-0.000 000 000 000 037 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 000 037(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
-0.000 000 000 000 037(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:

|-0.000 000 000 000 037| = 0.000 000 000 000 037


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: 0.000 000 000 000 037.

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) 0.000 000 000 000 037 × 2 = 0 + 0.000 000 000 000 074;
  • 2) 0.000 000 000 000 074 × 2 = 0 + 0.000 000 000 000 148;
  • 3) 0.000 000 000 000 148 × 2 = 0 + 0.000 000 000 000 296;
  • 4) 0.000 000 000 000 296 × 2 = 0 + 0.000 000 000 000 592;
  • 5) 0.000 000 000 000 592 × 2 = 0 + 0.000 000 000 001 184;
  • 6) 0.000 000 000 001 184 × 2 = 0 + 0.000 000 000 002 368;
  • 7) 0.000 000 000 002 368 × 2 = 0 + 0.000 000 000 004 736;
  • 8) 0.000 000 000 004 736 × 2 = 0 + 0.000 000 000 009 472;
  • 9) 0.000 000 000 009 472 × 2 = 0 + 0.000 000 000 018 944;
  • 10) 0.000 000 000 018 944 × 2 = 0 + 0.000 000 000 037 888;
  • 11) 0.000 000 000 037 888 × 2 = 0 + 0.000 000 000 075 776;
  • 12) 0.000 000 000 075 776 × 2 = 0 + 0.000 000 000 151 552;
  • 13) 0.000 000 000 151 552 × 2 = 0 + 0.000 000 000 303 104;
  • 14) 0.000 000 000 303 104 × 2 = 0 + 0.000 000 000 606 208;
  • 15) 0.000 000 000 606 208 × 2 = 0 + 0.000 000 001 212 416;
  • 16) 0.000 000 001 212 416 × 2 = 0 + 0.000 000 002 424 832;
  • 17) 0.000 000 002 424 832 × 2 = 0 + 0.000 000 004 849 664;
  • 18) 0.000 000 004 849 664 × 2 = 0 + 0.000 000 009 699 328;
  • 19) 0.000 000 009 699 328 × 2 = 0 + 0.000 000 019 398 656;
  • 20) 0.000 000 019 398 656 × 2 = 0 + 0.000 000 038 797 312;
  • 21) 0.000 000 038 797 312 × 2 = 0 + 0.000 000 077 594 624;
  • 22) 0.000 000 077 594 624 × 2 = 0 + 0.000 000 155 189 248;
  • 23) 0.000 000 155 189 248 × 2 = 0 + 0.000 000 310 378 496;
  • 24) 0.000 000 310 378 496 × 2 = 0 + 0.000 000 620 756 992;
  • 25) 0.000 000 620 756 992 × 2 = 0 + 0.000 001 241 513 984;
  • 26) 0.000 001 241 513 984 × 2 = 0 + 0.000 002 483 027 968;
  • 27) 0.000 002 483 027 968 × 2 = 0 + 0.000 004 966 055 936;
  • 28) 0.000 004 966 055 936 × 2 = 0 + 0.000 009 932 111 872;
  • 29) 0.000 009 932 111 872 × 2 = 0 + 0.000 019 864 223 744;
  • 30) 0.000 019 864 223 744 × 2 = 0 + 0.000 039 728 447 488;
  • 31) 0.000 039 728 447 488 × 2 = 0 + 0.000 079 456 894 976;
  • 32) 0.000 079 456 894 976 × 2 = 0 + 0.000 158 913 789 952;
  • 33) 0.000 158 913 789 952 × 2 = 0 + 0.000 317 827 579 904;
  • 34) 0.000 317 827 579 904 × 2 = 0 + 0.000 635 655 159 808;
  • 35) 0.000 635 655 159 808 × 2 = 0 + 0.001 271 310 319 616;
  • 36) 0.001 271 310 319 616 × 2 = 0 + 0.002 542 620 639 232;
  • 37) 0.002 542 620 639 232 × 2 = 0 + 0.005 085 241 278 464;
  • 38) 0.005 085 241 278 464 × 2 = 0 + 0.010 170 482 556 928;
  • 39) 0.010 170 482 556 928 × 2 = 0 + 0.020 340 965 113 856;
  • 40) 0.020 340 965 113 856 × 2 = 0 + 0.040 681 930 227 712;
  • 41) 0.040 681 930 227 712 × 2 = 0 + 0.081 363 860 455 424;
  • 42) 0.081 363 860 455 424 × 2 = 0 + 0.162 727 720 910 848;
  • 43) 0.162 727 720 910 848 × 2 = 0 + 0.325 455 441 821 696;
  • 44) 0.325 455 441 821 696 × 2 = 0 + 0.650 910 883 643 392;
  • 45) 0.650 910 883 643 392 × 2 = 1 + 0.301 821 767 286 784;
