0.000 000 000 000 242 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 0.000 000 000 000 242(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
0.000 000 000 000 242(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. 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;

2. 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)


3. Convert to binary (base 2) the fractional part: 0.000 000 000 000 242.

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) 0.000 000 000 000 242 × 2 = 0 + 0.000 000 000 000 484;
  • 2) 0.000 000 000 000 484 × 2 = 0 + 0.000 000 000 000 968;
  • 3) 0.000 000 000 000 968 × 2 = 0 + 0.000 000 000 001 936;
  • 4) 0.000 000 000 001 936 × 2 = 0 + 0.000 000 000 003 872;
  • 5) 0.000 000 000 003 872 × 2 = 0 + 0.000 000 000 007 744;
  • 6) 0.000 000 000 007 744 × 2 = 0 + 0.000 000 000 015 488;
  • 7) 0.000 000 000 015 488 × 2 = 0 + 0.000 000 000 030 976;
  • 8) 0.000 000 000 030 976 × 2 = 0 + 0.000 000 000 061 952;
  • 9) 0.000 000 000 061 952 × 2 = 0 + 0.000 000 000 123 904;
  • 10) 0.000 000 000 123 904 × 2 = 0 + 0.000 000 000 247 808;
  • 11) 0.000 000 000 247 808 × 2 = 0 + 0.000 000 000 495 616;
  • 12) 0.000 000 000 495 616 × 2 = 0 + 0.000 000 000 991 232;
  • 13) 0.000 000 000 991 232 × 2 = 0 + 0.000 000 001 982 464;
  • 14) 0.000 000 001 982 464 × 2 = 0 + 0.000 000 003 964 928;
  • 15) 0.000 000 003 964 928 × 2 = 0 + 0.000 000 007 929 856;
  • 16) 0.000 000 007 929 856 × 2 = 0 + 0.000 000 015 859 712;
  • 17) 0.000 000 015 859 712 × 2 = 0 + 0.000 000 031 719 424;
  • 18) 0.000 000 031 719 424 × 2 = 0 + 0.000 000 063 438 848;
  • 19) 0.000 000 063 438 848 × 2 = 0 + 0.000 000 126 877 696;
  • 20) 0.000 000 126 877 696 × 2 = 0 + 0.000 000 253 755 392;
  • 21) 0.000 000 253 755 392 × 2 = 0 + 0.000 000 507 510 784;
  • 22) 0.000 000 507 510 784 × 2 = 0 + 0.000 001 015 021 568;
  • 23) 0.000 001 015 021 568 × 2 = 0 + 0.000 002 030 043 136;
  • 24) 0.000 002 030 043 136 × 2 = 0 + 0.000 004 060 086 272;
  • 25) 0.000 004 060 086 272 × 2 = 0 + 0.000 008 120 172 544;
  • 26) 0.000 008 120 172 544 × 2 = 0 + 0.000 016 240 345 088;
  • 27) 0.000 016 240 345 088 × 2 = 0 + 0.000 032 480 690 176;
  • 28) 0.000 032 480 690 176 × 2 = 0 + 0.000 064 961 380 352;
  • 29) 0.000 064 961 380 352 × 2 = 0 + 0.000 129 922 760 704;
  • 30) 0.000 129 922 760 704 × 2 = 0 + 0.000 259 845 521 408;
  • 31) 0.000 259 845 521 408 × 2 = 0 + 0.000 519 691 042 816;
  • 32) 0.000 519 691 042 816 × 2 = 0 + 0.001 039 382 085 632;
  • 33) 0.001 039 382 085 632 × 2 = 0 + 0.002 078 764 171 264;
  • 34) 0.002 078 764 171 264 × 2 = 0 + 0.004 157 528 342 528;
  • 35) 0.004 157 528 342 528 × 2 = 0 + 0.008 315 056 685 056;
  • 36) 0.008 315 056 685 056 × 2 = 0 + 0.016 630 113 370 112;
  • 37) 0.016 630 113 370 112 × 2 = 0 + 0.033 260 226 740 224;
  • 38) 0.033 260 226 740 224 × 2 = 0 + 0.066 520 453 480 448;
  • 39) 0.066 520 453 480 448 × 2 = 0 + 0.133 040 906 960 896;
  • 40) 0.133 040 906 960 896 × 2 = 0 + 0.266 081 813 921 792;
  • 41) 0.266 081 813 921 792 × 2 = 0 + 0.532 163 627 843 584;
  • 42) 0.532 163 627 843 584 × 2 = 1 + 0.064 327 255 687 168;
  • 43) 0.064 327 255 687 168 × 2 = 0 + 0.128 654 511 374 336;
