-0.000 000 000 000 176 75 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 000 176 75(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
-0.000 000 000 000 176 75(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:

|-0.000 000 000 000 176 75| = 0.000 000 000 000 176 75


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: 0.000 000 000 000 176 75.

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) 0.000 000 000 000 176 75 × 2 = 0 + 0.000 000 000 000 353 5;
  • 2) 0.000 000 000 000 353 5 × 2 = 0 + 0.000 000 000 000 707;
  • 3) 0.000 000 000 000 707 × 2 = 0 + 0.000 000 000 001 414;
  • 4) 0.000 000 000 001 414 × 2 = 0 + 0.000 000 000 002 828;
  • 5) 0.000 000 000 002 828 × 2 = 0 + 0.000 000 000 005 656;
  • 6) 0.000 000 000 005 656 × 2 = 0 + 0.000 000 000 011 312;
  • 7) 0.000 000 000 011 312 × 2 = 0 + 0.000 000 000 022 624;
  • 8) 0.000 000 000 022 624 × 2 = 0 + 0.000 000 000 045 248;
  • 9) 0.000 000 000 045 248 × 2 = 0 + 0.000 000 000 090 496;
  • 10) 0.000 000 000 090 496 × 2 = 0 + 0.000 000 000 180 992;
  • 11) 0.000 000 000 180 992 × 2 = 0 + 0.000 000 000 361 984;
  • 12) 0.000 000 000 361 984 × 2 = 0 + 0.000 000 000 723 968;
  • 13) 0.000 000 000 723 968 × 2 = 0 + 0.000 000 001 447 936;
  • 14) 0.000 000 001 447 936 × 2 = 0 + 0.000 000 002 895 872;
  • 15) 0.000 000 002 895 872 × 2 = 0 + 0.000 000 005 791 744;
  • 16) 0.000 000 005 791 744 × 2 = 0 + 0.000 000 011 583 488;
  • 17) 0.000 000 011 583 488 × 2 = 0 + 0.000 000 023 166 976;
  • 18) 0.000 000 023 166 976 × 2 = 0 + 0.000 000 046 333 952;
  • 19) 0.000 000 046 333 952 × 2 = 0 + 0.000 000 092 667 904;
  • 20) 0.000 000 092 667 904 × 2 = 0 + 0.000 000 185 335 808;
  • 21) 0.000 000 185 335 808 × 2 = 0 + 0.000 000 370 671 616;
  • 22) 0.000 000 370 671 616 × 2 = 0 + 0.000 000 741 343 232;
  • 23) 0.000 000 741 343 232 × 2 = 0 + 0.000 001 482 686 464;
  • 24) 0.000 001 482 686 464 × 2 = 0 + 0.000 002 965 372 928;
  • 25) 0.000 002 965 372 928 × 2 = 0 + 0.000 005 930 745 856;
  • 26) 0.000 005 930 745 856 × 2 = 0 + 0.000 011 861 491 712;
  • 27) 0.000 011 861 491 712 × 2 = 0 + 0.000 023 722 983 424;
  • 28) 0.000 023 722 983 424 × 2 = 0 + 0.000 047 445 966 848;
  • 29) 0.000 047 445 966 848 × 2 = 0 + 0.000 094 891 933 696;
  • 30) 0.000 094 891 933 696 × 2 = 0 + 0.000 189 783 867 392;
  • 31) 0.000 189 783 867 392 × 2 = 0 + 0.000 379 567 734 784;
  • 32) 0.000 379 567 734 784 × 2 = 0 + 0.000 759 135 469 568;
  • 33) 0.000 759 135 469 568 × 2 = 0 + 0.001 518 270 939 136;
  • 34) 0.001 518 270 939 136 × 2 = 0 + 0.003 036 541 878 272;
  • 35) 0.003 036 541 878 272 × 2 = 0 + 0.006 073 083 756 544;
  • 36) 0.006 073 083 756 544 × 2 = 0 + 0.012 146 167 513 088;
  • 37) 0.012 146 167 513 088 × 2 = 0 + 0.024 292 335 026 176;
  • 38) 0.024 292 335 026 176 × 2 = 0 + 0.048 584 670 052 352;
  • 39) 0.048 584 670 052 352 × 2 = 0 + 0.097 169 340 104 704;
  • 40) 0.097 169 340 104 704 × 2 = 0 + 0.194 338 680 209 408;
  • 41) 0.194 338 680 209 408 × 2 = 0 + 0.388 677 360 418 816;
  • 42) 0.388 677 360 418 816 × 2 = 0 + 0.777 354 720 837 632;
  • 43) 0.777 354 720 837 632 × 2 = 1 + 0.554 709 441 675 264;
  • 44) 0.554 709 441 675 264 × 2 = 1 + 0.109 418 883 350 528;
