-0.000 000 019 557 719 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 019 557 719(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
-0.000 000 019 557 719(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:

|-0.000 000 019 557 719| = 0.000 000 019 557 719


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: 0.000 000 019 557 719.

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) 0.000 000 019 557 719 × 2 = 0 + 0.000 000 039 115 438;
  • 2) 0.000 000 039 115 438 × 2 = 0 + 0.000 000 078 230 876;
  • 3) 0.000 000 078 230 876 × 2 = 0 + 0.000 000 156 461 752;
  • 4) 0.000 000 156 461 752 × 2 = 0 + 0.000 000 312 923 504;
  • 5) 0.000 000 312 923 504 × 2 = 0 + 0.000 000 625 847 008;
  • 6) 0.000 000 625 847 008 × 2 = 0 + 0.000 001 251 694 016;
  • 7) 0.000 001 251 694 016 × 2 = 0 + 0.000 002 503 388 032;
  • 8) 0.000 002 503 388 032 × 2 = 0 + 0.000 005 006 776 064;
  • 9) 0.000 005 006 776 064 × 2 = 0 + 0.000 010 013 552 128;
  • 10) 0.000 010 013 552 128 × 2 = 0 + 0.000 020 027 104 256;
  • 11) 0.000 020 027 104 256 × 2 = 0 + 0.000 040 054 208 512;
  • 12) 0.000 040 054 208 512 × 2 = 0 + 0.000 080 108 417 024;
  • 13) 0.000 080 108 417 024 × 2 = 0 + 0.000 160 216 834 048;
  • 14) 0.000 160 216 834 048 × 2 = 0 + 0.000 320 433 668 096;
  • 15) 0.000 320 433 668 096 × 2 = 0 + 0.000 640 867 336 192;
  • 16) 0.000 640 867 336 192 × 2 = 0 + 0.001 281 734 672 384;
  • 17) 0.001 281 734 672 384 × 2 = 0 + 0.002 563 469 344 768;
  • 18) 0.002 563 469 344 768 × 2 = 0 + 0.005 126 938 689 536;
  • 19) 0.005 126 938 689 536 × 2 = 0 + 0.010 253 877 379 072;
  • 20) 0.010 253 877 379 072 × 2 = 0 + 0.020 507 754 758 144;
  • 21) 0.020 507 754 758 144 × 2 = 0 + 0.041 015 509 516 288;
  • 22) 0.041 015 509 516 288 × 2 = 0 + 0.082 031 019 032 576;
  • 23) 0.082 031 019 032 576 × 2 = 0 + 0.164 062 038 065 152;
  • 24) 0.164 062 038 065 152 × 2 = 0 + 0.328 124 076 130 304;
  • 25) 0.328 124 076 130 304 × 2 = 0 + 0.656 248 152 260 608;
  • 26) 0.656 248 152 260 608 × 2 = 1 + 0.312 496 304 521 216;
  • 27) 0.312 496 304 521 216 × 2 = 0 + 0.624 992 609 042 432;
  • 28) 0.624 992 609 042 432 × 2 = 1 + 0.249 985 218 084 864;
  • 29) 0.249 985 218 084 864 × 2 = 0 + 0.499 970 436 169 728;
  • 30) 0.499 970 436 169 728 × 2 = 0 + 0.999 940 872 339 456;
  • 31) 0.999 940 872 339 456 × 2 = 1 + 0.999 881 744 678 912;
  • 32) 0.999 881 744 678 912 × 2 = 1 + 0.999 763 489 357 824;
  • 33) 0.999 763 489 357 824 × 2 = 1 + 0.999 526 978 715 648;
  • 34) 0.999 526 978 715 648 × 2 = 1 + 0.999 053 957 431 296;
  • 35) 0.999 053 957 431 296 × 2 = 1 + 0.998 107 914 862 592;
  • 36) 0.998 107 914 862 592 × 2 = 1 + 0.996 215 829 725 184;
  • 37) 0.996 215 829 725 184 × 2 = 1 + 0.992 431 659 450 368;
  • 38) 0.992 431 659 450 368 × 2 = 1 + 0.984 863 318 900 736;
  • 39) 0.984 863 318 900 736 × 2 = 1 + 0.969 726 637 801 472;
  • 40) 0.969 726 637 801 472 × 2 = 1 + 0.939 453 275 602 944;
  • 41) 0.939 453 275 602 944 × 2 = 1 + 0.878 906 551 205 888;
  • 42) 0.878 906 551 205 888 × 2 = 1 + 0.757 813 102 411 776;
  • 43) 0.757 813 102 411 776 × 2 = 1 + 0.515 626 204 823 552;
  • 44) 0.515 626 204 823 552 × 2 = 1 + 0.031 252 409 647 104;
  • 45) 0.031 252 409 647 104 × 2 = 0 + 0.062 504 819 294 208;
  • 46) 0.062 504 819 294 208 × 2 = 0 + 0.125 009 638 588 416;
  • 47) 0.125 009 638 588 416 × 2 = 0 + 0.250 019 277 176 832;
  • 48) 0.250 019 277 176 832 × 2 = 0 + 0.500 038 554 353 664;
  • 49) 0.500 038 554 353 664 × 2 = 1 + 0.000 077 108 707 328;
  • 50) 0.000 077 108 707 328 × 2 = 0 + 0.000 154 217 414 656;
  • 51) 0.000 154 217 414 656 × 2 = 0 + 0.000 308 434 829 312;
  • 52) 0.000 308 434 829 312 × 2 = 0 + 0.000 616 869 658 624;
  • 53) 0.000 616 869 658 624 × 2 = 0 + 0.001 233 739 317 248;
  • 54) 0.001 233 739 317 248 × 2 = 0 + 0.002 467 478 634 496;
  • 55) 0.002 467 478 634 496 × 2 = 0 + 0.004 934 957 268 992;
  • 56) 0.004 934 957 268 992 × 2 = 0 + 0.009 869 914 537 984;
  • 57) 0.009 869 914 537 984 × 2 = 0 + 0.019 739 829 075 968;
  • 58) 0.019 739 829 075 968 × 2 = 0 + 0.039 479 658 151 936;
  • 59) 0.039 479 658 151 936 × 2 = 0 + 0.078 959 316 303 872;
  • 60) 0.078 959 316 303 872 × 2 = 0 + 0.157 918 632 607 744;
  • 61) 0.157 918 632 607 744 × 2 = 0 + 0.315 837 265 215 488;
  • 62) 0.315 837 265 215 488 × 2 = 0 + 0.631 674 530 430 976;
  • 63) 0.631 674 530 430 976 × 2 = 1 + 0.263 349 060 861 952;
  • 64) 0.263 349 060 861 952 × 2 = 0 + 0.526 698 121 723 904;
  • 65) 0.526 698 121 723 904 × 2 = 1 + 0.053 396 243 447 808;
  • 66) 0.053 396 243 447 808 × 2 = 0 + 0.106 792 486 895 616;
  • 67) 0.106 792 486 895 616 × 2 = 0 + 0.213 584 973 791 232;
  • 68) 0.213 584 973 791 232 × 2 = 0 + 0.427 169 947 582 464;
  • 69) 0.427 169 947 582 464 × 2 = 0 + 0.854 339 895 164 928;
  • 70) 0.854 339 895 164 928 × 2 = 1 + 0.708 679 790 329 856;
  • 71) 0.708 679 790 329 856 × 2 = 1 + 0.417 359 580 659 712;
  • 72) 0.417 359 580 659 712 × 2 = 0 + 0.834 719 161 319 424;
  • 73) 0.834 719 161 319 424 × 2 = 1 + 0.669 438 322 638 848;
  • 74) 0.669 438 322 638 848 × 2 = 1 + 0.338 876 645 277 696;
  • 75) 0.338 876 645 277 696 × 2 = 0 + 0.677 753 290 555 392;
  • 76) 0.677 753 290 555 392 × 2 = 1 + 0.355 506 581 110 784;
  • 77) 0.355 506 581 110 784 × 2 = 0 + 0.711 013 162 221 568;
  • 78) 0.711 013 162 221 568 × 2 = 1 + 0.422 026 324 443 136;

