-0.000 000 000 010 397 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 010 397(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
-0.000 000 000 010 397(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:

|-0.000 000 000 010 397| = 0.000 000 000 010 397


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: 0.000 000 000 010 397.

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) 0.000 000 000 010 397 × 2 = 0 + 0.000 000 000 020 794;
  • 2) 0.000 000 000 020 794 × 2 = 0 + 0.000 000 000 041 588;
  • 3) 0.000 000 000 041 588 × 2 = 0 + 0.000 000 000 083 176;
  • 4) 0.000 000 000 083 176 × 2 = 0 + 0.000 000 000 166 352;
  • 5) 0.000 000 000 166 352 × 2 = 0 + 0.000 000 000 332 704;
  • 6) 0.000 000 000 332 704 × 2 = 0 + 0.000 000 000 665 408;
  • 7) 0.000 000 000 665 408 × 2 = 0 + 0.000 000 001 330 816;
  • 8) 0.000 000 001 330 816 × 2 = 0 + 0.000 000 002 661 632;
  • 9) 0.000 000 002 661 632 × 2 = 0 + 0.000 000 005 323 264;
  • 10) 0.000 000 005 323 264 × 2 = 0 + 0.000 000 010 646 528;
  • 11) 0.000 000 010 646 528 × 2 = 0 + 0.000 000 021 293 056;
  • 12) 0.000 000 021 293 056 × 2 = 0 + 0.000 000 042 586 112;
  • 13) 0.000 000 042 586 112 × 2 = 0 + 0.000 000 085 172 224;
  • 14) 0.000 000 085 172 224 × 2 = 0 + 0.000 000 170 344 448;
  • 15) 0.000 000 170 344 448 × 2 = 0 + 0.000 000 340 688 896;
  • 16) 0.000 000 340 688 896 × 2 = 0 + 0.000 000 681 377 792;
  • 17) 0.000 000 681 377 792 × 2 = 0 + 0.000 001 362 755 584;
  • 18) 0.000 001 362 755 584 × 2 = 0 + 0.000 002 725 511 168;
  • 19) 0.000 002 725 511 168 × 2 = 0 + 0.000 005 451 022 336;
  • 20) 0.000 005 451 022 336 × 2 = 0 + 0.000 010 902 044 672;
  • 21) 0.000 010 902 044 672 × 2 = 0 + 0.000 021 804 089 344;
  • 22) 0.000 021 804 089 344 × 2 = 0 + 0.000 043 608 178 688;
  • 23) 0.000 043 608 178 688 × 2 = 0 + 0.000 087 216 357 376;
  • 24) 0.000 087 216 357 376 × 2 = 0 + 0.000 174 432 714 752;
  • 25) 0.000 174 432 714 752 × 2 = 0 + 0.000 348 865 429 504;
  • 26) 0.000 348 865 429 504 × 2 = 0 + 0.000 697 730 859 008;
  • 27) 0.000 697 730 859 008 × 2 = 0 + 0.001 395 461 718 016;
  • 28) 0.001 395 461 718 016 × 2 = 0 + 0.002 790 923 436 032;
  • 29) 0.002 790 923 436 032 × 2 = 0 + 0.005 581 846 872 064;
  • 30) 0.005 581 846 872 064 × 2 = 0 + 0.011 163 693 744 128;
  • 31) 0.011 163 693 744 128 × 2 = 0 + 0.022 327 387 488 256;
  • 32) 0.022 327 387 488 256 × 2 = 0 + 0.044 654 774 976 512;
  • 33) 0.044 654 774 976 512 × 2 = 0 + 0.089 309 549 953 024;
  • 34) 0.089 309 549 953 024 × 2 = 0 + 0.178 619 099 906 048;
  • 35) 0.178 619 099 906 048 × 2 = 0 + 0.357 238 199 812 096;
  • 36) 0.357 238 199 812 096 × 2 = 0 + 0.714 476 399 624 192;
  • 37) 0.714 476 399 624 192 × 2 = 1 + 0.428 952 799 248 384;
  • 38) 0.428 952 799 248 384 × 2 = 0 + 0.857 905 598 496 768;
  • 39) 0.857 905 598 496 768 × 2 = 1 + 0.715 811 196 993 536;
  • 40) 0.715 811 196 993 536 × 2 = 1 + 0.431 622 393 987 072;
  • 41) 0.431 622 393 987 072 × 2 = 0 + 0.863 244 787 974 144;
  • 42) 0.863 244 787 974 144 × 2 = 1 + 0.726 489 575 948 288;
  • 43) 0.726 489 575 948 288 × 2 = 1 + 0.452 979 151 896 576;
  • 44) 0.452 979 151 896 576 × 2 = 0 + 0.905 958 303 793 152;
  • 45) 0.905 958 303 793 152 × 2 = 1 + 0.811 916 607 586 304;
  • 46) 0.811 916 607 586 304 × 2 = 1 + 0.623 833 215 172 608;
  • 47) 0.623 833 215 172 608 × 2 = 1 + 0.247 666 430 345 216;
  • 48) 0.247 666 430 345 216 × 2 = 0 + 0.495 332 860 690 432;
  • 49) 0.495 332 860 690 432 × 2 = 0 + 0.990 665 721 380 864;
  • 50) 0.990 665 721 380 864 × 2 = 1 + 0.981 331 442 761 728;
  • 51) 0.981 331 442 761 728 × 2 = 1 + 0.962 662 885 523 456;
  • 52) 0.962 662 885 523 456 × 2 = 1 + 0.925 325 771 046 912;
  • 53) 0.925 325 771 046 912 × 2 = 1 + 0.850 651 542 093 824;
  • 54) 0.850 651 542 093 824 × 2 = 1 + 0.701 303 084 187 648;
  • 55) 0.701 303 084 187 648 × 2 = 1 + 0.402 606 168 375 296;
  • 56) 0.402 606 168 375 296 × 2 = 0 + 0.805 212 336 750 592;
  • 57) 0.805 212 336 750 592 × 2 = 1 + 0.610 424 673 501 184;
  • 58) 0.610 424 673 501 184 × 2 = 1 + 0.220 849 347 002 368;
  • 59) 0.220 849 347 002 368 × 2 = 0 + 0.441 698 694 004 736;
  • 60) 0.441 698 694 004 736 × 2 = 0 + 0.883 397 388 009 472;
  • 61) 0.883 397 388 009 472 × 2 = 1 + 0.766 794 776 018 944;
  • 62) 0.766 794 776 018 944 × 2 = 1 + 0.533 589 552 037 888;
  • 63) 0.533 589 552 037 888 × 2 = 1 + 0.067 179 104 075 776;
  • 64) 0.067 179 104 075 776 × 2 = 0 + 0.134 358 208 151 552;
  • 65) 0.134 358 208 151 552 × 2 = 0 + 0.268 716 416 303 104;
  • 66) 0.268 716 416 303 104 × 2 = 0 + 0.537 432 832 606 208;
  • 67) 0.537 432 832 606 208 × 2 = 1 + 0.074 865 665 212 416;
  • 68) 0.074 865 665 212 416 × 2 = 0 + 0.149 731 330 424 832;
  • 69) 0.149 731 330 424 832 × 2 = 0 + 0.299 462 660 849 664;
  • 70) 0.299 462 660 849 664 × 2 = 0 + 0.598 925 321 699 328;
  • 71) 0.598 925 321 699 328 × 2 = 1 + 0.197 850 643 398 656;
  • 72) 0.197 850 643 398 656 × 2 = 0 + 0.395 701 286 797 312;
  • 73) 0.395 701 286 797 312 × 2 = 0 + 0.791 402 573 594 624;
  • 74) 0.791 402 573 594 624 × 2 = 1 + 0.582 805 147 189 248;
  • 75) 0.582 805 147 189 248 × 2 = 1 + 0.165 610 294 378 496;
  • 76) 0.165 610 294 378 496 × 2 = 0 + 0.331 220 588 756 992;
  • 77) 0.331 220 588 756 992 × 2 = 0 + 0.662 441 177 513 984;
  • 78) 0.662 441 177 513 984 × 2 = 1 + 0.324 882 355 027 968;
  • 79) 0.324 882 355 027 968 × 2 = 0 + 0.649 764 710 055 936;
  • 80) 0.649 764 710 055 936 × 2 = 1 + 0.299 529 420 111 872;
  • 81) 0.299 529 420 111 872 × 2 = 0 + 0.599 058 840 223 744;
  • 82) 0.599 058 840 223 744 × 2 = 1 + 0.198 117 680 447 488;
  • 83) 0.198 117 680 447 488 × 2 = 0 + 0.396 235 360 894 976;
  • 84) 0.396 235 360 894 976 × 2 = 0 + 0.792 470 721 789 952;
  • 85) 0.792 470 721 789 952 × 2 = 1 + 0.584 941 443 579 904;
  • 86) 0.584 941 443 579 904 × 2 = 1 + 0.169 882 887 159 808;
  • 87) 0.169 882 887 159 808 × 2 = 0 + 0.339 765 774 319 616;
  • 88) 0.339 765 774 319 616 × 2 = 0 + 0.679 531 548 639 232;
  • 89) 0.679 531 548 639 232 × 2 = 1 + 0.359 063 097 278 464;

