-0.738 560 627 359 831 220 69 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.738 560 627 359 831 220 69(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
-0.738 560 627 359 831 220 69(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:

|-0.738 560 627 359 831 220 69| = 0.738 560 627 359 831 220 69


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: 0.738 560 627 359 831 220 69.

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) 0.738 560 627 359 831 220 69 × 2 = 1 + 0.477 121 254 719 662 441 38;
  • 2) 0.477 121 254 719 662 441 38 × 2 = 0 + 0.954 242 509 439 324 882 76;
  • 3) 0.954 242 509 439 324 882 76 × 2 = 1 + 0.908 485 018 878 649 765 52;
  • 4) 0.908 485 018 878 649 765 52 × 2 = 1 + 0.816 970 037 757 299 531 04;
  • 5) 0.816 970 037 757 299 531 04 × 2 = 1 + 0.633 940 075 514 599 062 08;
  • 6) 0.633 940 075 514 599 062 08 × 2 = 1 + 0.267 880 151 029 198 124 16;
  • 7) 0.267 880 151 029 198 124 16 × 2 = 0 + 0.535 760 302 058 396 248 32;
  • 8) 0.535 760 302 058 396 248 32 × 2 = 1 + 0.071 520 604 116 792 496 64;
  • 9) 0.071 520 604 116 792 496 64 × 2 = 0 + 0.143 041 208 233 584 993 28;
  • 10) 0.143 041 208 233 584 993 28 × 2 = 0 + 0.286 082 416 467 169 986 56;
  • 11) 0.286 082 416 467 169 986 56 × 2 = 0 + 0.572 164 832 934 339 973 12;
  • 12) 0.572 164 832 934 339 973 12 × 2 = 1 + 0.144 329 665 868 679 946 24;
  • 13) 0.144 329 665 868 679 946 24 × 2 = 0 + 0.288 659 331 737 359 892 48;
  • 14) 0.288 659 331 737 359 892 48 × 2 = 0 + 0.577 318 663 474 719 784 96;
  • 15) 0.577 318 663 474 719 784 96 × 2 = 1 + 0.154 637 326 949 439 569 92;
  • 16) 0.154 637 326 949 439 569 92 × 2 = 0 + 0.309 274 653 898 879 139 84;
  • 17) 0.309 274 653 898 879 139 84 × 2 = 0 + 0.618 549 307 797 758 279 68;
  • 18) 0.618 549 307 797 758 279 68 × 2 = 1 + 0.237 098 615 595 516 559 36;
  • 19) 0.237 098 615 595 516 559 36 × 2 = 0 + 0.474 197 231 191 033 118 72;
  • 20) 0.474 197 231 191 033 118 72 × 2 = 0 + 0.948 394 462 382 066 237 44;
  • 21) 0.948 394 462 382 066 237 44 × 2 = 1 + 0.896 788 924 764 132 474 88;
  • 22) 0.896 788 924 764 132 474 88 × 2 = 1 + 0.793 577 849 528 264 949 76;
  • 23) 0.793 577 849 528 264 949 76 × 2 = 1 + 0.587 155 699 056 529 899 52;
  • 24) 0.587 155 699 056 529 899 52 × 2 = 1 + 0.174 311 398 113 059 799 04;
  • 25) 0.174 311 398 113 059 799 04 × 2 = 0 + 0.348 622 796 226 119 598 08;
  • 26) 0.348 622 796 226 119 598 08 × 2 = 0 + 0.697 245 592 452 239 196 16;
  • 27) 0.697 245 592 452 239 196 16 × 2 = 1 + 0.394 491 184 904 478 392 32;
  • 28) 0.394 491 184 904 478 392 32 × 2 = 0 + 0.788 982 369 808 956 784 64;
  • 29) 0.788 982 369 808 956 784 64 × 2 = 1 + 0.577 964 739 617 913 569 28;
  • 30) 0.577 964 739 617 913 569 28 × 2 = 1 + 0.155 929 479 235 827 138 56;
  • 31) 0.155 929 479 235 827 138 56 × 2 = 0 + 0.311 858 958 471 654 277 12;
  • 32) 0.311 858 958 471 654 277 12 × 2 = 0 + 0.623 717 916 943 308 554 24;
  • 33) 0.623 717 916 943 308 554 24 × 2 = 1 + 0.247 435 833 886 617 108 48;
  • 34) 0.247 435 833 886 617 108 48 × 2 = 0 + 0.494 871 667 773 234 216 96;
  • 35) 0.494 871 667 773 234 216 96 × 2 = 0 + 0.989 743 335 546 468 433 92;
  • 36) 0.989 743 335 546 468 433 92 × 2 = 1 + 0.979 486 671 092 936 867 84;
  • 37) 0.979 486 671 092 936 867 84 × 2 = 1 + 0.958 973 342 185 873 735 68;
  • 38) 0.958 973 342 185 873 735 68 × 2 = 1 + 0.917 946 684 371 747 471 36;
  • 39) 0.917 946 684 371 747 471 36 × 2 = 1 + 0.835 893 368 743 494 942 72;
  • 40) 0.835 893 368 743 494 942 72 × 2 = 1 + 0.671 786 737 486 989 885 44;
  • 41) 0.671 786 737 486 989 885 44 × 2 = 1 + 0.343 573 474 973 979 770 88;
  • 42) 0.343 573 474 973 979 770 88 × 2 = 0 + 0.687 146 949 947 959 541 76;
  • 43) 0.687 146 949 947 959 541 76 × 2 = 1 + 0.374 293 899 895 919 083 52;
  • 44) 0.374 293 899 895 919 083 52 × 2 = 0 + 0.748 587 799 791 838 167 04;
  • 45) 0.748 587 799 791 838 167 04 × 2 = 1 + 0.497 175 599 583 676 334 08;
  • 46) 0.497 175 599 583 676 334 08 × 2 = 0 + 0.994 351 199 167 352 668 16;
  • 47) 0.994 351 199 167 352 668 16 × 2 = 1 + 0.988 702 398 334 705 336 32;
  • 48) 0.988 702 398 334 705 336 32 × 2 = 1 + 0.977 404 796 669 410 672 64;
  • 49) 0.977 404 796 669 410 672 64 × 2 = 1 + 0.954 809 593 338 821 345 28;
  • 50) 0.954 809 593 338 821 345 28 × 2 = 1 + 0.909 619 186 677 642 690 56;
  • 51) 0.909 619 186 677 642 690 56 × 2 = 1 + 0.819 238 373 355 285 381 12;
  • 52) 0.819 238 373 355 285 381 12 × 2 = 1 + 0.638 476 746 710 570 762 24;
  • 53) 0.638 476 746 710 570 762 24 × 2 = 1 + 0.276 953 493 421 141 524 48;

