-0.000 000 000 040 986 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 040 986(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
-0.000 000 000 040 986(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:

|-0.000 000 000 040 986| = 0.000 000 000 040 986


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: 0.000 000 000 040 986.

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) 0.000 000 000 040 986 × 2 = 0 + 0.000 000 000 081 972;
  • 2) 0.000 000 000 081 972 × 2 = 0 + 0.000 000 000 163 944;
  • 3) 0.000 000 000 163 944 × 2 = 0 + 0.000 000 000 327 888;
  • 4) 0.000 000 000 327 888 × 2 = 0 + 0.000 000 000 655 776;
  • 5) 0.000 000 000 655 776 × 2 = 0 + 0.000 000 001 311 552;
  • 6) 0.000 000 001 311 552 × 2 = 0 + 0.000 000 002 623 104;
  • 7) 0.000 000 002 623 104 × 2 = 0 + 0.000 000 005 246 208;
  • 8) 0.000 000 005 246 208 × 2 = 0 + 0.000 000 010 492 416;
  • 9) 0.000 000 010 492 416 × 2 = 0 + 0.000 000 020 984 832;
  • 10) 0.000 000 020 984 832 × 2 = 0 + 0.000 000 041 969 664;
  • 11) 0.000 000 041 969 664 × 2 = 0 + 0.000 000 083 939 328;
  • 12) 0.000 000 083 939 328 × 2 = 0 + 0.000 000 167 878 656;
  • 13) 0.000 000 167 878 656 × 2 = 0 + 0.000 000 335 757 312;
  • 14) 0.000 000 335 757 312 × 2 = 0 + 0.000 000 671 514 624;
  • 15) 0.000 000 671 514 624 × 2 = 0 + 0.000 001 343 029 248;
  • 16) 0.000 001 343 029 248 × 2 = 0 + 0.000 002 686 058 496;
  • 17) 0.000 002 686 058 496 × 2 = 0 + 0.000 005 372 116 992;
  • 18) 0.000 005 372 116 992 × 2 = 0 + 0.000 010 744 233 984;
  • 19) 0.000 010 744 233 984 × 2 = 0 + 0.000 021 488 467 968;
  • 20) 0.000 021 488 467 968 × 2 = 0 + 0.000 042 976 935 936;
  • 21) 0.000 042 976 935 936 × 2 = 0 + 0.000 085 953 871 872;
  • 22) 0.000 085 953 871 872 × 2 = 0 + 0.000 171 907 743 744;
  • 23) 0.000 171 907 743 744 × 2 = 0 + 0.000 343 815 487 488;
  • 24) 0.000 343 815 487 488 × 2 = 0 + 0.000 687 630 974 976;
  • 25) 0.000 687 630 974 976 × 2 = 0 + 0.001 375 261 949 952;
  • 26) 0.001 375 261 949 952 × 2 = 0 + 0.002 750 523 899 904;
  • 27) 0.002 750 523 899 904 × 2 = 0 + 0.005 501 047 799 808;
  • 28) 0.005 501 047 799 808 × 2 = 0 + 0.011 002 095 599 616;
  • 29) 0.011 002 095 599 616 × 2 = 0 + 0.022 004 191 199 232;
  • 30) 0.022 004 191 199 232 × 2 = 0 + 0.044 008 382 398 464;
  • 31) 0.044 008 382 398 464 × 2 = 0 + 0.088 016 764 796 928;
  • 32) 0.088 016 764 796 928 × 2 = 0 + 0.176 033 529 593 856;
  • 33) 0.176 033 529 593 856 × 2 = 0 + 0.352 067 059 187 712;
  • 34) 0.352 067 059 187 712 × 2 = 0 + 0.704 134 118 375 424;
  • 35) 0.704 134 118 375 424 × 2 = 1 + 0.408 268 236 750 848;
  • 36) 0.408 268 236 750 848 × 2 = 0 + 0.816 536 473 501 696;
  • 37) 0.816 536 473 501 696 × 2 = 1 + 0.633 072 947 003 392;
  • 38) 0.633 072 947 003 392 × 2 = 1 + 0.266 145 894 006 784;
  • 39) 0.266 145 894 006 784 × 2 = 0 + 0.532 291 788 013 568;
  • 40) 0.532 291 788 013 568 × 2 = 1 + 0.064 583 576 027 136;
  • 41) 0.064 583 576 027 136 × 2 = 0 + 0.129 167 152 054 272;
  • 42) 0.129 167 152 054 272 × 2 = 0 + 0.258 334 304 108 544;
  • 43) 0.258 334 304 108 544 × 2 = 0 + 0.516 668 608 217 088;
  • 44) 0.516 668 608 217 088 × 2 = 1 + 0.033 337 216 434 176;
  • 45) 0.033 337 216 434 176 × 2 = 0 + 0.066 674 432 868 352;
  • 46) 0.066 674 432 868 352 × 2 = 0 + 0.133 348 865 736 704;
  • 47) 0.133 348 865 736 704 × 2 = 0 + 0.266 697 731 473 408;
  • 48) 0.266 697 731 473 408 × 2 = 0 + 0.533 395 462 946 816;
  • 49) 0.533 395 462 946 816 × 2 = 1 + 0.066 790 925 893 632;
  • 50) 0.066 790 925 893 632 × 2 = 0 + 0.133 581 851 787 264;
  • 51) 0.133 581 851 787 264 × 2 = 0 + 0.267 163 703 574 528;
  • 52) 0.267 163 703 574 528 × 2 = 0 + 0.534 327 407 149 056;
  • 53) 0.534 327 407 149 056 × 2 = 1 + 0.068 654 814 298 112;
  • 54) 0.068 654 814 298 112 × 2 = 0 + 0.137 309 628 596 224;
  • 55) 0.137 309 628 596 224 × 2 = 0 + 0.274 619 257 192 448;
  • 56) 0.274 619 257 192 448 × 2 = 0 + 0.549 238 514 384 896;
  • 57) 0.549 238 514 384 896 × 2 = 1 + 0.098 477 028 769 792;
  • 58) 0.098 477 028 769 792 × 2 = 0 + 0.196 954 057 539 584;
  • 59) 0.196 954 057 539 584 × 2 = 0 + 0.393 908 115 079 168;
  • 60) 0.393 908 115 079 168 × 2 = 0 + 0.787 816 230 158 336;
  • 61) 0.787 816 230 158 336 × 2 = 1 + 0.575 632 460 316 672;
  • 62) 0.575 632 460 316 672 × 2 = 1 + 0.151 264 920 633 344;
  • 63) 0.151 264 920 633 344 × 2 = 0 + 0.302 529 841 266 688;
  • 64) 0.302 529 841 266 688 × 2 = 0 + 0.605 059 682 533 376;
  • 65) 0.605 059 682 533 376 × 2 = 1 + 0.210 119 365 066 752;
  • 66) 0.210 119 365 066 752 × 2 = 0 + 0.420 238 730 133 504;
  • 67) 0.420 238 730 133 504 × 2 = 0 + 0.840 477 460 267 008;
  • 68) 0.840 477 460 267 008 × 2 = 1 + 0.680 954 920 534 016;
  • 69) 0.680 954 920 534 016 × 2 = 1 + 0.361 909 841 068 032;
  • 70) 0.361 909 841 068 032 × 2 = 0 + 0.723 819 682 136 064;
  • 71) 0.723 819 682 136 064 × 2 = 1 + 0.447 639 364 272 128;
  • 72) 0.447 639 364 272 128 × 2 = 0 + 0.895 278 728 544 256;
  • 73) 0.895 278 728 544 256 × 2 = 1 + 0.790 557 457 088 512;
  • 74) 0.790 557 457 088 512 × 2 = 1 + 0.581 114 914 177 024;
  • 75) 0.581 114 914 177 024 × 2 = 1 + 0.162 229 828 354 048;
  • 76) 0.162 229 828 354 048 × 2 = 0 + 0.324 459 656 708 096;
  • 77) 0.324 459 656 708 096 × 2 = 0 + 0.648 919 313 416 192;
  • 78) 0.648 919 313 416 192 × 2 = 1 + 0.297 838 626 832 384;
  • 79) 0.297 838 626 832 384 × 2 = 0 + 0.595 677 253 664 768;
  • 80) 0.595 677 253 664 768 × 2 = 1 + 0.191 354 507 329 536;
  • 81) 0.191 354 507 329 536 × 2 = 0 + 0.382 709 014 659 072;
  • 82) 0.382 709 014 659 072 × 2 = 0 + 0.765 418 029 318 144;
  • 83) 0.765 418 029 318 144 × 2 = 1 + 0.530 836 058 636 288;
  • 84) 0.530 836 058 636 288 × 2 = 1 + 0.061 672 117 272 576;
  • 85) 0.061 672 117 272 576 × 2 = 0 + 0.123 344 234 545 152;
  • 86) 0.123 344 234 545 152 × 2 = 0 + 0.246 688 469 090 304;
  • 87) 0.246 688 469 090 304 × 2 = 0 + 0.493 376 938 180 608;