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:


0.000 000 000 010 397(10) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 1011 0110 1110 0111 1110 1100 1110 0010 0010 0110 0101 0100 1100 1(2)

6. Positive number before normalization:

0.000 000 000 010 397(10) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 1011 0110 1110 0111 1110 1100 1110 0010 0010 0110 0101 0100 1100 1(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:


0.000 000 000 010 397(10) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 1011 0110 1110 0111 1110 1100 1110 0010 0010 0110 0101 0100 1100 1(2) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 1011 0110 1110 0111 1110 1100 1110 0010 0010 0110 0101 0100 1100 1(2) × 20 =

1.0110 1101 1100 1111 1101 1001 1100 0100 0100 1100 1010 1001 1001(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 1101 1100 1111 1101 1001 1100 0100 0100 1100 1010 1001 1001


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 1101 1100 1111 1101 1001 1100 0100 0100 1100 1010 1001 1001 =

0110 1101 1100 1111 1101 1001 1100 0100 0100 1100 1010 1001 1001


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 1101 1100 1111 1101 1001 1100 0100 0100 1100 1010 1001 1001


Decimal number -0.000 000 000 010 397 converted to 64 bit double precision IEEE 754 binary floating point representation:

1 - 011 1101 1010 - 0110 1101 1100 1111 1101 1001 1100 0100 0100 1100 1010 1001 1001

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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