-7.769 489 691 002 61 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -7.769 489 691 002 61(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
-7.769 489 691 002 61(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:

|-7.769 489 691 002 61| = 7.769 489 691 002 61


2. First, convert to binary (in base 2) the integer part: 7.
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;
  • 7 ÷ 2 = 3 + 1;
  • 3 ÷ 2 = 1 + 1;
  • 1 ÷ 2 = 0 + 1;

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.

7(10) =

111(2)


4. Convert to binary (base 2) the fractional part: 0.769 489 691 002 61.

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) 0.769 489 691 002 61 × 2 = 1 + 0.538 979 382 005 22;
  • 2) 0.538 979 382 005 22 × 2 = 1 + 0.077 958 764 010 44;
  • 3) 0.077 958 764 010 44 × 2 = 0 + 0.155 917 528 020 88;
  • 4) 0.155 917 528 020 88 × 2 = 0 + 0.311 835 056 041 76;
  • 5) 0.311 835 056 041 76 × 2 = 0 + 0.623 670 112 083 52;
  • 6) 0.623 670 112 083 52 × 2 = 1 + 0.247 340 224 167 04;
  • 7) 0.247 340 224 167 04 × 2 = 0 + 0.494 680 448 334 08;
  • 8) 0.494 680 448 334 08 × 2 = 0 + 0.989 360 896 668 16;
  • 9) 0.989 360 896 668 16 × 2 = 1 + 0.978 721 793 336 32;
  • 10) 0.978 721 793 336 32 × 2 = 1 + 0.957 443 586 672 64;
  • 11) 0.957 443 586 672 64 × 2 = 1 + 0.914 887 173 345 28;
  • 12) 0.914 887 173 345 28 × 2 = 1 + 0.829 774 346 690 56;
  • 13) 0.829 774 346 690 56 × 2 = 1 + 0.659 548 693 381 12;
  • 14) 0.659 548 693 381 12 × 2 = 1 + 0.319 097 386 762 24;
  • 15) 0.319 097 386 762 24 × 2 = 0 + 0.638 194 773 524 48;
  • 16) 0.638 194 773 524 48 × 2 = 1 + 0.276 389 547 048 96;
  • 17) 0.276 389 547 048 96 × 2 = 0 + 0.552 779 094 097 92;
  • 18) 0.552 779 094 097 92 × 2 = 1 + 0.105 558 188 195 84;
  • 19) 0.105 558 188 195 84 × 2 = 0 + 0.211 116 376 391 68;
  • 20) 0.211 116 376 391 68 × 2 = 0 + 0.422 232 752 783 36;
  • 21) 0.422 232 752 783 36 × 2 = 0 + 0.844 465 505 566 72;
  • 22) 0.844 465 505 566 72 × 2 = 1 + 0.688 931 011 133 44;
  • 23) 0.688 931 011 133 44 × 2 = 1 + 0.377 862 022 266 88;
  • 24) 0.377 862 022 266 88 × 2 = 0 + 0.755 724 044 533 76;
  • 25) 0.755 724 044 533 76 × 2 = 1 + 0.511 448 089 067 52;
  • 26) 0.511 448 089 067 52 × 2 = 1 + 0.022 896 178 135 04;
  • 27) 0.022 896 178 135 04 × 2 = 0 + 0.045 792 356 270 08;
  • 28) 0.045 792 356 270 08 × 2 = 0 + 0.091 584 712 540 16;
  • 29) 0.091 584 712 540 16 × 2 = 0 + 0.183 169 425 080 32;
  • 30) 0.183 169 425 080 32 × 2 = 0 + 0.366 338 850 160 64;
  • 31) 0.366 338 850 160 64 × 2 = 0 + 0.732 677 700 321 28;
  • 32) 0.732 677 700 321 28 × 2 = 1 + 0.465 355 400 642 56;
  • 33) 0.465 355 400 642 56 × 2 = 0 + 0.930 710 801 285 12;
  • 34) 0.930 710 801 285 12 × 2 = 1 + 0.861 421 602 570 24;
  • 35) 0.861 421 602 570 24 × 2 = 1 + 0.722 843 205 140 48;
  • 36) 0.722 843 205 140 48 × 2 = 1 + 0.445 686 410 280 96;
  • 37) 0.445 686 410 280 96 × 2 = 0 + 0.891 372 820 561 92;
  • 38) 0.891 372 820 561 92 × 2 = 1 + 0.782 745 641 123 84;
  • 39) 0.782 745 641 123 84 × 2 = 1 + 0.565 491 282 247 68;
  • 40) 0.565 491 282 247 68 × 2 = 1 + 0.130 982 564 495 36;
  • 41) 0.130 982 564 495 36 × 2 = 0 + 0.261 965 128 990 72;
  • 42) 0.261 965 128 990 72 × 2 = 0 + 0.523 930 257 981 44;
  • 43) 0.523 930 257 981 44 × 2 = 1 + 0.047 860 515 962 88;
  • 44) 0.047 860 515 962 88 × 2 = 0 + 0.095 721 031 925 76;
  • 45) 0.095 721 031 925 76 × 2 = 0 + 0.191 442 063 851 52;
  • 46) 0.191 442 063 851 52 × 2 = 0 + 0.382 884 127 703 04;
  • 47) 0.382 884 127 703 04 × 2 = 0 + 0.765 768 255 406 08;
  • 48) 0.765 768 255 406 08 × 2 = 1 + 0.531 536 510 812 16;
  • 49) 0.531 536 510 812 16 × 2 = 1 + 0.063 073 021 624 32;
  • 50) 0.063 073 021 624 32 × 2 = 0 + 0.126 146 043 248 64;
  • 51) 0.126 146 043 248 64 × 2 = 0 + 0.252 292 086 497 28;
  • 52) 0.252 292 086 497 28 × 2 = 0 + 0.504 584 172 994 56;
  • 53) 0.504 584 172 994 56 × 2 = 1 + 0.009 168 345 989 12;

