0.000 046 473 601 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 0.000 046 473 601(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
0.000 046 473 601(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: 0.000 046 473 601.

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) 0.000 046 473 601 × 2 = 0 + 0.000 092 947 202;
  • 2) 0.000 092 947 202 × 2 = 0 + 0.000 185 894 404;
  • 3) 0.000 185 894 404 × 2 = 0 + 0.000 371 788 808;
  • 4) 0.000 371 788 808 × 2 = 0 + 0.000 743 577 616;
  • 5) 0.000 743 577 616 × 2 = 0 + 0.001 487 155 232;
  • 6) 0.001 487 155 232 × 2 = 0 + 0.002 974 310 464;
  • 7) 0.002 974 310 464 × 2 = 0 + 0.005 948 620 928;
  • 8) 0.005 948 620 928 × 2 = 0 + 0.011 897 241 856;
  • 9) 0.011 897 241 856 × 2 = 0 + 0.023 794 483 712;
  • 10) 0.023 794 483 712 × 2 = 0 + 0.047 588 967 424;
  • 11) 0.047 588 967 424 × 2 = 0 + 0.095 177 934 848;
  • 12) 0.095 177 934 848 × 2 = 0 + 0.190 355 869 696;
  • 13) 0.190 355 869 696 × 2 = 0 + 0.380 711 739 392;
  • 14) 0.380 711 739 392 × 2 = 0 + 0.761 423 478 784;
  • 15) 0.761 423 478 784 × 2 = 1 + 0.522 846 957 568;
  • 16) 0.522 846 957 568 × 2 = 1 + 0.045 693 915 136;
  • 17) 0.045 693 915 136 × 2 = 0 + 0.091 387 830 272;
  • 18) 0.091 387 830 272 × 2 = 0 + 0.182 775 660 544;
  • 19) 0.182 775 660 544 × 2 = 0 + 0.365 551 321 088;
  • 20) 0.365 551 321 088 × 2 = 0 + 0.731 102 642 176;
  • 21) 0.731 102 642 176 × 2 = 1 + 0.462 205 284 352;
  • 22) 0.462 205 284 352 × 2 = 0 + 0.924 410 568 704;
  • 23) 0.924 410 568 704 × 2 = 1 + 0.848 821 137 408;
  • 24) 0.848 821 137 408 × 2 = 1 + 0.697 642 274 816;
  • 25) 0.697 642 274 816 × 2 = 1 + 0.395 284 549 632;
  • 26) 0.395 284 549 632 × 2 = 0 + 0.790 569 099 264;
  • 27) 0.790 569 099 264 × 2 = 1 + 0.581 138 198 528;
  • 28) 0.581 138 198 528 × 2 = 1 + 0.162 276 397 056;
  • 29) 0.162 276 397 056 × 2 = 0 + 0.324 552 794 112;
  • 30) 0.324 552 794 112 × 2 = 0 + 0.649 105 588 224;
  • 31) 0.649 105 588 224 × 2 = 1 + 0.298 211 176 448;
  • 32) 0.298 211 176 448 × 2 = 0 + 0.596 422 352 896;
  • 33) 0.596 422 352 896 × 2 = 1 + 0.192 844 705 792;
  • 34) 0.192 844 705 792 × 2 = 0 + 0.385 689 411 584;
  • 35) 0.385 689 411 584 × 2 = 0 + 0.771 378 823 168;
  • 36) 0.771 378 823 168 × 2 = 1 + 0.542 757 646 336;
  • 37) 0.542 757 646 336 × 2 = 1 + 0.085 515 292 672;
  • 38) 0.085 515 292 672 × 2 = 0 + 0.171 030 585 344;
  • 39) 0.171 030 585 344 × 2 = 0 + 0.342 061 170 688;
  • 40) 0.342 061 170 688 × 2 = 0 + 0.684 122 341 376;
  • 41) 0.684 122 341 376 × 2 = 1 + 0.368 244 682 752;
  • 42) 0.368 244 682 752 × 2 = 0 + 0.736 489 365 504;
  • 43) 0.736 489 365 504 × 2 = 1 + 0.472 978 731 008;
  • 44) 0.472 978 731 008 × 2 = 0 + 0.945 957 462 016;
  • 45) 0.945 957 462 016 × 2 = 1 + 0.891 914 924 032;
  • 46) 0.891 914 924 032 × 2 = 1 + 0.783 829 848 064;
  • 47) 0.783 829 848 064 × 2 = 1 + 0.567 659 696 128;
  • 48) 0.567 659 696 128 × 2 = 1 + 0.135 319 392 256;
  • 49) 0.135 319 392 256 × 2 = 0 + 0.270 638 784 512;
  • 50) 0.270 638 784 512 × 2 = 0 + 0.541 277 569 024;
  • 51) 0.541 277 569 024 × 2 = 1 + 0.082 555 138 048;
  • 52) 0.082 555 138 048 × 2 = 0 + 0.165 110 276 096;
  • 53) 0.165 110 276 096 × 2 = 0 + 0.330 220 552 192;
  • 54) 0.330 220 552 192 × 2 = 0 + 0.660 441 104 384;
  • 55) 0.660 441 104 384 × 2 = 1 + 0.320 882 208 768;
  • 56) 0.320 882 208 768 × 2 = 0 + 0.641 764 417 536;
  • 57) 0.641 764 417 536 × 2 = 1 + 0.283 528 835 072;
  • 58) 0.283 528 835 072 × 2 = 0 + 0.567 057 670 144;
  • 59) 0.567 057 670 144 × 2 = 1 + 0.134 115 340 288;
  • 60) 0.134 115 340 288 × 2 = 0 + 0.268 230 680 576;
  • 61) 0.268 230 680 576 × 2 = 0 + 0.536 461 361 152;
  • 62) 0.536 461 361 152 × 2 = 1 + 0.072 922 722 304;
  • 63) 0.072 922 722 304 × 2 = 0 + 0.145 845 444 608;
  • 64) 0.145 845 444 608 × 2 = 0 + 0.291 690 889 216;
  • 65) 0.291 690 889 216 × 2 = 0 + 0.583 381 778 432;
  • 66) 0.583 381 778 432 × 2 = 1 + 0.166 763 556 864;
  • 67) 0.166 763 556 864 × 2 = 0 + 0.333 527 113 728;