  • 45) 0.109 418 883 350 528 × 2 = 0 + 0.218 837 766 701 056;
  • 46) 0.218 837 766 701 056 × 2 = 0 + 0.437 675 533 402 112;
  • 47) 0.437 675 533 402 112 × 2 = 0 + 0.875 351 066 804 224;
  • 48) 0.875 351 066 804 224 × 2 = 1 + 0.750 702 133 608 448;
  • 49) 0.750 702 133 608 448 × 2 = 1 + 0.501 404 267 216 896;
  • 50) 0.501 404 267 216 896 × 2 = 1 + 0.002 808 534 433 792;
  • 51) 0.002 808 534 433 792 × 2 = 0 + 0.005 617 068 867 584;
  • 52) 0.005 617 068 867 584 × 2 = 0 + 0.011 234 137 735 168;
  • 53) 0.011 234 137 735 168 × 2 = 0 + 0.022 468 275 470 336;
  • 54) 0.022 468 275 470 336 × 2 = 0 + 0.044 936 550 940 672;
  • 55) 0.044 936 550 940 672 × 2 = 0 + 0.089 873 101 881 344;
  • 56) 0.089 873 101 881 344 × 2 = 0 + 0.179 746 203 762 688;
  • 57) 0.179 746 203 762 688 × 2 = 0 + 0.359 492 407 525 376;
  • 58) 0.359 492 407 525 376 × 2 = 0 + 0.718 984 815 050 752;
  • 59) 0.718 984 815 050 752 × 2 = 1 + 0.437 969 630 101 504;
  • 60) 0.437 969 630 101 504 × 2 = 0 + 0.875 939 260 203 008;
  • 61) 0.875 939 260 203 008 × 2 = 1 + 0.751 878 520 406 016;
  • 62) 0.751 878 520 406 016 × 2 = 1 + 0.503 757 040 812 032;
  • 63) 0.503 757 040 812 032 × 2 = 1 + 0.007 514 081 624 064;
  • 64) 0.007 514 081 624 064 × 2 = 0 + 0.015 028 163 248 128;
  • 65) 0.015 028 163 248 128 × 2 = 0 + 0.030 056 326 496 256;
  • 66) 0.030 056 326 496 256 × 2 = 0 + 0.060 112 652 992 512;
  • 67) 0.060 112 652 992 512 × 2 = 0 + 0.120 225 305 985 024;
  • 68) 0.120 225 305 985 024 × 2 = 0 + 0.240 450 611 970 048;
  • 69) 0.240 450 611 970 048 × 2 = 0 + 0.480 901 223 940 096;
  • 70) 0.480 901 223 940 096 × 2 = 0 + 0.961 802 447 880 192;
  • 71) 0.961 802 447 880 192 × 2 = 1 + 0.923 604 895 760 384;
  • 72) 0.923 604 895 760 384 × 2 = 1 + 0.847 209 791 520 768;
  • 73) 0.847 209 791 520 768 × 2 = 1 + 0.694 419 583 041 536;
  • 74) 0.694 419 583 041 536 × 2 = 1 + 0.388 839 166 083 072;
  • 75) 0.388 839 166 083 072 × 2 = 0 + 0.777 678 332 166 144;
  • 76) 0.777 678 332 166 144 × 2 = 1 + 0.555 356 664 332 288;
  • 77) 0.555 356 664 332 288 × 2 = 1 + 0.110 713 328 664 576;
  • 78) 0.110 713 328 664 576 × 2 = 0 + 0.221 426 657 329 152;
  • 79) 0.221 426 657 329 152 × 2 = 0 + 0.442 853 314 658 304;
  • 80) 0.442 853 314 658 304 × 2 = 0 + 0.885 706 629 316 608;
  • 81) 0.885 706 629 316 608 × 2 = 1 + 0.771 413 258 633 216;
  • 82) 0.771 413 258 633 216 × 2 = 1 + 0.542 826 517 266 432;
  • 83) 0.542 826 517 266 432 × 2 = 1 + 0.085 653 034 532 864;
  • 84) 0.085 653 034 532 864 × 2 = 0 + 0.171 306 069 065 728;
  • 85) 0.171 306 069 065 728 × 2 = 0 + 0.342 612 138 131 456;
  • 86) 0.342 612 138 131 456 × 2 = 0 + 0.685 224 276 262 912;
  • 87) 0.685 224 276 262 912 × 2 = 1 + 0.370 448 552 525 824;
  • 88) 0.370 448 552 525 824 × 2 = 0 + 0.740 897 105 051 648;
  • 89) 0.740 897 105 051 648 × 2 = 1 + 0.481 794 210 103 296;
  • 90) 0.481 794 210 103 296 × 2 = 0 + 0.963 588 420 206 592;
  • 91) 0.963 588 420 206 592 × 2 = 1 + 0.927 176 840 413 184;
  • 92) 0.927 176 840 413 184 × 2 = 1 + 0.854 353 680 826 368;
  • 93) 0.854 353 680 826 368 × 2 = 1 + 0.708 707 361 652 736;
  • 94) 0.708 707 361 652 736 × 2 = 1 + 0.417 414 723 305 472;
  • 95) 0.417 414 723 305 472 × 2 = 0 + 0.834 829 446 610 944;

