0.000 000 000 000 144 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 0.000 000 000 000 144(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
0.000 000 000 000 144(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: 0.000 000 000 000 144.

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) 0.000 000 000 000 144 × 2 = 0 + 0.000 000 000 000 288;
  • 2) 0.000 000 000 000 288 × 2 = 0 + 0.000 000 000 000 576;
  • 3) 0.000 000 000 000 576 × 2 = 0 + 0.000 000 000 001 152;
  • 4) 0.000 000 000 001 152 × 2 = 0 + 0.000 000 000 002 304;
  • 5) 0.000 000 000 002 304 × 2 = 0 + 0.000 000 000 004 608;
  • 6) 0.000 000 000 004 608 × 2 = 0 + 0.000 000 000 009 216;
  • 7) 0.000 000 000 009 216 × 2 = 0 + 0.000 000 000 018 432;
  • 8) 0.000 000 000 018 432 × 2 = 0 + 0.000 000 000 036 864;
  • 9) 0.000 000 000 036 864 × 2 = 0 + 0.000 000 000 073 728;
  • 10) 0.000 000 000 073 728 × 2 = 0 + 0.000 000 000 147 456;
  • 11) 0.000 000 000 147 456 × 2 = 0 + 0.000 000 000 294 912;
  • 12) 0.000 000 000 294 912 × 2 = 0 + 0.000 000 000 589 824;
  • 13) 0.000 000 000 589 824 × 2 = 0 + 0.000 000 001 179 648;
  • 14) 0.000 000 001 179 648 × 2 = 0 + 0.000 000 002 359 296;
  • 15) 0.000 000 002 359 296 × 2 = 0 + 0.000 000 004 718 592;
  • 16) 0.000 000 004 718 592 × 2 = 0 + 0.000 000 009 437 184;
  • 17) 0.000 000 009 437 184 × 2 = 0 + 0.000 000 018 874 368;
  • 18) 0.000 000 018 874 368 × 2 = 0 + 0.000 000 037 748 736;
  • 19) 0.000 000 037 748 736 × 2 = 0 + 0.000 000 075 497 472;
  • 20) 0.000 000 075 497 472 × 2 = 0 + 0.000 000 150 994 944;
  • 21) 0.000 000 150 994 944 × 2 = 0 + 0.000 000 301 989 888;
  • 22) 0.000 000 301 989 888 × 2 = 0 + 0.000 000 603 979 776;
  • 23) 0.000 000 603 979 776 × 2 = 0 + 0.000 001 207 959 552;
  • 24) 0.000 001 207 959 552 × 2 = 0 + 0.000 002 415 919 104;
  • 25) 0.000 002 415 919 104 × 2 = 0 + 0.000 004 831 838 208;
  • 26) 0.000 004 831 838 208 × 2 = 0 + 0.000 009 663 676 416;
  • 27) 0.000 009 663 676 416 × 2 = 0 + 0.000 019 327 352 832;
  • 28) 0.000 019 327 352 832 × 2 = 0 + 0.000 038 654 705 664;
  • 29) 0.000 038 654 705 664 × 2 = 0 + 0.000 077 309 411 328;
  • 30) 0.000 077 309 411 328 × 2 = 0 + 0.000 154 618 822 656;
  • 31) 0.000 154 618 822 656 × 2 = 0 + 0.000 309 237 645 312;
  • 32) 0.000 309 237 645 312 × 2 = 0 + 0.000 618 475 290 624;
  • 33) 0.000 618 475 290 624 × 2 = 0 + 0.001 236 950 581 248;
  • 34) 0.001 236 950 581 248 × 2 = 0 + 0.002 473 901 162 496;
  • 35) 0.002 473 901 162 496 × 2 = 0 + 0.004 947 802 324 992;
  • 36) 0.004 947 802 324 992 × 2 = 0 + 0.009 895 604 649 984;
  • 37) 0.009 895 604 649 984 × 2 = 0 + 0.019 791 209 299 968;
  • 38) 0.019 791 209 299 968 × 2 = 0 + 0.039 582 418 599 936;
  • 39) 0.039 582 418 599 936 × 2 = 0 + 0.079 164 837 199 872;
  • 40) 0.079 164 837 199 872 × 2 = 0 + 0.158 329 674 399 744;
  • 41) 0.158 329 674 399 744 × 2 = 0 + 0.316 659 348 799 488;
  • 42) 0.316 659 348 799 488 × 2 = 0 + 0.633 318 697 598 976;
  • 43) 0.633 318 697 598 976 × 2 = 1 + 0.266 637 395 197 952;
  • 44) 0.266 637 395 197 952 × 2 = 0 + 0.533 274 790 395 904;
  • 45) 0.533 274 790 395 904 × 2 = 1 + 0.066 549 580 791 808;
  • 46) 0.066 549 580 791 808 × 2 = 0 + 0.133 099 161 583 616;
  • 47) 0.133 099 161 583 616 × 2 = 0 + 0.266 198 323 167 232;
  • 48) 0.266 198 323 167 232 × 2 = 0 + 0.532 396 646 334 464;
  • 49) 0.532 396 646 334 464 × 2 = 1 + 0.064 793 292 668 928;
  • 50) 0.064 793 292 668 928 × 2 = 0 + 0.129 586 585 337 856;
  • 51) 0.129 586 585 337 856 × 2 = 0 + 0.259 173 170 675 712;
  • 52) 0.259 173 170 675 712 × 2 = 0 + 0.518 346 341 351 424;
  • 53) 0.518 346 341 351 424 × 2 = 1 + 0.036 692 682 702 848;
  • 54) 0.036 692 682 702 848 × 2 = 0 + 0.073 385 365 405 696;
  • 55) 0.073 385 365 405 696 × 2 = 0 + 0.146 770 730 811 392;
  • 56) 0.146 770 730 811 392 × 2 = 0 + 0.293 541 461 622 784;
  • 57) 0.293 541 461 622 784 × 2 = 0 + 0.587 082 923 245 568;
  • 58) 0.587 082 923 245 568 × 2 = 1 + 0.174 165 846 491 136;
  • 59) 0.174 165 846 491 136 × 2 = 0 + 0.348 331 692 982 272;
  • 60) 0.348 331 692 982 272 × 2 = 0 + 0.696 663 385 964 544;
  • 61) 0.696 663 385 964 544 × 2 = 1 + 0.393 326 771 929 088;
  • 62) 0.393 326 771 929 088 × 2 = 0 + 0.786 653 543 858 176;
  • 63) 0.786 653 543 858 176 × 2 = 1 + 0.573 307 087 716 352;
  • 64) 0.573 307 087 716 352 × 2 = 1 + 0.146 614 175 432 704;
  • 65) 0.146 614 175 432 704 × 2 = 0 + 0.293 228 350 865 408;
  • 66) 0.293 228 350 865 408 × 2 = 0 + 0.586 456 701 730 816;
  • 67) 0.586 456 701 730 816 × 2 = 1 + 0.172 913 403 461 632;
  • 68) 0.172 913 403 461 632 × 2 = 0 + 0.345 826 806 923 264;
  • 69) 0.345 826 806 923 264 × 2 = 0 + 0.691 653 613 846 528;
  • 70) 0.691 653 613 846 528 × 2 = 1 + 0.383 307 227 693 056;
  • 71) 0.383 307 227 693 056 × 2 = 0 + 0.766 614 455 386 112;
  • 72) 0.766 614 455 386 112 × 2 = 1 + 0.533 228 910 772 224;
  • 73) 0.533 228 910 772 224 × 2 = 1 + 0.066 457 821 544 448;
  • 74) 0.066 457 821 544 448 × 2 = 0 + 0.132 915 643 088 896;
  • 75) 0.132 915 643 088 896 × 2 = 0 + 0.265 831 286 177 792;
  • 76) 0.265 831 286 177 792 × 2 = 0 + 0.531 662 572 355 584;
  • 77) 0.531 662 572 355 584 × 2 = 1 + 0.063 325 144 711 168;
  • 78) 0.063 325 144 711 168 × 2 = 0 + 0.126 650 289 422 336;
  • 79) 0.126 650 289 422 336 × 2 = 0 + 0.253 300 578 844 672;
  • 80) 0.253 300 578 844 672 × 2 = 0 + 0.506 601 157 689 344;
  • 81) 0.506 601 157 689 344 × 2 = 1 + 0.013 202 315 378 688;
  • 82) 0.013 202 315 378 688 × 2 = 0 + 0.026 404 630 757 376;
  • 83) 0.026 404 630 757 376 × 2 = 0 + 0.052 809 261 514 752;
  • 84) 0.052 809 261 514 752 × 2 = 0 + 0.105 618 523 029 504;
  • 85) 0.105 618 523 029 504 × 2 = 0 + 0.211 237 046 059 008;
  • 86) 0.211 237 046 059 008 × 2 = 0 + 0.422 474 092 118 016;
  • 87) 0.422 474 092 118 016 × 2 = 0 + 0.844 948 184 236 032;
  • 88) 0.844 948 184 236 032 × 2 = 1 + 0.689 896 368 472 064;
  • 89) 0.689 896 368 472 064 × 2 = 1 + 0.379 792 736 944 128;
  • 90) 0.379 792 736 944 128 × 2 = 0 + 0.759 585 473 888 256;
  • 91) 0.759 585 473 888 256 × 2 = 1 + 0.519 170 947 776 512;
  • 92) 0.519 170 947 776 512 × 2 = 1 + 0.038 341 895 553 024;
  • 93) 0.038 341 895 553 024 × 2 = 0 + 0.076 683 791 106 048;
  • 94) 0.076 683 791 106 048 × 2 = 0 + 0.153 367 582 212 096;
  • 95) 0.153 367 582 212 096 × 2 = 0 + 0.306 735 164 424 192;

