-0.000 000 000 010 072 7 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 010 072 7(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
-0.000 000 000 010 072 7(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:

|-0.000 000 000 010 072 7| = 0.000 000 000 010 072 7


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: 0.000 000 000 010 072 7.

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) 0.000 000 000 010 072 7 × 2 = 0 + 0.000 000 000 020 145 4;
  • 2) 0.000 000 000 020 145 4 × 2 = 0 + 0.000 000 000 040 290 8;
  • 3) 0.000 000 000 040 290 8 × 2 = 0 + 0.000 000 000 080 581 6;
  • 4) 0.000 000 000 080 581 6 × 2 = 0 + 0.000 000 000 161 163 2;
  • 5) 0.000 000 000 161 163 2 × 2 = 0 + 0.000 000 000 322 326 4;
  • 6) 0.000 000 000 322 326 4 × 2 = 0 + 0.000 000 000 644 652 8;
  • 7) 0.000 000 000 644 652 8 × 2 = 0 + 0.000 000 001 289 305 6;
  • 8) 0.000 000 001 289 305 6 × 2 = 0 + 0.000 000 002 578 611 2;
  • 9) 0.000 000 002 578 611 2 × 2 = 0 + 0.000 000 005 157 222 4;
  • 10) 0.000 000 005 157 222 4 × 2 = 0 + 0.000 000 010 314 444 8;
  • 11) 0.000 000 010 314 444 8 × 2 = 0 + 0.000 000 020 628 889 6;
  • 12) 0.000 000 020 628 889 6 × 2 = 0 + 0.000 000 041 257 779 2;
  • 13) 0.000 000 041 257 779 2 × 2 = 0 + 0.000 000 082 515 558 4;
  • 14) 0.000 000 082 515 558 4 × 2 = 0 + 0.000 000 165 031 116 8;
  • 15) 0.000 000 165 031 116 8 × 2 = 0 + 0.000 000 330 062 233 6;
  • 16) 0.000 000 330 062 233 6 × 2 = 0 + 0.000 000 660 124 467 2;
  • 17) 0.000 000 660 124 467 2 × 2 = 0 + 0.000 001 320 248 934 4;
  • 18) 0.000 001 320 248 934 4 × 2 = 0 + 0.000 002 640 497 868 8;
  • 19) 0.000 002 640 497 868 8 × 2 = 0 + 0.000 005 280 995 737 6;
  • 20) 0.000 005 280 995 737 6 × 2 = 0 + 0.000 010 561 991 475 2;
  • 21) 0.000 010 561 991 475 2 × 2 = 0 + 0.000 021 123 982 950 4;
  • 22) 0.000 021 123 982 950 4 × 2 = 0 + 0.000 042 247 965 900 8;
  • 23) 0.000 042 247 965 900 8 × 2 = 0 + 0.000 084 495 931 801 6;
  • 24) 0.000 084 495 931 801 6 × 2 = 0 + 0.000 168 991 863 603 2;
  • 25) 0.000 168 991 863 603 2 × 2 = 0 + 0.000 337 983 727 206 4;
  • 26) 0.000 337 983 727 206 4 × 2 = 0 + 0.000 675 967 454 412 8;
  • 27) 0.000 675 967 454 412 8 × 2 = 0 + 0.001 351 934 908 825 6;
  • 28) 0.001 351 934 908 825 6 × 2 = 0 + 0.002 703 869 817 651 2;
  • 29) 0.002 703 869 817 651 2 × 2 = 0 + 0.005 407 739 635 302 4;
  • 30) 0.005 407 739 635 302 4 × 2 = 0 + 0.010 815 479 270 604 8;
  • 31) 0.010 815 479 270 604 8 × 2 = 0 + 0.021 630 958 541 209 6;
  • 32) 0.021 630 958 541 209 6 × 2 = 0 + 0.043 261 917 082 419 2;
  • 33) 0.043 261 917 082 419 2 × 2 = 0 + 0.086 523 834 164 838 4;
  • 34) 0.086 523 834 164 838 4 × 2 = 0 + 0.173 047 668 329 676 8;
  • 35) 0.173 047 668 329 676 8 × 2 = 0 + 0.346 095 336 659 353 6;
  • 36) 0.346 095 336 659 353 6 × 2 = 0 + 0.692 190 673 318 707 2;
  • 37) 0.692 190 673 318 707 2 × 2 = 1 + 0.384 381 346 637 414 4;
  • 38) 0.384 381 346 637 414 4 × 2 = 0 + 0.768 762 693 274 828 8;
  • 39) 0.768 762 693 274 828 8 × 2 = 1 + 0.537 525 386 549 657 6;
  • 40) 0.537 525 386 549 657 6 × 2 = 1 + 0.075 050 773 099 315 2;
  • 41) 0.075 050 773 099 315 2 × 2 = 0 + 0.150 101 546 198 630 4;
  • 42) 0.150 101 546 198 630 4 × 2 = 0 + 0.300 203 092 397 260 8;
