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

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

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


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: 0.016 738 891 601 562 496 530 601.

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) 0.016 738 891 601 562 496 530 601 × 2 = 0 + 0.033 477 783 203 124 993 061 202;
  • 2) 0.033 477 783 203 124 993 061 202 × 2 = 0 + 0.066 955 566 406 249 986 122 404;
  • 3) 0.066 955 566 406 249 986 122 404 × 2 = 0 + 0.133 911 132 812 499 972 244 808;
  • 4) 0.133 911 132 812 499 972 244 808 × 2 = 0 + 0.267 822 265 624 999 944 489 616;
  • 5) 0.267 822 265 624 999 944 489 616 × 2 = 0 + 0.535 644 531 249 999 888 979 232;
  • 6) 0.535 644 531 249 999 888 979 232 × 2 = 1 + 0.071 289 062 499 999 777 958 464;
  • 7) 0.071 289 062 499 999 777 958 464 × 2 = 0 + 0.142 578 124 999 999 555 916 928;
  • 8) 0.142 578 124 999 999 555 916 928 × 2 = 0 + 0.285 156 249 999 999 111 833 856;
  • 9) 0.285 156 249 999 999 111 833 856 × 2 = 0 + 0.570 312 499 999 998 223 667 712;
  • 10) 0.570 312 499 999 998 223 667 712 × 2 = 1 + 0.140 624 999 999 996 447 335 424;
  • 11) 0.140 624 999 999 996 447 335 424 × 2 = 0 + 0.281 249 999 999 992 894 670 848;
  • 12) 0.281 249 999 999 992 894 670 848 × 2 = 0 + 0.562 499 999 999 985 789 341 696;
  • 13) 0.562 499 999 999 985 789 341 696 × 2 = 1 + 0.124 999 999 999 971 578 683 392;
  • 14) 0.124 999 999 999 971 578 683 392 × 2 = 0 + 0.249 999 999 999 943 157 366 784;
  • 15) 0.249 999 999 999 943 157 366 784 × 2 = 0 + 0.499 999 999 999 886 314 733 568;
  • 16) 0.499 999 999 999 886 314 733 568 × 2 = 0 + 0.999 999 999 999 772 629 467 136;
  • 17) 0.999 999 999 999 772 629 467 136 × 2 = 1 + 0.999 999 999 999 545 258 934 272;
  • 18) 0.999 999 999 999 545 258 934 272 × 2 = 1 + 0.999 999 999 999 090 517 868 544;
  • 19) 0.999 999 999 999 090 517 868 544 × 2 = 1 + 0.999 999 999 998 181 035 737 088;
  • 20) 0.999 999 999 998 181 035 737 088 × 2 = 1 + 0.999 999 999 996 362 071 474 176;
  • 21) 0.999 999 999 996 362 071 474 176 × 2 = 1 + 0.999 999 999 992 724 142 948 352;
  • 22) 0.999 999 999 992 724 142 948 352 × 2 = 1 + 0.999 999 999 985 448 285 896 704;
  • 23) 0.999 999 999 985 448 285 896 704 × 2 = 1 + 0.999 999 999 970 896 571 793 408;
  • 24) 0.999 999 999 970 896 571 793 408 × 2 = 1 + 0.999 999 999 941 793 143 586 816;
  • 25) 0.999 999 999 941 793 143 586 816 × 2 = 1 + 0.999 999 999 883 586 287 173 632;
  • 26) 0.999 999 999 883 586 287 173 632 × 2 = 1 + 0.999 999 999 767 172 574 347 264;
  • 27) 0.999 999 999 767 172 574 347 264 × 2 = 1 + 0.999 999 999 534 345 148 694 528;
  • 28) 0.999 999 999 534 345 148 694 528 × 2 = 1 + 0.999 999 999 068 690 297 389 056;
  • 29) 0.999 999 999 068 690 297 389 056 × 2 = 1 + 0.999 999 998 137 380 594 778 112;