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:


0.000 046 473 601(10) =

0.0000 0000 0000 0011 0000 1011 1011 0010 1001 1000 1010 1111 0010 0010 1010 0100 010(2)

5. Positive number before normalization:

0.000 046 473 601(10) =

0.0000 0000 0000 0011 0000 1011 1011 0010 1001 1000 1010 1111 0010 0010 1010 0100 010(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:


0.000 046 473 601(10) =

0.0000 0000 0000 0011 0000 1011 1011 0010 1001 1000 1010 1111 0010 0010 1010 0100 010(2) =

0.0000 0000 0000 0011 0000 1011 1011 0010 1001 1000 1010 1111 0010 0010 1010 0100 010(2) × 20 =

1.1000 0101 1101 1001 0100 1100 0101 0111 1001 0001 0101 0010 0010(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):
1.1000 0101 1101 1001 0100 1100 0101 0111 1001 0001 0101 0010 0010


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. 1000 0101 1101 1001 0100 1100 0101 0111 1001 0001 0101 0010 0010 =

1000 0101 1101 1001 0100 1100 0101 0111 1001 0001 0101 0010 0010


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) =
1000 0101 1101 1001 0100 1100 0101 0111 1001 0001 0101 0010 0010


Decimal number 0.000 046 473 601 converted to 64 bit double precision IEEE 754 binary floating point representation:

0 - 011 1111 0000 - 1000 0101 1101 1001 0100 1100 0101 0111 1001 0001 0101 0010 0010

» Convert decimal 0.000 046 473 646 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