-0.000 035 666 835 299 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 035 666 835 299(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
-0.000 035 666 835 299(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:

|-0.000 035 666 835 299| = 0.000 035 666 835 299


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: 0.000 035 666 835 299.

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) 0.000 035 666 835 299 × 2 = 0 + 0.000 071 333 670 598;
  • 2) 0.000 071 333 670 598 × 2 = 0 + 0.000 142 667 341 196;
  • 3) 0.000 142 667 341 196 × 2 = 0 + 0.000 285 334 682 392;
  • 4) 0.000 285 334 682 392 × 2 = 0 + 0.000 570 669 364 784;
  • 5) 0.000 570 669 364 784 × 2 = 0 + 0.001 141 338 729 568;
  • 6) 0.001 141 338 729 568 × 2 = 0 + 0.002 282 677 459 136;
  • 7) 0.002 282 677 459 136 × 2 = 0 + 0.004 565 354 918 272;
  • 8) 0.004 565 354 918 272 × 2 = 0 + 0.009 130 709 836 544;
  • 9) 0.009 130 709 836 544 × 2 = 0 + 0.018 261 419 673 088;
  • 10) 0.018 261 419 673 088 × 2 = 0 + 0.036 522 839 346 176;
  • 11) 0.036 522 839 346 176 × 2 = 0 + 0.073 045 678 692 352;
  • 12) 0.073 045 678 692 352 × 2 = 0 + 0.146 091 357 384 704;
  • 13) 0.146 091 357 384 704 × 2 = 0 + 0.292 182 714 769 408;
  • 14) 0.292 182 714 769 408 × 2 = 0 + 0.584 365 429 538 816;
  • 15) 0.584 365 429 538 816 × 2 = 1 + 0.168 730 859 077 632;
  • 16) 0.168 730 859 077 632 × 2 = 0 + 0.337 461 718 155 264;
  • 17) 0.337 461 718 155 264 × 2 = 0 + 0.674 923 436 310 528;
  • 18) 0.674 923 436 310 528 × 2 = 1 + 0.349 846 872 621 056;
  • 19) 0.349 846 872 621 056 × 2 = 0 + 0.699 693 745 242 112;
  • 20) 0.699 693 745 242 112 × 2 = 1 + 0.399 387 490 484 224;
  • 21) 0.399 387 490 484 224 × 2 = 0 + 0.798 774 980 968 448;
  • 22) 0.798 774 980 968 448 × 2 = 1 + 0.597 549 961 936 896;
  • 23) 0.597 549 961 936 896 × 2 = 1 + 0.195 099 923 873 792;
  • 24) 0.195 099 923 873 792 × 2 = 0 + 0.390 199 847 747 584;
  • 25) 0.390 199 847 747 584 × 2 = 0 + 0.780 399 695 495 168;
  • 26) 0.780 399 695 495 168 × 2 = 1 + 0.560 799 390 990 336;
  • 27) 0.560 799 390 990 336 × 2 = 1 + 0.121 598 781 980 672;
  • 28) 0.121 598 781 980 672 × 2 = 0 + 0.243 197 563 961 344;
  • 29) 0.243 197 563 961 344 × 2 = 0 + 0.486 395 127 922 688;
  • 30) 0.486 395 127 922 688 × 2 = 0 + 0.972 790 255 845 376;
  • 31) 0.972 790 255 845 376 × 2 = 1 + 0.945 580 511 690 752;
  • 32) 0.945 580 511 690 752 × 2 = 1 + 0.891 161 023 381 504;
  • 33) 0.891 161 023 381 504 × 2 = 1 + 0.782 322 046 763 008;
  • 34) 0.782 322 046 763 008 × 2 = 1 + 0.564 644 093 526 016;
  • 35) 0.564 644 093 526 016 × 2 = 1 + 0.129 288 187 052 032;
  • 36) 0.129 288 187 052 032 × 2 = 0 + 0.258 576 374 104 064;
  • 37) 0.258 576 374 104 064 × 2 = 0 + 0.517 152 748 208 128;
  • 38) 0.517 152 748 208 128 × 2 = 1 + 0.034 305 496 416 256;
  • 39) 0.034 305 496 416 256 × 2 = 0 + 0.068 610 992 832 512;
  • 40) 0.068 610 992 832 512 × 2 = 0 + 0.137 221 985 665 024;
  • 41) 0.137 221 985 665 024 × 2 = 0 + 0.274 443 971 330 048;
  • 42) 0.274 443 971 330 048 × 2 = 0 + 0.548 887 942 660 096;
  • 43) 0.548 887 942 660 096 × 2 = 1 + 0.097 775 885 320 192;
  • 44) 0.097 775 885 320 192 × 2 = 0 + 0.195 551 770 640 384;
  • 45) 0.195 551 770 640 384 × 2 = 0 + 0.391 103 541 280 768;
  • 46) 0.391 103 541 280 768 × 2 = 0 + 0.782 207 082 561 536;
  • 47) 0.782 207 082 561 536 × 2 = 1 + 0.564 414 165 123 072;
  • 48) 0.564 414 165 123 072 × 2 = 1 + 0.128 828 330 246 144;
  • 49) 0.128 828 330 246 144 × 2 = 0 + 0.257 656 660 492 288;
  • 50) 0.257 656 660 492 288 × 2 = 0 + 0.515 313 320 984 576;
  • 51) 0.515 313 320 984 576 × 2 = 1 + 0.030 626 641 969 152;
  • 52) 0.030 626 641 969 152 × 2 = 0 + 0.061 253 283 938 304;
  • 53) 0.061 253 283 938 304 × 2 = 0 + 0.122 506 567 876 608;
  • 54) 0.122 506 567 876 608 × 2 = 0 + 0.245 013 135 753 216;
  • 55) 0.245 013 135 753 216 × 2 = 0 + 0.490 026 271 506 432;
  • 56) 0.490 026 271 506 432 × 2 = 0 + 0.980 052 543 012 864;
  • 57) 0.980 052 543 012 864 × 2 = 1 + 0.960 105 086 025 728;
  • 58) 0.960 105 086 025 728 × 2 = 1 + 0.920 210 172 051 456;
  • 59) 0.920 210 172 051 456 × 2 = 1 + 0.840 420 344 102 912;
  • 60) 0.840 420 344 102 912 × 2 = 1 + 0.680 840 688 205 824;
  • 61) 0.680 840 688 205 824 × 2 = 1 + 0.361 681 376 411 648;
  • 62) 0.361 681 376 411 648 × 2 = 0 + 0.723 362 752 823 296;
  • 63) 0.723 362 752 823 296 × 2 = 1 + 0.446 725 505 646 592;
  • 64) 0.446 725 505 646 592 × 2 = 0 + 0.893 451 011 293 184;
  • 65) 0.893 451 011 293 184 × 2 = 1 + 0.786 902 022 586 368;
  • 66) 0.786 902 022 586 368 × 2 = 1 + 0.573 804 045 172 736;
  • 67) 0.573 804 045 172 736 × 2 = 1 + 0.147 608 090 345 472;