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:


0.000 000 000 040 986(10) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0010 1101 0001 0000 1000 1000 1000 1100 1001 1010 1110 0101 0011 000(2)

6. Positive number before normalization:

0.000 000 000 040 986(10) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0010 1101 0001 0000 1000 1000 1000 1100 1001 1010 1110 0101 0011 000(2)

7. Normalize the binary representation of the number.

Shift the decimal mark 35 positions to the right, so that only one non zero digit remains to the left of it:


0.000 000 000 040 986(10) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0010 1101 0001 0000 1000 1000 1000 1100 1001 1010 1110 0101 0011 000(2) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0010 1101 0001 0000 1000 1000 1000 1100 1001 1010 1110 0101 0011 000(2) × 20 =

1.0110 1000 1000 0100 0100 0100 0110 0100 1101 0111 0010 1001 1000(2) × 2-35


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): -35

Mantissa (not normalized):
1.0110 1000 1000 0100 0100 0100 0110 0100 1101 0111 0010 1001 1000


9. Adjust the exponent.

Use the 11 bit excess/bias notation:


Exponent (adjusted) =

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

-35 + 2(11-1) - 1 =

(-35 + 1 023)(10) =

988(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;
  • 988 ÷ 2 = 494 + 0;
  • 494 ÷ 2 = 247 + 0;
  • 247 ÷ 2 = 123 + 1;
  • 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) =

988(10) =

011 1101 1100(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 1000 1000 0100 0100 0100 0110 0100 1101 0111 0010 1001 1000 =

0110 1000 1000 0100 0100 0100 0110 0100 1101 0111 0010 1001 1000


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 1100

Mantissa (52 bits) =
0110 1000 1000 0100 0100 0100 0110 0100 1101 0111 0010 1001 1000


Decimal number -0.000 000 000 040 986 converted to 64 bit double precision IEEE 754 binary floating point representation:

1 - 011 1101 1100 - 0110 1000 1000 0100 0100 0100 0110 0100 1101 0111 0010 1001 1000

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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