0.000 000 000 000 182 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 0.000 000 000 000 182(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
0.000 000 000 000 182(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: 0.000 000 000 000 182.

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) 0.000 000 000 000 182 × 2 = 0 + 0.000 000 000 000 364;
  • 2) 0.000 000 000 000 364 × 2 = 0 + 0.000 000 000 000 728;
  • 3) 0.000 000 000 000 728 × 2 = 0 + 0.000 000 000 001 456;
  • 4) 0.000 000 000 001 456 × 2 = 0 + 0.000 000 000 002 912;
  • 5) 0.000 000 000 002 912 × 2 = 0 + 0.000 000 000 005 824;
  • 6) 0.000 000 000 005 824 × 2 = 0 + 0.000 000 000 011 648;
  • 7) 0.000 000 000 011 648 × 2 = 0 + 0.000 000 000 023 296;
  • 8) 0.000 000 000 023 296 × 2 = 0 + 0.000 000 000 046 592;
  • 9) 0.000 000 000 046 592 × 2 = 0 + 0.000 000 000 093 184;
  • 10) 0.000 000 000 093 184 × 2 = 0 + 0.000 000 000 186 368;
  • 11) 0.000 000 000 186 368 × 2 = 0 + 0.000 000 000 372 736;
  • 12) 0.000 000 000 372 736 × 2 = 0 + 0.000 000 000 745 472;
  • 13) 0.000 000 000 745 472 × 2 = 0 + 0.000 000 001 490 944;
  • 14) 0.000 000 001 490 944 × 2 = 0 + 0.000 000 002 981 888;
  • 15) 0.000 000 002 981 888 × 2 = 0 + 0.000 000 005 963 776;
  • 16) 0.000 000 005 963 776 × 2 = 0 + 0.000 000 011 927 552;
  • 17) 0.000 000 011 927 552 × 2 = 0 + 0.000 000 023 855 104;
  • 18) 0.000 000 023 855 104 × 2 = 0 + 0.000 000 047 710 208;
  • 19) 0.000 000 047 710 208 × 2 = 0 + 0.000 000 095 420 416;
  • 20) 0.000 000 095 420 416 × 2 = 0 + 0.000 000 190 840 832;
  • 21) 0.000 000 190 840 832 × 2 = 0 + 0.000 000 381 681 664;
  • 22) 0.000 000 381 681 664 × 2 = 0 + 0.000 000 763 363 328;
  • 23) 0.000 000 763 363 328 × 2 = 0 + 0.000 001 526 726 656;
  • 24) 0.000 001 526 726 656 × 2 = 0 + 0.000 003 053 453 312;
  • 25) 0.000 003 053 453 312 × 2 = 0 + 0.000 006 106 906 624;
  • 26) 0.000 006 106 906 624 × 2 = 0 + 0.000 012 213 813 248;
  • 27) 0.000 012 213 813 248 × 2 = 0 + 0.000 024 427 626 496;
  • 28) 0.000 024 427 626 496 × 2 = 0 + 0.000 048 855 252 992;
  • 29) 0.000 048 855 252 992 × 2 = 0 + 0.000 097 710 505 984;
  • 30) 0.000 097 710 505 984 × 2 = 0 + 0.000 195 421 011 968;
  • 31) 0.000 195 421 011 968 × 2 = 0 + 0.000 390 842 023 936;
  • 32) 0.000 390 842 023 936 × 2 = 0 + 0.000 781 684 047 872;
  • 33) 0.000 781 684 047 872 × 2 = 0 + 0.001 563 368 095 744;
  • 34) 0.001 563 368 095 744 × 2 = 0 + 0.003 126 736 191 488;
  • 35) 0.003 126 736 191 488 × 2 = 0 + 0.006 253 472 382 976;
  • 36) 0.006 253 472 382 976 × 2 = 0 + 0.012 506 944 765 952;
  • 37) 0.012 506 944 765 952 × 2 = 0 + 0.025 013 889 531 904;
  • 38) 0.025 013 889 531 904 × 2 = 0 + 0.050 027 779 063 808;
  • 39) 0.050 027 779 063 808 × 2 = 0 + 0.100 055 558 127 616;
  • 40) 0.100 055 558 127 616 × 2 = 0 + 0.200 111 116 255 232;
  • 41) 0.200 111 116 255 232 × 2 = 0 + 0.400 222 232 510 464;
  • 42) 0.400 222 232 510 464 × 2 = 0 + 0.800 444 465 020 928;
  • 43) 0.800 444 465 020 928 × 2 = 1 + 0.600 888 930 041 856;
  • 44) 0.600 888 930 041 856 × 2 = 1 + 0.201 777 860 083 712;
  • 45) 0.201 777 860 083 712 × 2 = 0 + 0.403 555 720 167 424;