  • 44) 0.128 654 511 374 336 × 2 = 0 + 0.257 309 022 748 672;
  • 45) 0.257 309 022 748 672 × 2 = 0 + 0.514 618 045 497 344;
  • 46) 0.514 618 045 497 344 × 2 = 1 + 0.029 236 090 994 688;
  • 47) 0.029 236 090 994 688 × 2 = 0 + 0.058 472 181 989 376;
  • 48) 0.058 472 181 989 376 × 2 = 0 + 0.116 944 363 978 752;
  • 49) 0.116 944 363 978 752 × 2 = 0 + 0.233 888 727 957 504;
  • 50) 0.233 888 727 957 504 × 2 = 0 + 0.467 777 455 915 008;
  • 51) 0.467 777 455 915 008 × 2 = 0 + 0.935 554 911 830 016;
  • 52) 0.935 554 911 830 016 × 2 = 1 + 0.871 109 823 660 032;
  • 53) 0.871 109 823 660 032 × 2 = 1 + 0.742 219 647 320 064;
  • 54) 0.742 219 647 320 064 × 2 = 1 + 0.484 439 294 640 128;
  • 55) 0.484 439 294 640 128 × 2 = 0 + 0.968 878 589 280 256;
  • 56) 0.968 878 589 280 256 × 2 = 1 + 0.937 757 178 560 512;
  • 57) 0.937 757 178 560 512 × 2 = 1 + 0.875 514 357 121 024;
  • 58) 0.875 514 357 121 024 × 2 = 1 + 0.751 028 714 242 048;
  • 59) 0.751 028 714 242 048 × 2 = 1 + 0.502 057 428 484 096;
  • 60) 0.502 057 428 484 096 × 2 = 1 + 0.004 114 856 968 192;
  • 61) 0.004 114 856 968 192 × 2 = 0 + 0.008 229 713 936 384;
  • 62) 0.008 229 713 936 384 × 2 = 0 + 0.016 459 427 872 768;
  • 63) 0.016 459 427 872 768 × 2 = 0 + 0.032 918 855 745 536;
  • 64) 0.032 918 855 745 536 × 2 = 0 + 0.065 837 711 491 072;
  • 65) 0.065 837 711 491 072 × 2 = 0 + 0.131 675 422 982 144;
  • 66) 0.131 675 422 982 144 × 2 = 0 + 0.263 350 845 964 288;
  • 67) 0.263 350 845 964 288 × 2 = 0 + 0.526 701 691 928 576;
  • 68) 0.526 701 691 928 576 × 2 = 1 + 0.053 403 383 857 152;
  • 69) 0.053 403 383 857 152 × 2 = 0 + 0.106 806 767 714 304;
  • 70) 0.106 806 767 714 304 × 2 = 0 + 0.213 613 535 428 608;
  • 71) 0.213 613 535 428 608 × 2 = 0 + 0.427 227 070 857 216;
  • 72) 0.427 227 070 857 216 × 2 = 0 + 0.854 454 141 714 432;
  • 73) 0.854 454 141 714 432 × 2 = 1 + 0.708 908 283 428 864;
  • 74) 0.708 908 283 428 864 × 2 = 1 + 0.417 816 566 857 728;
  • 75) 0.417 816 566 857 728 × 2 = 0 + 0.835 633 133 715 456;
  • 76) 0.835 633 133 715 456 × 2 = 1 + 0.671 266 267 430 912;
  • 77) 0.671 266 267 430 912 × 2 = 1 + 0.342 532 534 861 824;
  • 78) 0.342 532 534 861 824 × 2 = 0 + 0.685 065 069 723 648;
  • 79) 0.685 065 069 723 648 × 2 = 1 + 0.370 130 139 447 296;
  • 80) 0.370 130 139 447 296 × 2 = 0 + 0.740 260 278 894 592;
  • 81) 0.740 260 278 894 592 × 2 = 1 + 0.480 520 557 789 184;
  • 82) 0.480 520 557 789 184 × 2 = 0 + 0.961 041 115 578 368;
  • 83) 0.961 041 115 578 368 × 2 = 1 + 0.922 082 231 156 736;
  • 84) 0.922 082 231 156 736 × 2 = 1 + 0.844 164 462 313 472;
  • 85) 0.844 164 462 313 472 × 2 = 1 + 0.688 328 924 626 944;
  • 86) 0.688 328 924 626 944 × 2 = 1 + 0.376 657 849 253 888;
  • 87) 0.376 657 849 253 888 × 2 = 0 + 0.753 315 698 507 776;
  • 88) 0.753 315 698 507 776 × 2 = 1 + 0.506 631 397 015 552;
  • 89) 0.506 631 397 015 552 × 2 = 1 + 0.013 262 794 031 104;
  • 90) 0.013 262 794 031 104 × 2 = 0 + 0.026 525 588 062 208;
  • 91) 0.026 525 588 062 208 × 2 = 0 + 0.053 051 176 124 416;
  • 92) 0.053 051 176 124 416 × 2 = 0 + 0.106 102 352 248 832;
  • 93) 0.106 102 352 248 832 × 2 = 0 + 0.212 204 704 497 664;
  • 94) 0.212 204 704 497 664 × 2 = 0 + 0.424 409 408 995 328;

