-0.000 000 160 584 794 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 160 584 794(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
-0.000 000 160 584 794(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:

|-0.000 000 160 584 794| = 0.000 000 160 584 794


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: 0.000 000 160 584 794.

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) 0.000 000 160 584 794 × 2 = 0 + 0.000 000 321 169 588;
  • 2) 0.000 000 321 169 588 × 2 = 0 + 0.000 000 642 339 176;
  • 3) 0.000 000 642 339 176 × 2 = 0 + 0.000 001 284 678 352;
  • 4) 0.000 001 284 678 352 × 2 = 0 + 0.000 002 569 356 704;
  • 5) 0.000 002 569 356 704 × 2 = 0 + 0.000 005 138 713 408;
  • 6) 0.000 005 138 713 408 × 2 = 0 + 0.000 010 277 426 816;
  • 7) 0.000 010 277 426 816 × 2 = 0 + 0.000 020 554 853 632;
  • 8) 0.000 020 554 853 632 × 2 = 0 + 0.000 041 109 707 264;
  • 9) 0.000 041 109 707 264 × 2 = 0 + 0.000 082 219 414 528;
  • 10) 0.000 082 219 414 528 × 2 = 0 + 0.000 164 438 829 056;
  • 11) 0.000 164 438 829 056 × 2 = 0 + 0.000 328 877 658 112;
  • 12) 0.000 328 877 658 112 × 2 = 0 + 0.000 657 755 316 224;
  • 13) 0.000 657 755 316 224 × 2 = 0 + 0.001 315 510 632 448;
  • 14) 0.001 315 510 632 448 × 2 = 0 + 0.002 631 021 264 896;
  • 15) 0.002 631 021 264 896 × 2 = 0 + 0.005 262 042 529 792;
  • 16) 0.005 262 042 529 792 × 2 = 0 + 0.010 524 085 059 584;
  • 17) 0.010 524 085 059 584 × 2 = 0 + 0.021 048 170 119 168;
  • 18) 0.021 048 170 119 168 × 2 = 0 + 0.042 096 340 238 336;
  • 19) 0.042 096 340 238 336 × 2 = 0 + 0.084 192 680 476 672;
  • 20) 0.084 192 680 476 672 × 2 = 0 + 0.168 385 360 953 344;
  • 21) 0.168 385 360 953 344 × 2 = 0 + 0.336 770 721 906 688;
  • 22) 0.336 770 721 906 688 × 2 = 0 + 0.673 541 443 813 376;
  • 23) 0.673 541 443 813 376 × 2 = 1 + 0.347 082 887 626 752;
  • 24) 0.347 082 887 626 752 × 2 = 0 + 0.694 165 775 253 504;
  • 25) 0.694 165 775 253 504 × 2 = 1 + 0.388 331 550 507 008;
  • 26) 0.388 331 550 507 008 × 2 = 0 + 0.776 663 101 014 016;
  • 27) 0.776 663 101 014 016 × 2 = 1 + 0.553 326 202 028 032;
  • 28) 0.553 326 202 028 032 × 2 = 1 + 0.106 652 404 056 064;
  • 29) 0.106 652 404 056 064 × 2 = 0 + 0.213 304 808 112 128;
  • 30) 0.213 304 808 112 128 × 2 = 0 + 0.426 609 616 224 256;
  • 31) 0.426 609 616 224 256 × 2 = 0 + 0.853 219 232 448 512;
  • 32) 0.853 219 232 448 512 × 2 = 1 + 0.706 438 464 897 024;
  • 33) 0.706 438 464 897 024 × 2 = 1 + 0.412 876 929 794 048;
  • 34) 0.412 876 929 794 048 × 2 = 0 + 0.825 753 859 588 096;
  • 35) 0.825 753 859 588 096 × 2 = 1 + 0.651 507 719 176 192;
  • 36) 0.651 507 719 176 192 × 2 = 1 + 0.303 015 438 352 384;
  • 37) 0.303 015 438 352 384 × 2 = 0 + 0.606 030 876 704 768;
  • 38) 0.606 030 876 704 768 × 2 = 1 + 0.212 061 753 409 536;
  • 39) 0.212 061 753 409 536 × 2 = 0 + 0.424 123 506 819 072;
  • 40) 0.424 123 506 819 072 × 2 = 0 + 0.848 247 013 638 144;
  • 41) 0.848 247 013 638 144 × 2 = 1 + 0.696 494 027 276 288;
  • 42) 0.696 494 027 276 288 × 2 = 1 + 0.392 988 054 552 576;
  • 43) 0.392 988 054 552 576 × 2 = 0 + 0.785 976 109 105 152;
  • 44) 0.785 976 109 105 152 × 2 = 1 + 0.571 952 218 210 304;
  • 45) 0.571 952 218 210 304 × 2 = 1 + 0.143 904 436 420 608;
  • 46) 0.143 904 436 420 608 × 2 = 0 + 0.287 808 872 841 216;
  • 47) 0.287 808 872 841 216 × 2 = 0 + 0.575 617 745 682 432;
  • 48) 0.575 617 745 682 432 × 2 = 1 + 0.151 235 491 364 864;
  • 49) 0.151 235 491 364 864 × 2 = 0 + 0.302 470 982 729 728;
  • 50) 0.302 470 982 729 728 × 2 = 0 + 0.604 941 965 459 456;
  • 51) 0.604 941 965 459 456 × 2 = 1 + 0.209 883 930 918 912;
  • 52) 0.209 883 930 918 912 × 2 = 0 + 0.419 767 861 837 824;
  • 53) 0.419 767 861 837 824 × 2 = 0 + 0.839 535 723 675 648;
  • 54) 0.839 535 723 675 648 × 2 = 1 + 0.679 071 447 351 296;
  • 55) 0.679 071 447 351 296 × 2 = 1 + 0.358 142 894 702 592;
  • 56) 0.358 142 894 702 592 × 2 = 0 + 0.716 285 789 405 184;
  • 57) 0.716 285 789 405 184 × 2 = 1 + 0.432 571 578 810 368;
  • 58) 0.432 571 578 810 368 × 2 = 0 + 0.865 143 157 620 736;
  • 59) 0.865 143 157 620 736 × 2 = 1 + 0.730 286 315 241 472;
  • 60) 0.730 286 315 241 472 × 2 = 1 + 0.460 572 630 482 944;
  • 61) 0.460 572 630 482 944 × 2 = 0 + 0.921 145 260 965 888;
  • 62) 0.921 145 260 965 888 × 2 = 1 + 0.842 290 521 931 776;
  • 63) 0.842 290 521 931 776 × 2 = 1 + 0.684 581 043 863 552;
  • 64) 0.684 581 043 863 552 × 2 = 1 + 0.369 162 087 727 104;
  • 65) 0.369 162 087 727 104 × 2 = 0 + 0.738 324 175 454 208;
  • 66) 0.738 324 175 454 208 × 2 = 1 + 0.476 648 350 908 416;
  • 67) 0.476 648 350 908 416 × 2 = 0 + 0.953 296 701 816 832;
  • 68) 0.953 296 701 816 832 × 2 = 1 + 0.906 593 403 633 664;
  • 69) 0.906 593 403 633 664 × 2 = 1 + 0.813 186 807 267 328;
  • 70) 0.813 186 807 267 328 × 2 = 1 + 0.626 373 614 534 656;
  • 71) 0.626 373 614 534 656 × 2 = 1 + 0.252 747 229 069 312;
  • 72) 0.252 747 229 069 312 × 2 = 0 + 0.505 494 458 138 624;
  • 73) 0.505 494 458 138 624 × 2 = 1 + 0.010 988 916 277 248;
  • 74) 0.010 988 916 277 248 × 2 = 0 + 0.021 977 832 554 496;
  • 75) 0.021 977 832 554 496 × 2 = 0 + 0.043 955 665 108 992;