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:


0.738 560 627 359 831 220 69(10) =

0.1011 1101 0001 0010 0100 1111 0010 1100 1001 1111 1010 1011 1111 1(2)

6. Positive number before normalization:

0.738 560 627 359 831 220 69(10) =

0.1011 1101 0001 0010 0100 1111 0010 1100 1001 1111 1010 1011 1111 1(2)

7. Normalize the binary representation of the number.

Shift the decimal mark 1 positions to the right, so that only one non zero digit remains to the left of it:


0.738 560 627 359 831 220 69(10) =

0.1011 1101 0001 0010 0100 1111 0010 1100 1001 1111 1010 1011 1111 1(2) =

0.1011 1101 0001 0010 0100 1111 0010 1100 1001 1111 1010 1011 1111 1(2) × 20 =

1.0111 1010 0010 0100 1001 1110 0101 1001 0011 1111 0101 0111 1111(2) × 2-1


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): -1

Mantissa (not normalized):
1.0111 1010 0010 0100 1001 1110 0101 1001 0011 1111 0101 0111 1111


9. Adjust the exponent.

Use the 11 bit excess/bias notation:


Exponent (adjusted) =

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

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

(-1 + 1 023)(10) =

1 022(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;
  • 1 022 ÷ 2 = 511 + 0;
  • 511 ÷ 2 = 255 + 1;
  • 255 ÷ 2 = 127 + 1;
  • 127 ÷ 2 = 63 + 1;
  • 63 ÷ 2 = 31 + 1;
  • 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) =

1022(10) =

011 1111 1110(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. 0111 1010 0010 0100 1001 1110 0101 1001 0011 1111 0101 0111 1111 =

0111 1010 0010 0100 1001 1110 0101 1001 0011 1111 0101 0111 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 1111 1110

Mantissa (52 bits) =
0111 1010 0010 0100 1001 1110 0101 1001 0011 1111 0101 0111 1111


Decimal number -0.738 560 627 359 831 220 69 converted to 64 bit double precision IEEE 754 binary floating point representation:

1 - 011 1111 1110 - 0111 1010 0010 0100 1001 1110 0101 1001 0011 1111 0101 0111 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