304 704 × 2 = 1 + 0.999 999 999 985 118 609 408;
  • 23) 0.999 999 999 985 118 609 408 × 2 = 1 + 0.999 999 999 970 237 218 816;
  • 24) 0.999 999 999 970 237 218 816 × 2 = 1 + 0.999 999 999 940 474 437 632;
  • 25) 0.999 999 999 940 474 437 632 × 2 = 1 + 0.999 999 999 880 948 875 264;
  • 26) 0.999 999 999 880 948 875 264 × 2 = 1 + 0.999 999 999 761 897 750 528;
  • 27) 0.999 999 999 761 897 750 528 × 2 = 1 + 0.999 999 999 523 795 501 056;
  • 28) 0.999 999 999 523 795 501 056 × 2 = 1 + 0.999 999 999 047 591 002 112;
  • 29) 0.999 999 999 047 591 002 112 × 2 = 1 + 0.999 999 998 095 182 004 224;
  • 30) 0.999 999 998 095 182 004 224 × 2 = 1 + 0.999 999 996 190 364 008 448;
  • 31) 0.999 999 996 190 364 008 448 × 2 = 1 + 0.999 999 992 380 728 016 896;
  • 32) 0.999 999 992 380 728 016 896 × 2 = 1 + 0.999 999 984 761 456 033 792;
  • 33) 0.999 999 984 761 456 033 792 × 2 = 1 + 0.999 999 969 522 912 067 584;
  • 34) 0.999 999 969 522 912 067 584 × 2 = 1 + 0.999 999 939 045 824 135 168;
  • 35) 0.999 999 939 045 824 135 168 × 2 = 1 + 0.999 999 878 091 648 270 336;
  • 36) 0.999 999 878 091 648 270 336 × 2 = 1 + 0.999 999 756 183 296 540 672;
  • 37) 0.999 999 756 183 296 540 672 × 2 = 1 + 0.999 999 512 366 593 081 344;
  • 38) 0.999 999 512 366 593 081 344 × 2 = 1 + 0.999 999 024 733 186 162 688;
  • 39) 0.999 999 024 733 186 162 688 × 2 = 1 + 0.999 998 049 466 372 325 376;
  • 40) 0.999 998 049 466 372 325 376 × 2 = 1 + 0.999 996 098 932 744 650 752;
  • 41) 0.999 996 098 932 744 650 752 × 2 = 1 + 0.999 992 197 865 489 301 504;
  • 42) 0.999 992 197 865 489 301 504 × 2 = 1 + 0.999 984 395 730 978 603 008;
  • 43) 0.999 984 395 730 978 603 008 × 2 = 1 + 0.999 968 791 461 957 206 016;
  • 44) 0.999 968 791 461 957 206 016 × 2 = 1 + 0.999 937 582 923 914 412 032;
  • 45) 0.999 937 582 923 914 412 032 × 2 = 1 + 0.999 875 165 847 828 824 064;
  • 46) 0.999 875 165 847 828 824 064 × 2 = 1 + 0.999 750 331 695 657 648 128;
  • 47) 0.999 750 331 695 657 648 128 × 2 = 1 + 0.999 500 663 391 315 296 256;
  • 48) 0.999 500 663 391 315 296 256 × 2 = 1 + 0.999 001 326 782 630 592 512;
  • 49) 0.999 001 326 782 630 592 512 × 2 = 1 + 0.998 002 653 565 261 185 024;
  • 50) 0.998 002 653 565 261 185 024 × 2 = 1 + 0.996 005 307 130 522 370 048;
  • 51) 0.996 005 307 130 522 370 048 × 2 = 1 + 0.992 010 614 261 044 740 096;
  • 52) 0.992 010 614 261 044 740 096 × 2 = 1 + 0.984 021 228 522 089 480 192;
  • 53) 0.984 021 228 522 089 480 192 × 2 = 1 + 0.968 042 457 044 178 960 384;
  • 54) 0.968 042 457 044 178 960 384 × 2 = 1 + 0.936 084 914 088 357 920 768;
  • 55) 0.936 084 914 088 357 920 768 × 2 = 1 + 0.872 169 828 176 715 841 536;
  • 56) 0.872 169 828 176 715 841 536 × 2 = 1 + 0.744 339 656 353 431 683 072;
  • 57) 0.744 339 656 353 431 683 072 × 2 = 1 + 0.488 679 312 706 863 366 144;
  • 58) 0.488 679 312 706 863 366 144 × 2 = 0 + 0.977 358 625 413 726 732 288;

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:


0.016 738 891 601 562 496 452(10) =

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

6. Positive number before normalization:

0.016 738 891 601 562 496 452(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:


0.016 738 891 601 562 496 452(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 -0.016 738 891 601 562 496 452 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