-0.000 000 165 751 5 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 165 751 5(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
-0.000 000 165 751 5(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:

|-0.000 000 165 751 5| = 0.000 000 165 751 5


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: 0.000 000 165 751 5.

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) 0.000 000 165 751 5 × 2 = 0 + 0.000 000 331 503;
  • 2) 0.000 000 331 503 × 2 = 0 + 0.000 000 663 006;
  • 3) 0.000 000 663 006 × 2 = 0 + 0.000 001 326 012;
  • 4) 0.000 001 326 012 × 2 = 0 + 0.000 002 652 024;
  • 5) 0.000 002 652 024 × 2 = 0 + 0.000 005 304 048;
  • 6) 0.000 005 304 048 × 2 = 0 + 0.000 010 608 096;
  • 7) 0.000 010 608 096 × 2 = 0 + 0.000 021 216 192;
  • 8) 0.000 021 216 192 × 2 = 0 + 0.000 042 432 384;
  • 9) 0.000 042 432 384 × 2 = 0 + 0.000 084 864 768;
  • 10) 0.000 084 864 768 × 2 = 0 + 0.000 169 729 536;
  • 11) 0.000 169 729 536 × 2 = 0 + 0.000 339 459 072;
  • 12) 0.000 339 459 072 × 2 = 0 + 0.000 678 918 144;
  • 13) 0.000 678 918 144 × 2 = 0 + 0.001 357 836 288;
  • 14) 0.001 357 836 288 × 2 = 0 + 0.002 715 672 576;
  • 15) 0.002 715 672 576 × 2 = 0 + 0.005 431 345 152;
  • 16) 0.005 431 345 152 × 2 = 0 + 0.010 862 690 304;
  • 17) 0.010 862 690 304 × 2 = 0 + 0.021 725 380 608;
  • 18) 0.021 725 380 608 × 2 = 0 + 0.043 450 761 216;
  • 19) 0.043 450 761 216 × 2 = 0 + 0.086 901 522 432;
  • 20) 0.086 901 522 432 × 2 = 0 + 0.173 803 044 864;
  • 21) 0.173 803 044 864 × 2 = 0 + 0.347 606 089 728;
  • 22) 0.347 606 089 728 × 2 = 0 + 0.695 212 179 456;
  • 23) 0.695 212 179 456 × 2 = 1 + 0.390 424 358 912;
  • 24) 0.390 424 358 912 × 2 = 0 + 0.780 848 717 824;
  • 25) 0.780 848 717 824 × 2 = 1 + 0.561 697 435 648;
  • 26) 0.561 697 435 648 × 2 = 1 + 0.123 394 871 296;
  • 27) 0.123 394 871 296 × 2 = 0 + 0.246 789 742 592;
  • 28) 0.246 789 742 592 × 2 = 0 + 0.493 579 485 184;
  • 29) 0.493 579 485 184 × 2 = 0 + 0.987 158 970 368;
  • 30) 0.987 158 970 368 × 2 = 1 + 0.974 317 940 736;
  • 31) 0.974 317 940 736 × 2 = 1 + 0.948 635 881 472;
  • 32) 0.948 635 881 472 × 2 = 1 + 0.897 271 762 944;
  • 33) 0.897 271 762 944 × 2 = 1 + 0.794 543 525 888;
  • 34) 0.794 543 525 888 × 2 = 1 + 0.589 087 051 776;
  • 35) 0.589 087 051 776 × 2 = 1 + 0.178 174 103 552;
  • 36) 0.178 174 103 552 × 2 = 0 + 0.356 348 207 104;
  • 37) 0.356 348 207 104 × 2 = 0 + 0.712 696 414 208;
  • 38) 0.712 696 414 208 × 2 = 1 + 0.425 392 828 416;
  • 39) 0.425 392 828 416 × 2 = 0 + 0.850 785 656 832;
  • 40) 0.850 785 656 832 × 2 = 1 + 0.701 571 313 664;
  • 41) 0.701 571 313 664 × 2 = 1 + 0.403 142 627 328;
  • 42) 0.403 142 627 328 × 2 = 0 + 0.806 285 254 656;
  • 43) 0.806 285 254 656 × 2 = 1 + 0.612 570 509 312;
  • 44) 0.612 570 509 312 × 2 = 1 + 0.225 141 018 624;
  • 45) 0.225 141 018 624 × 2 = 0 + 0.450 282 037 248;
  • 46) 0.450 282 037 248 × 2 = 0 + 0.900 564 074 496;
  • 47) 0.900 564 074 496 × 2 = 1 + 0.801 128 148 992;
  • 48) 0.801 128 148 992 × 2 = 1 + 0.602 256 297 984;
  • 49) 0.602 256 297 984 × 2 = 1 + 0.204 512 595 968;
  • 50) 0.204 512 595 968 × 2 = 0 + 0.409 025 191 936;
  • 51) 0.409 025 191 936 × 2 = 0 + 0.818 050 383 872;
  • 52) 0.818 050 383 872 × 2 = 1 + 0.636 100 767 744;
  • 53) 0.636 100 767 744 × 2 = 1 + 0.272 201 535 488;
  • 54) 0.272 201 535 488 × 2 = 0 + 0.544 403 070 976;
  • 55) 0.544 403 070 976 × 2 = 1 + 0.088 806 141 952;
  • 56) 0.088 806 141 952 × 2 = 0 + 0.177 612 283 904;
  • 57) 0.177 612 283 904 × 2 = 0 + 0.355 224 567 808;
  • 58) 0.355 224 567 808 × 2 = 0 + 0.710 449 135 616;
  • 59) 0.710 449 135 616 × 2 = 1 + 0.420 898 271 232;
  • 60) 0.420 898 271 232 × 2 = 0 + 0.841 796 542 464;
  • 61) 0.841 796 542 464 × 2 = 1 + 0.683 593 084 928;
  • 62) 0.683 593 084 928 × 2 = 1 + 0.367 186 169 856;
  • 63) 0.367 186 169 856 × 2 = 0 + 0.734 372 339 712;
  • 64) 0.734 372 339 712 × 2 = 1 + 0.468 744 679 424;
  • 65) 0.468 744 679 424 × 2 = 0 + 0.937 489 358 848;
  • 66) 0.937 489 358 848 × 2 = 1 + 0.874 978 717 696;
  • 67) 0.874 978 717 696 × 2 = 1 + 0.749 957 435 392;
  • 68) 0.749 957 435 392 × 2 = 1 + 0.499 914 870 784;
  • 69) 0.499 914 870 784 × 2 = 0 + 0.999 829 741 568;
  • 70) 0.999 829 741 568 × 2 = 1 + 0.999 659 483 136;
  • 71) 0.999 659 483 136 × 2 = 1 + 0.999 318 966 272;
  • 72) 0.999 318 966 272 × 2 = 1 + 0.998 637 932 544;
  • 73) 0.998 637 932 544 × 2 = 1 + 0.997 275 865 088;
  • 74) 0.997 275 865 088 × 2 = 1 + 0.994 551 730 176;
  • 75) 0.994 551 730 176 × 2 = 1 + 0.989 103 460 352;

