-0.016 738 891 601 562 496 530 544 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.016 738 891 601 562 496 530 544(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
-0.016 738 891 601 562 496 530 544(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:

|-0.016 738 891 601 562 496 530 544| = 0.016 738 891 601 562 496 530 544


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: 0.016 738 891 601 562 496 530 544.

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) 0.016 738 891 601 562 496 530 544 × 2 = 0 + 0.033 477 783 203 124 993 061 088;
  • 2) 0.033 477 783 203 124 993 061 088 × 2 = 0 + 0.066 955 566 406 249 986 122 176;
  • 3) 0.066 955 566 406 249 986 122 176 × 2 = 0 + 0.133 911 132 812 499 972 244 352;
  • 4) 0.133 911 132 812 499 972 244 352 × 2 = 0 + 0.267 822 265 624 999 944 488 704;
  • 5) 0.267 822 265 624 999 944 488 704 × 2 = 0 + 0.535 644 531 249 999 888 977 408;
  • 6) 0.535 644 531 249 999 888 977 408 × 2 = 1 + 0.071 289 062 499 999 777 954 816;
  • 7) 0.071 289 062 499 999 777 954 816 × 2 = 0 + 0.142 578 124 999 999 555 909 632;
  • 8) 0.142 578 124 999 999 555 909 632 × 2 = 0 + 0.285 156 249 999 999 111 819 264;
  • 9) 0.285 156 249 999 999 111 819 264 × 2 = 0 + 0.570 312 499 999 998 223 638 528;
  • 10) 0.570 312 499 999 998 223 638 528 × 2 = 1 + 0.140 624 999 999 996 447 277 056;
  • 11) 0.140 624 999 999 996 447 277 056 × 2 = 0 + 0.281 249 999 999 992 894 554 112;
  • 12) 0.281 249 999 999 992 894 554 112 × 2 = 0 + 0.562 499 999 999 985 789 108 224;
  • 13) 0.562 499 999 999 985 789 108 224 × 2 = 1 + 0.124 999 999 999 971 578 216 448;
  • 14) 0.124 999 999 999 971 578 216 448 × 2 = 0 + 0.249 999 999 999 943 156 432 896;
  • 15) 0.249 999 999 999 943 156 432 896 × 2 = 0 + 0.499 999 999 999 886 312 865 792;
  • 16) 0.499 999 999 999 886 312 865 792 × 2 = 0 + 0.999 999 999 999 772 625 731 584;
  • 17) 0.999 999 999 999 772 625 731 584 × 2 = 1 + 0.999 999 999 999 545 251 463 168;
  • 18) 0.999 999 999 999 545 251 463 168 × 2 = 1 + 0.999 999 999 999 090 502 926 336;
  • 19) 0.999 999 999 999 090 502 926 336 × 2 = 1 + 0.999 999 999 998 181 005 852 672;
  • 20) 0.999 999 999 998 181 005 852 672 × 2 = 1 + 0.999 999 999 996 362 011 705 344;
  • 21) 0.999 999 999 996 362 011 705 344 × 2 = 1 + 0.999 999 999 992 724 023 410 688;
  • 22) 0.999 999 999 992 724 023 410 688 × 2 = 1 + 0.999 999 999 985 448 046 821 376;
  • 23) 0.999 999 999 985 448 046 821 376 × 2 = 1 + 0.999 999 999 970 896 093 642 752;
  • 24) 0.999 999 999 970 896 093 642 752 × 2 = 1 + 0.999 999 999 941 792 187 285 504;
  • 25) 0.999 999 999 941 792 187 285 504 × 2 = 1 + 0.999 999 999 883 584 374 571 008;
  • 26) 0.999 999 999 883 584 374 571 008 × 2 = 1 + 0.999 999 999 767 168 749 142 016;
  • 27) 0.999 999 999 767 168 749 142 016 × 2 = 1 + 0.999 999 999 534 337 498 284 032;
  • 28) 0.999 999 999 534 337 498 284 032 × 2 = 1 + 0.999 999 999 068 674 996 568 064;
  • 29) 0.999 999 999 068 674 996 568 064 × 2 = 1 + 0.999 999 998 137 349 993 136 128;