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).


4. 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:


0.000 000 000 000 242(10) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0100 0100 0001 1101 1111 0000 0001 0000 1101 1010 1011 1101 1000 00(2)

5. Positive number before normalization:

0.000 000 000 000 242(10) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0100 0100 0001 1101 1111 0000 0001 0000 1101 1010 1011 1101 1000 00(2)

6. Normalize the binary representation of the number.

Shift the decimal mark 42 positions to the right, so that only one non zero digit remains to the left of it:


0.000 000 000 000 242(10) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0100 0100 0001 1101 1111 0000 0001 0000 1101 1010 1011 1101 1000 00(2) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0100 0100 0001 1101 1111 0000 0001 0000 1101 1010 1011 1101 1000 00(2) × 20 =

1.0001 0000 0111 0111 1100 0000 0100 0011 0110 1010 1111 0110 0000(2) × 2-42


7. 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 0 (a positive number)

Exponent (unadjusted): -42

Mantissa (not normalized):
1.0001 0000 0111 0111 1100 0000 0100 0011 0110 1010 1111 0110 0000


8. Adjust the exponent.

Use the 11 bit excess/bias notation:


Exponent (adjusted) =

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

-42 + 2(11-1) - 1 =

(-42 + 1 023)(10) =

981(10)


9. 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;
  • 981 ÷ 2 = 490 + 1;
  • 490 ÷ 2 = 245 + 0;
  • 245 ÷ 2 = 122 + 1;
  • 122 ÷ 2 = 61 + 0;
  • 61 ÷ 2 = 30 + 1;
  • 30 ÷ 2 = 15 + 0;
  • 15 ÷ 2 = 7 + 1;
  • 7 ÷ 2 = 3 + 1;
  • 3 ÷ 2 = 1 + 1;
  • 1 ÷ 2 = 0 + 1;

10. Construct the base 2 representation of the adjusted exponent.

Take all the remainders starting from the bottom of the list constructed above.


Exponent (adjusted) =

981(10) =

011 1101 0101(2)


11. 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. 0001 0000 0111 0111 1100 0000 0100 0011 0110 1010 1111 0110 0000 =

0001 0000 0111 0111 1100 0000 0100 0011 0110 1010 1111 0110 0000


12. The three elements that make up the number's 64 bit double precision IEEE 754 binary floating point representation:

Sign (1 bit) =
0 (a positive number)

Exponent (11 bits) =
011 1101 0101

Mantissa (52 bits) =
0001 0000 0111 0111 1100 0000 0100 0011 0110 1010 1111 0110 0000


Decimal number 0.000 000 000 000 242 converted to 64 bit double precision IEEE 754 binary floating point representation:

0 - 011 1101 0101 - 0001 0000 0111 0111 1100 0000 0100 0011 0110 1010 1111 0110 0000

Share

» Convert decimal 0.000 000 000 000 261 to 64 bit double precision IEEE 754 binary floating point representation standard

» Calculations Performed by Our Visitors: Decimal Numbers Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard. Data organized on a Monthly Basis

» Month 06, 2026 [June]: Decimal Numbers Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard. Calculations performed during the month of: June - by our visitors

» Improve your social skills: The Hidden Requirement in Every Single Job Description. NotebookLM YouTube Video of 8.15 minutes

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