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:


0.000 000 000 000 176 75(10) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0011 0001 1100 0000 0010 1110 0000 0011 1101 1000 1110 0010 1011 110(2)

6. Positive number before normalization:

0.000 000 000 000 176 75(10) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0011 0001 1100 0000 0010 1110 0000 0011 1101 1000 1110 0010 1011 110(2)

7. 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:


0.000 000 000 000 176 75(10) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0011 0001 1100 0000 0010 1110 0000 0011 1101 1000 1110 0010 1011 110(2) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0011 0001 1100 0000 0010 1110 0000 0011 1101 1000 1110 0010 1011 110(2) × 20 =

1.1000 1110 0000 0001 0111 0000 0001 1110 1100 0111 0001 0101 1110(2) × 2-43


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): -43

Mantissa (not normalized):
1.1000 1110 0000 0001 0111 0000 0001 1110 1100 0111 0001 0101 1110


9. 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)


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;
  • 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;

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) =

980(10) =

011 1101 0100(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. 1000 1110 0000 0001 0111 0000 0001 1110 1100 0111 0001 0101 1110 =

1000 1110 0000 0001 0111 0000 0001 1110 1100 0111 0001 0101 1110


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 0100

Mantissa (52 bits) =
1000 1110 0000 0001 0111 0000 0001 1110 1100 0111 0001 0101 1110


Decimal number -0.000 000 000 000 176 75 converted to 64 bit double precision IEEE 754 binary floating point representation:

1 - 011 1101 0100 - 1000 1110 0000 0001 0111 0000 0001 1110 1100 0111 0001 0101 1110

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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