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:


0.000 000 019 557 719(10) =

0.0000 0000 0000 0000 0000 0000 0101 0011 1111 1111 1111 0000 1000 0000 0000 0010 1000 0110 1101 01(2)

6. Positive number before normalization:

0.000 000 019 557 719(10) =

0.0000 0000 0000 0000 0000 0000 0101 0011 1111 1111 1111 0000 1000 0000 0000 0010 1000 0110 1101 01(2)

7. Normalize the binary representation of the number.

Shift the decimal mark 26 positions to the right, so that only one non zero digit remains to the left of it:


0.000 000 019 557 719(10) =

0.0000 0000 0000 0000 0000 0000 0101 0011 1111 1111 1111 0000 1000 0000 0000 0010 1000 0110 1101 01(2) =

0.0000 0000 0000 0000 0000 0000 0101 0011 1111 1111 1111 0000 1000 0000 0000 0010 1000 0110 1101 01(2) × 20 =

1.0100 1111 1111 1111 1100 0010 0000 0000 0000 1010 0001 1011 0101(2) × 2-26


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): -26

Mantissa (not normalized):
1.0100 1111 1111 1111 1100 0010 0000 0000 0000 1010 0001 1011 0101


9. Adjust the exponent.

Use the 11 bit excess/bias notation:


Exponent (adjusted) =

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

-26 + 2(11-1) - 1 =

(-26 + 1 023)(10) =

997(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;
  • 997 ÷ 2 = 498 + 1;
  • 498 ÷ 2 = 249 + 0;
  • 249 ÷ 2 = 124 + 1;
  • 124 ÷ 2 = 62 + 0;
  • 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) =

997(10) =

011 1110 0101(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 1111 1111 1111 1100 0010 0000 0000 0000 1010 0001 1011 0101 =

0100 1111 1111 1111 1100 0010 0000 0000 0000 1010 0001 1011 0101


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 0101

Mantissa (52 bits) =
0100 1111 1111 1111 1100 0010 0000 0000 0000 1010 0001 1011 0101


Decimal number -0.000 000 019 557 719 converted to 64 bit double precision IEEE 754 binary floating point representation:

1 - 011 1110 0101 - 0100 1111 1111 1111 1100 0010 0000 0000 0000 1010 0001 1011 0101

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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