  • 43) 0.300 203 092 397 260 8 × 2 = 0 + 0.600 406 184 794 521 6;
  • 44) 0.600 406 184 794 521 6 × 2 = 1 + 0.200 812 369 589 043 2;
  • 45) 0.200 812 369 589 043 2 × 2 = 0 + 0.401 624 739 178 086 4;
  • 46) 0.401 624 739 178 086 4 × 2 = 0 + 0.803 249 478 356 172 8;
  • 47) 0.803 249 478 356 172 8 × 2 = 1 + 0.606 498 956 712 345 6;
  • 48) 0.606 498 956 712 345 6 × 2 = 1 + 0.212 997 913 424 691 2;
  • 49) 0.212 997 913 424 691 2 × 2 = 0 + 0.425 995 826 849 382 4;
  • 50) 0.425 995 826 849 382 4 × 2 = 0 + 0.851 991 653 698 764 8;
  • 51) 0.851 991 653 698 764 8 × 2 = 1 + 0.703 983 307 397 529 6;
  • 52) 0.703 983 307 397 529 6 × 2 = 1 + 0.407 966 614 795 059 2;
  • 53) 0.407 966 614 795 059 2 × 2 = 0 + 0.815 933 229 590 118 4;
  • 54) 0.815 933 229 590 118 4 × 2 = 1 + 0.631 866 459 180 236 8;
  • 55) 0.631 866 459 180 236 8 × 2 = 1 + 0.263 732 918 360 473 6;
  • 56) 0.263 732 918 360 473 6 × 2 = 0 + 0.527 465 836 720 947 2;
  • 57) 0.527 465 836 720 947 2 × 2 = 1 + 0.054 931 673 441 894 4;
  • 58) 0.054 931 673 441 894 4 × 2 = 0 + 0.109 863 346 883 788 8;
  • 59) 0.109 863 346 883 788 8 × 2 = 0 + 0.219 726 693 767 577 6;
  • 60) 0.219 726 693 767 577 6 × 2 = 0 + 0.439 453 387 535 155 2;
  • 61) 0.439 453 387 535 155 2 × 2 = 0 + 0.878 906 775 070 310 4;
  • 62) 0.878 906 775 070 310 4 × 2 = 1 + 0.757 813 550 140 620 8;
  • 63) 0.757 813 550 140 620 8 × 2 = 1 + 0.515 627 100 281 241 6;
  • 64) 0.515 627 100 281 241 6 × 2 = 1 + 0.031 254 200 562 483 2;
  • 65) 0.031 254 200 562 483 2 × 2 = 0 + 0.062 508 401 124 966 4;
  • 66) 0.062 508 401 124 966 4 × 2 = 0 + 0.125 016 802 249 932 8;
  • 67) 0.125 016 802 249 932 8 × 2 = 0 + 0.250 033 604 499 865 6;
  • 68) 0.250 033 604 499 865 6 × 2 = 0 + 0.500 067 208 999 731 2;
  • 69) 0.500 067 208 999 731 2 × 2 = 1 + 0.000 134 417 999 462 4;
  • 70) 0.000 134 417 999 462 4 × 2 = 0 + 0.000 268 835 998 924 8;
  • 71) 0.000 268 835 998 924 8 × 2 = 0 + 0.000 537 671 997 849 6;
  • 72) 0.000 537 671 997 849 6 × 2 = 0 + 0.001 075 343 995 699 2;
  • 73) 0.001 075 343 995 699 2 × 2 = 0 + 0.002 150 687 991 398 4;
  • 74) 0.002 150 687 991 398 4 × 2 = 0 + 0.004 301 375 982 796 8;
  • 75) 0.004 301 375 982 796 8 × 2 = 0 + 0.008 602 751 965 593 6;
  • 76) 0.008 602 751 965 593 6 × 2 = 0 + 0.017 205 503 931 187 2;
  • 77) 0.017 205 503 931 187 2 × 2 = 0 + 0.034 411 007 862 374 4;
  • 78) 0.034 411 007 862 374 4 × 2 = 0 + 0.068 822 015 724 748 8;
  • 79) 0.068 822 015 724 748 8 × 2 = 0 + 0.137 644 031 449 497 6;
  • 80) 0.137 644 031 449 497 6 × 2 = 0 + 0.275 288 062 898 995 2;
  • 81) 0.275 288 062 898 995 2 × 2 = 0 + 0.550 576 125 797 990 4;
  • 82) 0.550 576 125 797 990 4 × 2 = 1 + 0.101 152 251 595 980 8;
  • 83) 0.101 152 251 595 980 8 × 2 = 0 + 0.202 304 503 191 961 6;
  • 84) 0.202 304 503 191 961 6 × 2 = 0 + 0.404 609 006 383 923 2;
  • 85) 0.404 609 006 383 923 2 × 2 = 0 + 0.809 218 012 767 846 4;
  • 86) 0.809 218 012 767 846 4 × 2 = 1 + 0.618 436 025 535 692 8;
  • 87) 0.618 436 025 535 692 8 × 2 = 1 + 0.236 872 051 071 385 6;
  • 88) 0.236 872 051 071 385 6 × 2 = 0 + 0.473 744 102 142 771 2;
  • 89) 0.473 744 102 142 771 2 × 2 = 0 + 0.947 488 204 285 542 4;