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:


0.000 035 666 835 299(10) =

0.0000 0000 0000 0010 0101 0110 0110 0011 1110 0100 0010 0011 0010 0000 1111 1010 111(2)

6. Positive number before normalization:

0.000 035 666 835 299(10) =

0.0000 0000 0000 0010 0101 0110 0110 0011 1110 0100 0010 0011 0010 0000 1111 1010 111(2)

7. Normalize the binary representation of the number.

Shift the decimal mark 15 positions to the right, so that only one non zero digit remains to the left of it:


0.000 035 666 835 299(10) =

0.0000 0000 0000 0010 0101 0110 0110 0011 1110 0100 0010 0011 0010 0000 1111 1010 111(2) =

0.0000 0000 0000 0010 0101 0110 0110 0011 1110 0100 0010 0011 0010 0000 1111 1010 111(2) × 20 =

1.0010 1011 0011 0001 1111 0010 0001 0001 1001 0000 0111 1101 0111(2) × 2-15


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): -15

Mantissa (not normalized):
1.0010 1011 0011 0001 1111 0010 0001 0001 1001 0000 0111 1101 0111


9. Adjust the exponent.

Use the 11 bit excess/bias notation:


Exponent (adjusted) =

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

-15 + 2(11-1) - 1 =

(-15 + 1 023)(10) =

1 008(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;
  • 1 008 ÷ 2 = 504 + 0;
  • 504 ÷ 2 = 252 + 0;
  • 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) =

1008(10) =

011 1111 0000(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. 0010 1011 0011 0001 1111 0010 0001 0001 1001 0000 0111 1101 0111 =

0010 1011 0011 0001 1111 0010 0001 0001 1001 0000 0111 1101 0111


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 0000

Mantissa (52 bits) =
0010 1011 0011 0001 1111 0010 0001 0001 1001 0000 0111 1101 0111


Decimal number -0.000 035 666 835 299 converted to 64 bit double precision IEEE 754 binary floating point representation:

1 - 011 1111 0000 - 0010 1011 0011 0001 1111 0010 0001 0001 1001 0000 0111 1101 0111

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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