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:


0.769 489 691 002 61(10) =

0.1100 0100 1111 1101 0100 0110 1100 0001 0111 0111 0010 0001 1000 1(2)

6. Positive number before normalization:

7.769 489 691 002 61(10) =

111.1100 0100 1111 1101 0100 0110 1100 0001 0111 0111 0010 0001 1000 1(2)

7. Normalize the binary representation of the number.

Shift the decimal mark 2 positions to the left, so that only one non zero digit remains to the left of it:


7.769 489 691 002 61(10) =

111.1100 0100 1111 1101 0100 0110 1100 0001 0111 0111 0010 0001 1000 1(2) =

111.1100 0100 1111 1101 0100 0110 1100 0001 0111 0111 0010 0001 1000 1(2) × 20 =

1.1111 0001 0011 1111 0101 0001 1011 0000 0101 1101 1100 1000 0110 001(2) × 22


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): 2

Mantissa (not normalized):
1.1111 0001 0011 1111 0101 0001 1011 0000 0101 1101 1100 1000 0110 001


9. Adjust the exponent.

Use the 11 bit excess/bias notation:


Exponent (adjusted) =

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

2 + 2(11-1) - 1 =

(2 + 1 023)(10) =

1 025(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;
  • 1 025 ÷ 2 = 512 + 1;
  • 512 ÷ 2 = 256 + 0;
  • 256 ÷ 2 = 128 + 0;
  • 128 ÷ 2 = 64 + 0;
  • 64 ÷ 2 = 32 + 0;
  • 32 ÷ 2 = 16 + 0;
  • 16 ÷ 2 = 8 + 0;
  • 8 ÷ 2 = 4 + 0;
  • 4 ÷ 2 = 2 + 0;
  • 2 ÷ 2 = 1 + 0;
  • 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) =

1025(10) =

100 0000 0001(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, by removing the excess bits, from the right (if any of the excess bits is set on 1, we are losing precision...).


Mantissa (normalized) =

1. 1111 0001 0011 1111 0101 0001 1011 0000 0101 1101 1100 1000 0110 001 =

1111 0001 0011 1111 0101 0001 1011 0000 0101 1101 1100 1000 0110


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) =
100 0000 0001

Mantissa (52 bits) =
1111 0001 0011 1111 0101 0001 1011 0000 0101 1101 1100 1000 0110


Decimal number -7.769 489 691 002 61 converted to 64 bit double precision IEEE 754 binary floating point representation:

1 - 100 0000 0001 - 1111 0001 0011 1111 0101 0001 1011 0000 0101 1101 1100 1000 0110

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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