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:


0.000 000 000 010 072 7(10) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 1011 0001 0011 0011 0110 1000 0111 0000 1000 0000 0000 0100 0110 0(2)

6. Positive number before normalization:

0.000 000 000 010 072 7(10) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 1011 0001 0011 0011 0110 1000 0111 0000 1000 0000 0000 0100 0110 0(2)

7. Normalize the binary representation of the number.

Shift the decimal mark 37 positions to the right, so that only one non zero digit remains to the left of it:


0.000 000 000 010 072 7(10) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 1011 0001 0011 0011 0110 1000 0111 0000 1000 0000 0000 0100 0110 0(2) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 1011 0001 0011 0011 0110 1000 0111 0000 1000 0000 0000 0100 0110 0(2) × 20 =

1.0110 0010 0110 0110 1101 0000 1110 0001 0000 0000 0000 1000 1100(2) × 2-37


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): -37

Mantissa (not normalized):
1.0110 0010 0110 0110 1101 0000 1110 0001 0000 0000 0000 1000 1100


9. Adjust the exponent.

Use the 11 bit excess/bias notation:


Exponent (adjusted) =

Exponent (unadjusted) + 2(11-1) - 1 =

-37 + 2(11-1) - 1 =

(-37 + 1 023)(10) =

986(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;
  • 986 ÷ 2 = 493 + 0;
  • 493 ÷ 2 = 246 + 1;
  • 246 ÷ 2 = 123 + 0;
  • 123 ÷ 2 = 61 + 1;
  • 61 ÷ 2 = 30 + 1;
  • 30 ÷ 2 = 15 + 0;
  • 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) =

986(10) =

011 1101 1010(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. 0110 0010 0110 0110 1101 0000 1110 0001 0000 0000 0000 1000 1100 =

0110 0010 0110 0110 1101 0000 1110 0001 0000 0000 0000 1000 1100


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 1101 1010

Mantissa (52 bits) =
0110 0010 0110 0110 1101 0000 1110 0001 0000 0000 0000 1000 1100


Decimal number -0.000 000 000 010 072 7 converted to 64 bit double precision IEEE 754 binary floating point representation:

1 - 011 1101 1010 - 0110 0010 0110 0110 1101 0000 1110 0001 0000 0000 0000 1000 1100

» Convert decimal -0.000 000 000 010 075 2 to 64 bit double precision IEEE 754 binary floating point representation standard

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» Month 10, 2026 [October]: Decimal Numbers Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard. Calculations performed during the month of: October - by our visitors

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