-0.000 164 779 749 346 817 9 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 164 779 749 346 817 9(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
-0.000 164 779 749 346 817 9(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:

|-0.000 164 779 749 346 817 9| = 0.000 164 779 749 346 817 9


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: 0.000 164 779 749 346 817 9.

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) 0.000 164 779 749 346 817 9 × 2 = 0 + 0.000 329 559 498 693 635 8;
  • 2) 0.000 329 559 498 693 635 8 × 2 = 0 + 0.000 659 118 997 387 271 6;
  • 3) 0.000 659 118 997 387 271 6 × 2 = 0 + 0.001 318 237 994 774 543 2;
  • 4) 0.001 318 237 994 774 543 2 × 2 = 0 + 0.002 636 475 989 549 086 4;
  • 5) 0.002 636 475 989 549 086 4 × 2 = 0 + 0.005 272 951 979 098 172 8;
  • 6) 0.005 272 951 979 098 172 8 × 2 = 0 + 0.010 545 903 958 196 345 6;
  • 7) 0.010 545 903 958 196 345 6 × 2 = 0 + 0.021 091 807 916 392 691 2;
  • 8) 0.021 091 807 916 392 691 2 × 2 = 0 + 0.042 183 615 832 785 382 4;
  • 9) 0.042 183 615 832 785 382 4 × 2 = 0 + 0.084 367 231 665 570 764 8;
  • 10) 0.084 367 231 665 570 764 8 × 2 = 0 + 0.168 734 463 331 141 529 6;
  • 11) 0.168 734 463 331 141 529 6 × 2 = 0 + 0.337 468 926 662 283 059 2;
  • 12) 0.337 468 926 662 283 059 2 × 2 = 0 + 0.674 937 853 324 566 118 4;
  • 13) 0.674 937 853 324 566 118 4 × 2 = 1 + 0.349 875 706 649 132 236 8;
  • 14) 0.349 875 706 649 132 236 8 × 2 = 0 + 0.699 751 413 298 264 473 6;
  • 15) 0.699 751 413 298 264 473 6 × 2 = 1 + 0.399 502 826 596 528 947 2;
  • 16) 0.399 502 826 596 528 947 2 × 2 = 0 + 0.799 005 653 193 057 894 4;
  • 17) 0.799 005 653 193 057 894 4 × 2 = 1 + 0.598 011 306 386 115 788 8;
  • 18) 0.598 011 306 386 115 788 8 × 2 = 1 + 0.196 022 612 772 231 577 6;
  • 19) 0.196 022 612 772 231 577 6 × 2 = 0 + 0.392 045 225 544 463 155 2;
  • 20) 0.392 045 225 544 463 155 2 × 2 = 0 + 0.784 090 451 088 926 310 4;
  • 21) 0.784 090 451 088 926 310 4 × 2 = 1 + 0.568 180 902 177 852 620 8;
  • 22) 0.568 180 902 177 852 620 8 × 2 = 1 + 0.136 361 804 355 705 241 6;
  • 23) 0.136 361 804 355 705 241 6 × 2 = 0 + 0.272 723 608 711 410 483 2;
  • 24) 0.272 723 608 711 410 483 2 × 2 = 0 + 0.545 447 217 422 820 966 4;
  • 25) 0.545 447 217 422 820 966 4 × 2 = 1 + 0.090 894 434 845 641 932 8;
  • 26) 0.090 894 434 845 641 932 8 × 2 = 0 + 0.181 788 869 691 283 865 6;
  • 27) 0.181 788 869 691 283 865 6 × 2 = 0 + 0.363 577 739 382 567 731 2;
  • 28) 0.363 577 739 382 567 731 2 × 2 = 0 + 0.727 155 478 765 135 462 4;
  • 29) 0.727 155 478 765 135 462 4 × 2 = 1 + 0.454 310 957 530 270 924 8;
  • 30) 0.454 310 957 530 270 924 8 × 2 = 0 + 0.908 621 915 060 541 849 6;
  • 31) 0.908 621 915 060 541 849 6 × 2 = 1 + 0.817 243 830 121 083 699 2;
  • 32) 0.817 243 830 121 083 699 2 × 2 = 1 + 0.634 487 660 242 167 398 4;
  • 33) 0.634 487 660 242 167 398 4 × 2 = 1 + 0.268 975 320 484 334 796 8;
  • 34) 0.268 975 320 484 334 796 8 × 2 = 0 + 0.537 950 640 968 669 593 6;
  • 35) 0.537 950 640 968 669 593 6 × 2 = 1 + 0.075 901 281 937 339 187 2;
  • 36) 0.075 901 281 937 339 187 2 × 2 = 0 + 0.151 802 563 874 678 374 4;
  • 37) 0.151 802 563 874 678 374 4 × 2 = 0 + 0.303 605 127 749 356 748 8;
  • 38) 0.303 605 127 749 356 748 8 × 2 = 0 + 0.607 210 255 498 713 497 6;
  • 39) 0.607 210 255 498 713 497 6 × 2 = 1 + 0.214 420 510 997 426 995 2;
  • 40) 0.214 420 510 997 426 995 2 × 2 = 0 + 0.428 841 021 994 853 990 4;
  • 41) 0.428 841 021 994 853 990 4 × 2 = 0 + 0.857 682 043 989 707 980 8;
  • 42) 0.857 682 043 989 707 980 8 × 2 = 1 + 0.715 364 087 979 415 961 6;
  • 43) 0.715 364 087 979 415 961 6 × 2 = 1 + 0.430 728 175 958 831 923 2;
  • 44) 0.430 728 175 958 831 923 2 × 2 = 0 + 0.861 456 351 917 663 846 4;
  • 45) 0.861 456 351 917 663 846 4 × 2 = 1 + 0.722 912 703 835 327 692 8;
  • 46) 0.722 912 703 835 327 692 8 × 2 = 1 + 0.445 825 407 670 655 385 6;
  • 47) 0.445 825 407 670 655 385 6 × 2 = 0 + 0.891 650 815 341 310 771 2;
  • 48) 0.891 650 815 341 310 771 2 × 2 = 1 + 0.783 301 630 682 621 542 4;
  • 49) 0.783 301 630 682 621 542 4 × 2 = 1 + 0.566 603 261 365 243 084 8;
  • 50) 0.566 603 261 365 243 084 8 × 2 = 1 + 0.133 206 522 730 486 169 6;
  • 51) 0.133 206 522 730 486 169 6 × 2 = 0 + 0.266 413 045 460 972 339 2;
  • 52) 0.266 413 045 460 972 339 2 × 2 = 0 + 0.532 826 090 921 944 678 4;
  • 53) 0.532 826 090 921 944 678 4 × 2 = 1 + 0.065 652 181 843 889 356 8;
  • 54) 0.065 652 181 843 889 356 8 × 2 = 0 + 0.131 304 363 687 778 713 6;
  • 55) 0.131 304 363 687 778 713 6 × 2 = 0 + 0.262 608 727 375 557 427 2;
  • 56) 0.262 608 727 375 557 427 2 × 2 = 0 + 0.525 217 454 751 114 854 4;
  • 57) 0.525 217 454 751 114 854 4 × 2 = 1 + 0.050 434 909 502 229 708 8;
  • 58) 0.050 434 909 502 229 708 8 × 2 = 0 + 0.100 869 819 004 459 417 6;
  • 59) 0.100 869 819 004 459 417 6 × 2 = 0 + 0.201 739 638 008 918 835 2;
  • 60) 0.201 739 638 008 918 835 2 × 2 = 0 + 0.403 479 276 017 837 670 4;
  • 61) 0.403 479 276 017 837 670 4 × 2 = 0 + 0.806 958 552 035 675 340 8;
  • 62) 0.806 958 552 035 675 340 8 × 2 = 1 + 0.613 917 104 071 350 681 6;
  • 63) 0.613 917 104 071 350 681 6 × 2 = 1 + 0.227 834 208 142 701 363 2;
  • 64) 0.227 834 208 142 701 363 2 × 2 = 0 + 0.455 668 416 285 402 726 4;
  • 65) 0.455 668 416 285 402 726 4 × 2 = 0 + 0.911 336 832 570 805 452 8;