  • 46) 0.403 555 720 167 424 × 2 = 0 + 0.807 111 440 334 848;
  • 47) 0.807 111 440 334 848 × 2 = 1 + 0.614 222 880 669 696;
  • 48) 0.614 222 880 669 696 × 2 = 1 + 0.228 445 761 339 392;
  • 49) 0.228 445 761 339 392 × 2 = 0 + 0.456 891 522 678 784;
  • 50) 0.456 891 522 678 784 × 2 = 0 + 0.913 783 045 357 568;
  • 51) 0.913 783 045 357 568 × 2 = 1 + 0.827 566 090 715 136;
  • 52) 0.827 566 090 715 136 × 2 = 1 + 0.655 132 181 430 272;
  • 53) 0.655 132 181 430 272 × 2 = 1 + 0.310 264 362 860 544;
  • 54) 0.310 264 362 860 544 × 2 = 0 + 0.620 528 725 721 088;
  • 55) 0.620 528 725 721 088 × 2 = 1 + 0.241 057 451 442 176;
  • 56) 0.241 057 451 442 176 × 2 = 0 + 0.482 114 902 884 352;
  • 57) 0.482 114 902 884 352 × 2 = 0 + 0.964 229 805 768 704;
  • 58) 0.964 229 805 768 704 × 2 = 1 + 0.928 459 611 537 408;
  • 59) 0.928 459 611 537 408 × 2 = 1 + 0.856 919 223 074 816;
  • 60) 0.856 919 223 074 816 × 2 = 1 + 0.713 838 446 149 632;
  • 61) 0.713 838 446 149 632 × 2 = 1 + 0.427 676 892 299 264;
  • 62) 0.427 676 892 299 264 × 2 = 0 + 0.855 353 784 598 528;
  • 63) 0.855 353 784 598 528 × 2 = 1 + 0.710 707 569 197 056;
  • 64) 0.710 707 569 197 056 × 2 = 1 + 0.421 415 138 394 112;
  • 65) 0.421 415 138 394 112 × 2 = 0 + 0.842 830 276 788 224;
  • 66) 0.842 830 276 788 224 × 2 = 1 + 0.685 660 553 576 448;
  • 67) 0.685 660 553 576 448 × 2 = 1 + 0.371 321 107 152 896;
  • 68) 0.371 321 107 152 896 × 2 = 0 + 0.742 642 214 305 792;
  • 69) 0.742 642 214 305 792 × 2 = 1 + 0.485 284 428 611 584;
  • 70) 0.485 284 428 611 584 × 2 = 0 + 0.970 568 857 223 168;
  • 71) 0.970 568 857 223 168 × 2 = 1 + 0.941 137 714 446 336;
  • 72) 0.941 137 714 446 336 × 2 = 1 + 0.882 275 428 892 672;
  • 73) 0.882 275 428 892 672 × 2 = 1 + 0.764 550 857 785 344;
  • 74) 0.764 550 857 785 344 × 2 = 1 + 0.529 101 715 570 688;
  • 75) 0.529 101 715 570 688 × 2 = 1 + 0.058 203 431 141 376;
  • 76) 0.058 203 431 141 376 × 2 = 0 + 0.116 406 862 282 752;
  • 77) 0.116 406 862 282 752 × 2 = 0 + 0.232 813 724 565 504;
  • 78) 0.232 813 724 565 504 × 2 = 0 + 0.465 627 449 131 008;
  • 79) 0.465 627 449 131 008 × 2 = 0 + 0.931 254 898 262 016;
  • 80) 0.931 254 898 262 016 × 2 = 1 + 0.862 509 796 524 032;
  • 81) 0.862 509 796 524 032 × 2 = 1 + 0.725 019 593 048 064;
  • 82) 0.725 019 593 048 064 × 2 = 1 + 0.450 039 186 096 128;
  • 83) 0.450 039 186 096 128 × 2 = 0 + 0.900 078 372 192 256;
  • 84) 0.900 078 372 192 256 × 2 = 1 + 0.800 156 744 384 512;
  • 85) 0.800 156 744 384 512 × 2 = 1 + 0.600 313 488 769 024;
  • 86) 0.600 313 488 769 024 × 2 = 1 + 0.200 626 977 538 048;
  • 87) 0.200 626 977 538 048 × 2 = 0 + 0.401 253 955 076 096;
  • 88) 0.401 253 955 076 096 × 2 = 0 + 0.802 507 910 152 192;
  • 89) 0.802 507 910 152 192 × 2 = 1 + 0.605 015 820 304 384;
  • 90) 0.605 015 820 304 384 × 2 = 1 + 0.210 031 640 608 768;
  • 91) 0.210 031 640 608 768 × 2 = 0 + 0.420 063 281 217 536;
  • 92) 0.420 063 281 217 536 × 2 = 0 + 0.840 126 562 435 072;
  • 93) 0.840 126 562 435 072 × 2 = 1 + 0.680 253 124 870 144;
  • 94) 0.680 253 124 870 144 × 2 = 1 + 0.360 506 249 740 288;
  • 95) 0.360 506 249 740 288 × 2 = 0 + 0.721 012 499 480 576;

