0.000 030 959 679 4 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 0.000 030 959 679 4(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
0.000 030 959 679 4(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: 0.000 030 959 679 4.

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) 0.000 030 959 679 4 × 2 = 0 + 0.000 061 919 358 8;
  • 2) 0.000 061 919 358 8 × 2 = 0 + 0.000 123 838 717 6;
  • 3) 0.000 123 838 717 6 × 2 = 0 + 0.000 247 677 435 2;
  • 4) 0.000 247 677 435 2 × 2 = 0 + 0.000 495 354 870 4;
  • 5) 0.000 495 354 870 4 × 2 = 0 + 0.000 990 709 740 8;
  • 6) 0.000 990 709 740 8 × 2 = 0 + 0.001 981 419 481 6;
  • 7) 0.001 981 419 481 6 × 2 = 0 + 0.003 962 838 963 2;
  • 8) 0.003 962 838 963 2 × 2 = 0 + 0.007 925 677 926 4;
  • 9) 0.007 925 677 926 4 × 2 = 0 + 0.015 851 355 852 8;
  • 10) 0.015 851 355 852 8 × 2 = 0 + 0.031 702 711 705 6;
  • 11) 0.031 702 711 705 6 × 2 = 0 + 0.063 405 423 411 2;
  • 12) 0.063 405 423 411 2 × 2 = 0 + 0.126 810 846 822 4;
  • 13) 0.126 810 846 822 4 × 2 = 0 + 0.253 621 693 644 8;
  • 14) 0.253 621 693 644 8 × 2 = 0 + 0.507 243 387 289 6;
  • 15) 0.507 243 387 289 6 × 2 = 1 + 0.014 486 774 579 2;
  • 16) 0.014 486 774 579 2 × 2 = 0 + 0.028 973 549 158 4;
  • 17) 0.028 973 549 158 4 × 2 = 0 + 0.057 947 098 316 8;
  • 18) 0.057 947 098 316 8 × 2 = 0 + 0.115 894 196 633 6;
  • 19) 0.115 894 196 633 6 × 2 = 0 + 0.231 788 393 267 2;
  • 20) 0.231 788 393 267 2 × 2 = 0 + 0.463 576 786 534 4;
  • 21) 0.463 576 786 534 4 × 2 = 0 + 0.927 153 573 068 8;
  • 22) 0.927 153 573 068 8 × 2 = 1 + 0.854 307 146 137 6;
  • 23) 0.854 307 146 137 6 × 2 = 1 + 0.708 614 292 275 2;
  • 24) 0.708 614 292 275 2 × 2 = 1 + 0.417 228 584 550 4;
  • 25) 0.417 228 584 550 4 × 2 = 0 + 0.834 457 169 100 8;
  • 26) 0.834 457 169 100 8 × 2 = 1 + 0.668 914 338 201 6;
  • 27) 0.668 914 338 201 6 × 2 = 1 + 0.337 828 676 403 2;
  • 28) 0.337 828 676 403 2 × 2 = 0 + 0.675 657 352 806 4;
  • 29) 0.675 657 352 806 4 × 2 = 1 + 0.351 314 705 612 8;
  • 30) 0.351 314 705 612 8 × 2 = 0 + 0.702 629 411 225 6;
  • 31) 0.702 629 411 225 6 × 2 = 1 + 0.405 258 822 451 2;
  • 32) 0.405 258 822 451 2 × 2 = 0 + 0.810 517 644 902 4;
  • 33) 0.810 517 644 902 4 × 2 = 1 + 0.621 035 289 804 8;
  • 34) 0.621 035 289 804 8 × 2 = 1 + 0.242 070 579 609 6;
  • 35) 0.242 070 579 609 6 × 2 = 0 + 0.484 141 159 219 2;
  • 36) 0.484 141 159 219 2 × 2 = 0 + 0.968 282 318 438 4;
  • 37) 0.968 282 318 438 4 × 2 = 1 + 0.936 564 636 876 8;
  • 38) 0.936 564 636 876 8 × 2 = 1 + 0.873 129 273 753 6;
  • 39) 0.873 129 273 753 6 × 2 = 1 + 0.746 258 547 507 2;
  • 40) 0.746 258 547 507 2 × 2 = 1 + 0.492 517 095 014 4;
  • 41) 0.492 517 095 014 4 × 2 = 0 + 0.985 034 190 028 8;
  • 42) 0.985 034 190 028 8 × 2 = 1 + 0.970 068 380 057 6;
  • 43) 0.970 068 380 057 6 × 2 = 1 + 0.940 136 760 115 2;
  • 44) 0.940 136 760 115 2 × 2 = 1 + 0.880 273 520 230 4;
  • 45) 0.880 273 520 230 4 × 2 = 1 + 0.760 547 040 460 8;
  • 46) 0.760 547 040 460 8 × 2 = 1 + 0.521 094 080 921 6;
  • 47) 0.521 094 080 921 6 × 2 = 1 + 0.042 188 161 843 2;
  • 48) 0.042 188 161 843 2 × 2 = 0 + 0.084 376 323 686 4;
  • 49) 0.084 376 323 686 4 × 2 = 0 + 0.168 752 647 372 8;
  • 50) 0.168 752 647 372 8 × 2 = 0 + 0.337 505 294 745 6;
  • 51) 0.337 505 294 745 6 × 2 = 0 + 0.675 010 589 491 2;
  • 52) 0.675 010 589 491 2 × 2 = 1 + 0.350 021 178 982 4;
  • 53) 0.350 021 178 982 4 × 2 = 0 + 0.700 042 357 964 8;
  • 54) 0.700 042 357 964 8 × 2 = 1 + 0.400 084 715 929 6;
  • 55) 0.400 084 715 929 6 × 2 = 0 + 0.800 169 431 859 2;
  • 56) 0.800 169 431 859 2 × 2 = 1 + 0.600 338 863 718 4;
  • 57) 0.600 338 863 718 4 × 2 = 1 + 0.200 677 727 436 8;
  • 58) 0.200 677 727 436 8 × 2 = 0 + 0.401 355 454 873 6;
  • 59) 0.401 355 454 873 6 × 2 = 0 + 0.802 710 909 747 2;
  • 60) 0.802 710 909 747 2 × 2 = 1 + 0.605 421 819 494 4;
  • 61) 0.605 421 819 494 4 × 2 = 1 + 0.210 843 638 988 8;
  • 62) 0.210 843 638 988 8 × 2 = 0 + 0.421 687 277 977 6;
  • 63) 0.421 687 277 977 6 × 2 = 0 + 0.843 374 555 955 2;
  • 64) 0.843 374 555 955 2 × 2 = 1 + 0.686 749 111 910 4;
  • 65) 0.686 749 111 910 4 × 2 = 1 + 0.373 498 223 820 8;
  • 66) 0.373 498 223 820 8 × 2 = 0 + 0.746 996 447 641 6;
  • 67) 0.746 996 447 641 6 × 2 = 1 + 0.493 992 895 283 2;