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:


0.000 164 779 749 346 817 9(10) =

0.0000 0000 0000 1010 1100 1100 1000 1011 1010 0010 0110 1101 1100 1000 1000 0110 0(2)

6. Positive number before normalization:

0.000 164 779 749 346 817 9(10) =

0.0000 0000 0000 1010 1100 1100 1000 1011 1010 0010 0110 1101 1100 1000 1000 0110 0(2)

7. Normalize the binary representation of the number.

Shift the decimal mark 13 positions to the right, so that only one non zero digit remains to the left of it:


0.000 164 779 749 346 817 9(10) =

0.0000 0000 0000 1010 1100 1100 1000 1011 1010 0010 0110 1101 1100 1000 1000 0110 0(2) =

0.0000 0000 0000 1010 1100 1100 1000 1011 1010 0010 0110 1101 1100 1000 1000 0110 0(2) × 20 =

1.0101 1001 1001 0001 0111 0100 0100 1101 1011 1001 0001 0000 1100(2) × 2-13


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): -13

Mantissa (not normalized):
1.0101 1001 1001 0001 0111 0100 0100 1101 1011 1001 0001 0000 1100


9. Adjust the exponent.

Use the 11 bit excess/bias notation:


Exponent (adjusted) =

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

-13 + 2(11-1) - 1 =

(-13 + 1 023)(10) =

1 010(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;
  • 1 010 ÷ 2 = 505 + 0;
  • 505 ÷ 2 = 252 + 1;
  • 252 ÷ 2 = 126 + 0;
  • 126 ÷ 2 = 63 + 0;
  • 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) =

1010(10) =

011 1111 0010(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. 0101 1001 1001 0001 0111 0100 0100 1101 1011 1001 0001 0000 1100 =

0101 1001 1001 0001 0111 0100 0100 1101 1011 1001 0001 0000 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 1111 0010

Mantissa (52 bits) =
0101 1001 1001 0001 0111 0100 0100 1101 1011 1001 0001 0000 1100


Decimal number -0.000 164 779 749 346 817 9 converted to 64 bit double precision IEEE 754 binary floating point representation:

1 - 011 1111 0010 - 0101 1001 1001 0001 0111 0100 0100 1101 1011 1001 0001 0000 1100

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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