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:


0.000 000 000 000 144(10) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0010 1000 1000 1000 0100 1011 0010 0101 1000 1000 1000 0001 1011 000(2)

5. Positive number before normalization:

0.000 000 000 000 144(10) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0010 1000 1000 1000 0100 1011 0010 0101 1000 1000 1000 0001 1011 000(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:


0.000 000 000 000 144(10) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0010 1000 1000 1000 0100 1011 0010 0101 1000 1000 1000 0001 1011 000(2) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0010 1000 1000 1000 0100 1011 0010 0101 1000 1000 1000 0001 1011 000(2) × 20 =

1.0100 0100 0100 0010 0101 1001 0010 1100 0100 0100 0000 1101 1000(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):
1.0100 0100 0100 0010 0101 1001 0010 1100 0100 0100 0000 1101 1000


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. 0100 0100 0100 0010 0101 1001 0010 1100 0100 0100 0000 1101 1000 =

0100 0100 0100 0010 0101 1001 0010 1100 0100 0100 0000 1101 1000


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) =
0100 0100 0100 0010 0101 1001 0010 1100 0100 0100 0000 1101 1000


Decimal number 0.000 000 000 000 144 converted to 64 bit double precision IEEE 754 binary floating point representation:

0 - 011 1101 0100 - 0100 0100 0100 0010 0101 1001 0010 1100 0100 0100 0000 1101 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