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:


0.000 030 959 679 4(10) =

0.0000 0000 0000 0010 0000 0111 0110 1010 1100 1111 0111 1110 0001 0101 1001 1001 101(2)

5. Positive number before normalization:

0.000 030 959 679 4(10) =

0.0000 0000 0000 0010 0000 0111 0110 1010 1100 1111 0111 1110 0001 0101 1001 1001 101(2)

6. Normalize the binary representation of the number.

Shift the decimal mark 15 positions to the right, so that only one non zero digit remains to the left of it:


0.000 030 959 679 4(10) =

0.0000 0000 0000 0010 0000 0111 0110 1010 1100 1111 0111 1110 0001 0101 1001 1001 101(2) =

0.0000 0000 0000 0010 0000 0111 0110 1010 1100 1111 0111 1110 0001 0101 1001 1001 101(2) × 20 =

1.0000 0011 1011 0101 0110 0111 1011 1111 0000 1010 1100 1100 1101(2) × 2-15


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): -15

Mantissa (not normalized):
1.0000 0011 1011 0101 0110 0111 1011 1111 0000 1010 1100 1100 1101


8. Adjust the exponent.

Use the 11 bit excess/bias notation:


Exponent (adjusted) =

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

-15 + 2(11-1) - 1 =

(-15 + 1 023)(10) =

1 008(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;
  • 1 008 ÷ 2 = 504 + 0;
  • 504 ÷ 2 = 252 + 0;
  • 252 ÷ 2 = 126 + 0;
  • 126 ÷ 2 = 63 + 0;
  • 63 ÷ 2 = 31 + 1;
  • 31 ÷ 2 = 15 + 1;
  • 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) =

1008(10) =

011 1111 0000(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. 0000 0011 1011 0101 0110 0111 1011 1111 0000 1010 1100 1100 1101 =

0000 0011 1011 0101 0110 0111 1011 1111 0000 1010 1100 1100 1101


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 1111 0000

Mantissa (52 bits) =
0000 0011 1011 0101 0110 0111 1011 1111 0000 1010 1100 1100 1101


Decimal number 0.000 030 959 679 4 converted to 64 bit double precision IEEE 754 binary floating point representation:

0 - 011 1111 0000 - 0000 0011 1011 0101 0110 0111 1011 1111 0000 1010 1100 1100 1101

» Convert decimal 0.000 030 959 672 3 to 64 bit double precision IEEE 754 binary floating point representation standard

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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