  • 30) 0.999 999 998 137 380 594 778 112 × 2 = 1 + 0.999 999 996 274 761 189 556 224;
  • 31) 0.999 999 996 274 761 189 556 224 × 2 = 1 + 0.999 999 992 549 522 379 112 448;
  • 32) 0.999 999 992 549 522 379 112 448 × 2 = 1 + 0.999 999 985 099 044 758 224 896;
  • 33) 0.999 999 985 099 044 758 224 896 × 2 = 1 + 0.999 999 970 198 089 516 449 792;
  • 34) 0.999 999 970 198 089 516 449 792 × 2 = 1 + 0.999 999 940 396 179 032 899 584;
  • 35) 0.999 999 940 396 179 032 899 584 × 2 = 1 + 0.999 999 880 792 358 065 799 168;
  • 36) 0.999 999 880 792 358 065 799 168 × 2 = 1 + 0.999 999 761 584 716 131 598 336;
  • 37) 0.999 999 761 584 716 131 598 336 × 2 = 1 + 0.999 999 523 169 432 263 196 672;
  • 38) 0.999 999 523 169 432 263 196 672 × 2 = 1 + 0.999 999 046 338 864 526 393 344;
  • 39) 0.999 999 046 338 864 526 393 344 × 2 = 1 + 0.999 998 092 677 729 052 786 688;
  • 40) 0.999 998 092 677 729 052 786 688 × 2 = 1 + 0.999 996 185 355 458 105 573 376;
  • 41) 0.999 996 185 355 458 105 573 376 × 2 = 1 + 0.999 992 370 710 916 211 146 752;
  • 42) 0.999 992 370 710 916 211 146 752 × 2 = 1 + 0.999 984 741 421 832 422 293 504;
  • 43) 0.999 984 741 421 832 422 293 504 × 2 = 1 + 0.999 969 482 843 664 844 587 008;
  • 44) 0.999 969 482 843 664 844 587 008 × 2 = 1 + 0.999 938 965 687 329 689 174 016;
  • 45) 0.999 938 965 687 329 689 174 016 × 2 = 1 + 0.999 877 931 374 659 378 348 032;
  • 46) 0.999 877 931 374 659 378 348 032 × 2 = 1 + 0.999 755 862 749 318 756 696 064;
  • 47) 0.999 755 862 749 318 756 696 064 × 2 = 1 + 0.999 511 725 498 637 513 392 128;
  • 48) 0.999 511 725 498 637 513 392 128 × 2 = 1 + 0.999 023 450 997 275 026 784 256;
  • 49) 0.999 023 450 997 275 026 784 256 × 2 = 1 + 0.998 046 901 994 550 053 568 512;
  • 50) 0.998 046 901 994 550 053 568 512 × 2 = 1 + 0.996 093 803 989 100 107 137 024;
  • 51) 0.996 093 803 989 100 107 137 024 × 2 = 1 + 0.992 187 607 978 200 214 274 048;
  • 52) 0.992 187 607 978 200 214 274 048 × 2 = 1 + 0.984 375 215 956 400 428 548 096;
  • 53) 0.984 375 215 956 400 428 548 096 × 2 = 1 + 0.968 750 431 912 800 857 096 192;
  • 54) 0.968 750 431 912 800 857 096 192 × 2 = 1 + 0.937 500 863 825 601 714 192 384;
  • 55) 0.937 500 863 825 601 714 192 384 × 2 = 1 + 0.875 001 727 651 203 428 384 768;
  • 56) 0.875 001 727 651 203 428 384 768 × 2 = 1 + 0.750 003 455 302 406 856 769 536;
  • 57) 0.750 003 455 302 406 856 769 536 × 2 = 1 + 0.500 006 910 604 813 713 539 072;
  • 58) 0.500 006 910 604 813 713 539 072 × 2 = 1 + 0.000 013 821 209 627 427 078 144;

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:


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

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

6. Positive number before normalization:

0.016 738 891 601 562 496 530 601(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:


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