  • 46) 0.301 821 767 286 784 × 2 = 0 + 0.603 643 534 573 568;
  • 47) 0.603 643 534 573 568 × 2 = 1 + 0.207 287 069 147 136;
  • 48) 0.207 287 069 147 136 × 2 = 0 + 0.414 574 138 294 272;
  • 49) 0.414 574 138 294 272 × 2 = 0 + 0.829 148 276 588 544;
  • 50) 0.829 148 276 588 544 × 2 = 1 + 0.658 296 553 177 088;
  • 51) 0.658 296 553 177 088 × 2 = 1 + 0.316 593 106 354 176;
  • 52) 0.316 593 106 354 176 × 2 = 0 + 0.633 186 212 708 352;
  • 53) 0.633 186 212 708 352 × 2 = 1 + 0.266 372 425 416 704;
  • 54) 0.266 372 425 416 704 × 2 = 0 + 0.532 744 850 833 408;
  • 55) 0.532 744 850 833 408 × 2 = 1 + 0.065 489 701 666 816;
  • 56) 0.065 489 701 666 816 × 2 = 0 + 0.130 979 403 333 632;
  • 57) 0.130 979 403 333 632 × 2 = 0 + 0.261 958 806 667 264;
  • 58) 0.261 958 806 667 264 × 2 = 0 + 0.523 917 613 334 528;
  • 59) 0.523 917 613 334 528 × 2 = 1 + 0.047 835 226 669 056;
  • 60) 0.047 835 226 669 056 × 2 = 0 + 0.095 670 453 338 112;
  • 61) 0.095 670 453 338 112 × 2 = 0 + 0.191 340 906 676 224;
  • 62) 0.191 340 906 676 224 × 2 = 0 + 0.382 681 813 352 448;
  • 63) 0.382 681 813 352 448 × 2 = 0 + 0.765 363 626 704 896;
  • 64) 0.765 363 626 704 896 × 2 = 1 + 0.530 727 253 409 792;
  • 65) 0.530 727 253 409 792 × 2 = 1 + 0.061 454 506 819 584;
  • 66) 0.061 454 506 819 584 × 2 = 0 + 0.122 909 013 639 168;
  • 67) 0.122 909 013 639 168 × 2 = 0 + 0.245 818 027 278 336;
  • 68) 0.245 818 027 278 336 × 2 = 0 + 0.491 636 054 556 672;
  • 69) 0.491 636 054 556 672 × 2 = 0 + 0.983 272 109 113 344;
  • 70) 0.983 272 109 113 344 × 2 = 1 + 0.966 544 218 226 688;
  • 71) 0.966 544 218 226 688 × 2 = 1 + 0.933 088 436 453 376;
  • 72) 0.933 088 436 453 376 × 2 = 1 + 0.866 176 872 906 752;
  • 73) 0.866 176 872 906 752 × 2 = 1 + 0.732 353 745 813 504;
  • 74) 0.732 353 745 813 504 × 2 = 1 + 0.464 707 491 627 008;
  • 75) 0.464 707 491 627 008 × 2 = 0 + 0.929 414 983 254 016;
  • 76) 0.929 414 983 254 016 × 2 = 1 + 0.858 829 966 508 032;
  • 77) 0.858 829 966 508 032 × 2 = 1 + 0.717 659 933 016 064;
  • 78) 0.717 659 933 016 064 × 2 = 1 + 0.435 319 866 032 128;
  • 79) 0.435 319 866 032 128 × 2 = 0 + 0.870 639 732 064 256;
  • 80) 0.870 639 732 064 256 × 2 = 1 + 0.741 279 464 128 512;
  • 81) 0.741 279 464 128 512 × 2 = 1 + 0.482 558 928 257 024;
  • 82) 0.482 558 928 257 024 × 2 = 0 + 0.965 117 856 514 048;
  • 83) 0.965 117 856 514 048 × 2 = 1 + 0.930 235 713 028 096;
  • 84) 0.930 235 713 028 096 × 2 = 1 + 0.860 471 426 056 192;
  • 85) 0.860 471 426 056 192 × 2 = 1 + 0.720 942 852 112 384;
  • 86) 0.720 942 852 112 384 × 2 = 1 + 0.441 885 704 224 768;
  • 87) 0.441 885 704 224 768 × 2 = 0 + 0.883 771 408 449 536;
  • 88) 0.883 771 408 449 536 × 2 = 1 + 0.767 542 816 899 072;
  • 89) 0.767 542 816 899 072 × 2 = 1 + 0.535 085 633 798 144;
  • 90) 0.535 085 633 798 144 × 2 = 1 + 0.070 171 267 596 288;
  • 91) 0.070 171 267 596 288 × 2 = 0 + 0.140 342 535 192 576;
  • 92) 0.140 342 535 192 576 × 2 = 0 + 0.280 685 070 385 152;
  • 93) 0.280 685 070 385 152 × 2 = 0 + 0.561 370 140 770 304;
  • 94) 0.561 370 140 770 304 × 2 = 1 + 0.122 740 281 540 608;
  • 95) 0.122 740 281 540 608 × 2 = 0 + 0.245 480 563 081 216;
  • 96) 0.245 480 563 081 216 × 2 = 0 + 0.490 961 126 162 432;
  • 97) 0.490 961 126 162 432 × 2 = 0 + 0.981 922 252 324 864;