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:


0.000 000 160 584 794(10) =

0.0000 0000 0000 0000 0000 0010 1011 0001 1011 0100 1101 1001 0010 0110 1011 0111 0101 1110 100(2)

6. Positive number before normalization:

0.000 000 160 584 794(10) =

0.0000 0000 0000 0000 0000 0010 1011 0001 1011 0100 1101 1001 0010 0110 1011 0111 0101 1110 100(2)

7. Normalize the binary representation of the number.

Shift the decimal mark 23 positions to the right, so that only one non zero digit remains to the left of it:


0.000 000 160 584 794(10) =

0.0000 0000 0000 0000 0000 0010 1011 0001 1011 0100 1101 1001 0010 0110 1011 0111 0101 1110 100(2) =

0.0000 0000 0000 0000 0000 0010 1011 0001 1011 0100 1101 1001 0010 0110 1011 0111 0101 1110 100(2) × 20 =

1.0101 1000 1101 1010 0110 1100 1001 0011 0101 1011 1010 1111 0100(2) × 2-23


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): -23

Mantissa (not normalized):
1.0101 1000 1101 1010 0110 1100 1001 0011 0101 1011 1010 1111 0100


9. Adjust the exponent.

Use the 11 bit excess/bias notation:


Exponent (adjusted) =

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

-23 + 2(11-1) - 1 =

(-23 + 1 023)(10) =

1 000(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;
  • 1 000 ÷ 2 = 500 + 0;
  • 500 ÷ 2 = 250 + 0;
  • 250 ÷ 2 = 125 + 0;
  • 125 ÷ 2 = 62 + 1;
  • 62 ÷ 2 = 31 + 0;
  • 31 ÷ 2 = 15 + 1;
  • 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) =

1000(10) =

011 1110 1000(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. 0101 1000 1101 1010 0110 1100 1001 0011 0101 1011 1010 1111 0100 =

0101 1000 1101 1010 0110 1100 1001 0011 0101 1011 1010 1111 0100


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 1110 1000

Mantissa (52 bits) =
0101 1000 1101 1010 0110 1100 1001 0011 0101 1011 1010 1111 0100


Decimal number -0.000 000 160 584 794 converted to 64 bit double precision IEEE 754 binary floating point representation:

1 - 011 1110 1000 - 0101 1000 1101 1010 0110 1100 1001 0011 0101 1011 1010 1111 0100

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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