  • 30) 0.999 999 998 137 349 993 136 128 × 2 = 1 + 0.999 999 996 274 699 986 272 256;
  • 31) 0.999 999 996 274 699 986 272 256 × 2 = 1 + 0.999 999 992 549 399 972 544 512;
  • 32) 0.999 999 992 549 399 972 544 512 × 2 = 1 + 0.999 999 985 098 799 945 089 024;
  • 33) 0.999 999 985 098 799 945 089 024 × 2 = 1 + 0.999 999 970 197 599 890 178 048;
  • 34) 0.999 999 970 197 599 890 178 048 × 2 = 1 + 0.999 999 940 395 199 780 356 096;
  • 35) 0.999 999 940 395 199 780 356 096 × 2 = 1 + 0.999 999 880 790 399 560 712 192;
  • 36) 0.999 999 880 790 399 560 712 192 × 2 = 1 + 0.999 999 761 580 799 121 424 384;
  • 37) 0.999 999 761 580 799 121 424 384 × 2 = 1 + 0.999 999 523 161 598 242 848 768;
  • 38) 0.999 999 523 161 598 242 848 768 × 2 = 1 + 0.999 999 046 323 196 485 697 536;
  • 39) 0.999 999 046 323 196 485 697 536 × 2 = 1 + 0.999 998 092 646 392 971 395 072;
  • 40) 0.999 998 092 646 392 971 395 072 × 2 = 1 + 0.999 996 185 292 785 942 790 144;
  • 41) 0.999 996 185 292 785 942 790 144 × 2 = 1 + 0.999 992 370 585 571 885 580 288;
  • 42) 0.999 992 370 585 571 885 580 288 × 2 = 1 + 0.999 984 741 171 143 771 160 576;
  • 43) 0.999 984 741 171 143 771 160 576 × 2 = 1 + 0.999 969 482 342 287 542 321 152;
  • 44) 0.999 969 482 342 287 542 321 152 × 2 = 1 + 0.999 938 964 684 575 084 642 304;
  • 45) 0.999 938 964 684 575 084 642 304 × 2 = 1 + 0.999 877 929 369 150 169 284 608;
  • 46) 0.999 877 929 369 150 169 284 608 × 2 = 1 + 0.999 755 858 738 300 338 569 216;
  • 47) 0.999 755 858 738 300 338 569 216 × 2 = 1 + 0.999 511 717 476 600 677 138 432;
  • 48) 0.999 511 717 476 600 677 138 432 × 2 = 1 + 0.999 023 434 953 201 354 276 864;
  • 49) 0.999 023 434 953 201 354 276 864 × 2 = 1 + 0.998 046 869 906 402 708 553 728;
  • 50) 0.998 046 869 906 402 708 553 728 × 2 = 1 + 0.996 093 739 812 805 417 107 456;
  • 51) 0.996 093 739 812 805 417 107 456 × 2 = 1 + 0.992 187 479 625 610 834 214 912;
  • 52) 0.992 187 479 625 610 834 214 912 × 2 = 1 + 0.984 374 959 251 221 668 429 824;
  • 53) 0.984 374 959 251 221 668 429 824 × 2 = 1 + 0.968 749 918 502 443 336 859 648;
  • 54) 0.968 749 918 502 443 336 859 648 × 2 = 1 + 0.937 499 837 004 886 673 719 296;
  • 55) 0.937 499 837 004 886 673 719 296 × 2 = 1 + 0.874 999 674 009 773 347 438 592;
  • 56) 0.874 999 674 009 773 347 438 592 × 2 = 1 + 0.749 999 348 019 546 694 877 184;
  • 57) 0.749 999 348 019 546 694 877 184 × 2 = 1 + 0.499 998 696 039 093 389 754 368;
  • 58) 0.499 998 696 039 093 389 754 368 × 2 = 0 + 0.999 997 392 078 186 779 508 736;

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:


0.016 738 891 601 562 496 530 544(10) =

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

6. Positive number before normalization:

0.016 738 891 601 562 496 530 544(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:


0.016 738 891 601 562 496 530 544(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 -0.016 738 891 601 562 496 530 544 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