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:


0.000 000 000 000 037(10) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 1010 0110 1010 0010 0001 1000 0111 1101 1101 1011 1101 1100 0100 0(2)

6. Positive number before normalization:

0.000 000 000 000 037(10) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 1010 0110 1010 0010 0001 1000 0111 1101 1101 1011 1101 1100 0100 0(2)

7. Normalize the binary representation of the number.

Shift the decimal mark 45 positions to the right, so that only one non zero digit remains to the left of it:


0.000 000 000 000 037(10) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 1010 0110 1010 0010 0001 1000 0111 1101 1101 1011 1101 1100 0100 0(2) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 1010 0110 1010 0010 0001 1000 0111 1101 1101 1011 1101 1100 0100 0(2) × 20 =

1.0100 1101 0100 0100 0011 0000 1111 1011 1011 0111 1011 1000 1000(2) × 2-45


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): -45

Mantissa (not normalized):
1.0100 1101 0100 0100 0011 0000 1111 1011 1011 0111 1011 1000 1000


9. Adjust the exponent.

Use the 11 bit excess/bias notation:


Exponent (adjusted) =

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

-45 + 2(11-1) - 1 =

(-45 + 1 023)(10) =

978(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;
  • 978 ÷ 2 = 489 + 0;
  • 489 ÷ 2 = 244 + 1;
  • 244 ÷ 2 = 122 + 0;
  • 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) =

978(10) =

011 1101 0010(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. 0100 1101 0100 0100 0011 0000 1111 1011 1011 0111 1011 1000 1000 =

0100 1101 0100 0100 0011 0000 1111 1011 1011 0111 1011 1000 1000


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 0010

Mantissa (52 bits) =
0100 1101 0100 0100 0011 0000 1111 1011 1011 0111 1011 1000 1000


Decimal number -0.000 000 000 000 037 converted to 64 bit double precision IEEE 754 binary floating point representation:

1 - 011 1101 0010 - 0100 1101 0100 0100 0011 0000 1111 1011 1011 0111 1011 1000 1000

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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