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:


0.000 000 165 751 5(10) =

0.0000 0000 0000 0000 0000 0010 1100 0111 1110 0101 1011 0011 1001 1010 0010 1101 0111 0111 111(2)

6. Positive number before normalization:

0.000 000 165 751 5(10) =

0.0000 0000 0000 0000 0000 0010 1100 0111 1110 0101 1011 0011 1001 1010 0010 1101 0111 0111 111(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:


0.000 000 165 751 5(10) =

0.0000 0000 0000 0000 0000 0010 1100 0111 1110 0101 1011 0011 1001 1010 0010 1101 0111 0111 111(2) =

0.0000 0000 0000 0000 0000 0010 1100 0111 1110 0101 1011 0011 1001 1010 0010 1101 0111 0111 111(2) × 20 =

1.0110 0011 1111 0010 1101 1001 1100 1101 0001 0110 1011 1011 1111(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):
1.0110 0011 1111 0010 1101 1001 1100 1101 0001 0110 1011 1011 1111


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. 0110 0011 1111 0010 1101 1001 1100 1101 0001 0110 1011 1011 1111 =

0110 0011 1111 0010 1101 1001 1100 1101 0001 0110 1011 1011 1111


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) =
0110 0011 1111 0010 1101 1001 1100 1101 0001 0110 1011 1011 1111


Decimal number -0.000 000 165 751 5 converted to 64 bit double precision IEEE 754 binary floating point representation:

1 - 011 1110 1000 - 0110 0011 1111 0010 1101 1001 1100 1101 0001 0110 1011 1011 1111

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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