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:


0.000 000 000 000 182(10) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0011 0011 0011 1010 0111 1011 0110 1011 1110 0001 1101 1100 1100 110(2)

5. Positive number before normalization:

0.000 000 000 000 182(10) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0011 0011 0011 1010 0111 1011 0110 1011 1110 0001 1101 1100 1100 110(2)

6. Normalize the binary representation of the number.

Shift the decimal mark 43 positions to the right, so that only one non zero digit remains to the left of it:


0.000 000 000 000 182(10) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0011 0011 0011 1010 0111 1011 0110 1011 1110 0001 1101 1100 1100 110(2) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0011 0011 0011 1010 0111 1011 0110 1011 1110 0001 1101 1100 1100 110(2) × 20 =

1.1001 1001 1101 0011 1101 1011 0101 1111 0000 1110 1110 0110 0110(2) × 2-43


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): -43

Mantissa (not normalized):
1.1001 1001 1101 0011 1101 1011 0101 1111 0000 1110 1110 0110 0110


8. Adjust the exponent.

Use the 11 bit excess/bias notation:


Exponent (adjusted) =

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

-43 + 2(11-1) - 1 =

(-43 + 1 023)(10) =

980(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;
  • 980 ÷ 2 = 490 + 0;
  • 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) =

980(10) =

011 1101 0100(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. 1001 1001 1101 0011 1101 1011 0101 1111 0000 1110 1110 0110 0110 =

1001 1001 1101 0011 1101 1011 0101 1111 0000 1110 1110 0110 0110


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 0100

Mantissa (52 bits) =
1001 1001 1101 0011 1101 1011 0101 1111 0000 1110 1110 0110 0110


Decimal number 0.000 000 000 000 182 converted to 64 bit double precision IEEE 754 binary floating point representation:

0 - 011 1101 0100 - 1001 1001 1101 0011 1101 1011 0101 1111 0000 1110 1110 0110 0110

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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