-0.738 560 627 359 831 220 74 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.738 560 627 359 831 220 74(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 74(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 74| = 0.738 560 627 359 831 220 74


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 74.

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 74 × 2 = 1 + 0.477 121 254 719 662 441 48;
  • 2) 0.477 121 254 719 662 441 48 × 2 = 0 + 0.954 242 509 439 324 882 96;
  • 3) 0.954 242 509 439 324 882 96 × 2 = 1 + 0.908 485 018 878 649 765 92;
  • 4) 0.908 485 018 878 649 765 92 × 2 = 1 + 0.816 970 037 757 299 531 84;
  • 5) 0.816 970 037 757 299 531 84 × 2 = 1 + 0.633 940 075 514 599 063 68;
  • 6) 0.633 940 075 514 599 063 68 × 2 = 1 + 0.267 880 151 029 198 127 36;
  • 7) 0.267 880 151 029 198 127 36 × 2 = 0 + 0.535 760 302 058 396 254 72;
  • 8) 0.535 760 302 058 396 254 72 × 2 = 1 + 0.071 520 604 116 792 509 44;
  • 9) 0.071 520 604 116 792 509 44 × 2 = 0 + 0.143 041 208 233 585 018 88;
  • 10) 0.143 041 208 233 585 018 88 × 2 = 0 + 0.286 082 416 467 170 037 76;
  • 11) 0.286 082 416 467 170 037 76 × 2 = 0 + 0.572 164 832 934 340 075 52;
  • 12) 0.572 164 832 934 340 075 52 × 2 = 1 + 0.144 329 665 868 680 151 04;
  • 13) 0.144 329 665 868 680 151 04 × 2 = 0 + 0.288 659 331 737 360 302 08;
  • 14) 0.288 659 331 737 360 302 08 × 2 = 0 + 0.577 318 663 474 720 604 16;
  • 15) 0.577 318 663 474 720 604 16 × 2 = 1 + 0.154 637 326 949 441 208 32;
  • 16) 0.154 637 326 949 441 208 32 × 2 = 0 + 0.309 274 653 898 882 416 64;
  • 17) 0.309 274 653 898 882 416 64 × 2 = 0 + 0.618 549 307 797 764 833 28;
  • 18) 0.618 549 307 797 764 833 28 × 2 = 1 + 0.237 098 615 595 529 666 56;
  • 19) 0.237 098 615 595 529 666 56 × 2 = 0 + 0.474 197 231 191 059 333 12;
  • 20) 0.474 197 231 191 059 333 12 × 2 = 0 + 0.948 394 462 382 118 666 24;
  • 21) 0.948 394 462 382 118 666 24 × 2 = 1 + 0.896 788 924 764 237 332 48;
  • 22) 0.896 788 924 764 237 332 48 × 2 = 1 + 0.793 577 849 528 474 664 96;
  • 23) 0.793 577 849 528 474 664 96 × 2 = 1 + 0.587 155 699 056 949 329 92;
  • 24) 0.587 155 699 056 949 329 92 × 2 = 1 + 0.174 311 398 113 898 659 84;
  • 25) 0.174 311 398 113 898 659 84 × 2 = 0 + 0.348 622 796 227 797 319 68;
  • 26) 0.348 622 796 227 797 319 68 × 2 = 0 + 0.697 245 592 455 594 639 36;
  • 27) 0.697 245 592 455 594 639 36 × 2 = 1 + 0.394 491 184 911 189 278 72;
  • 28) 0.394 491 184 911 189 278 72 × 2 = 0 + 0.788 982 369 822 378 557 44;
  • 29) 0.788 982 369 822 378 557 44 × 2 = 1 + 0.577 964 739 644 757 114 88;
  • 30) 0.577 964 739 644 757 114 88 × 2 = 1 + 0.155 929 479 289 514 229 76;
  • 31) 0.155 929 479 289 514 229 76 × 2 = 0 + 0.311 858 958 579 028 459 52;
  • 32) 0.311 858 958 579 028 459 52 × 2 = 0 + 0.623 717 917 158 056 919 04;
  • 33) 0.623 717 917 158 056 919 04 × 2 = 1 + 0.247 435 834 316 113 838 08;
  • 34) 0.247 435 834 316 113 838 08 × 2 = 0 + 0.494 871 668 632 227 676 16;
  • 35) 0.494 871 668 632 227 676 16 × 2 = 0 + 0.989 743 337 264 455 352 32;
  • 36) 0.989 743 337 264 455 352 32 × 2 = 1 + 0.979 486 674 528 910 704 64;
  • 37) 0.979 486 674 528 910 704 64 × 2 = 1 + 0.958 973 349 057 821 409 28;
  • 38) 0.958 973 349 057 821 409 28 × 2 = 1 + 0.917 946 698 115 642 818 56;
  • 39) 0.917 946 698 115 642 818 56 × 2 = 1 + 0.835 893 396 231 285 637 12;
  • 40) 0.835 893 396 231 285 637 12 × 2 = 1 + 0.671 786 792 462 571 274 24;
  • 41) 0.671 786 792 462 571 274 24 × 2 = 1 + 0.343 573 584 925 142 548 48;
  • 42) 0.343 573 584 925 142 548 48 × 2 = 0 + 0.687 147 169 850 285 096 96;
  • 43) 0.687 147 169 850 285 096 96 × 2 = 1 + 0.374 294 339 700 570 193 92;
  • 44) 0.374 294 339 700 570 193 92 × 2 = 0 + 0.748 588 679 401 140 387 84;
  • 45) 0.748 588 679 401 140 387 84 × 2 = 1 + 0.497 177 358 802 280 775 68;
  • 46) 0.497 177 358 802 280 775 68 × 2 = 0 + 0.994 354 717 604 561 551 36;
  • 47) 0.994 354 717 604 561 551 36 × 2 = 1 + 0.988 709 435 209 123 102 72;
  • 48) 0.988 709 435 209 123 102 72 × 2 = 1 + 0.977 418 870 418 246 205 44;
  • 49) 0.977 418 870 418 246 205 44 × 2 = 1 + 0.954 837 740 836 492 410 88;
  • 50) 0.954 837 740 836 492 410 88 × 2 = 1 + 0.909 675 481 672 984 821 76;
  • 51) 0.909 675 481 672 984 821 76 × 2 = 1 + 0.819 350 963 345 969 643 52;
  • 52) 0.819 350 963 345 969 643 52 × 2 = 1 + 0.638 701 926 691 939 287 04;
  • 53) 0.638 701 926 691 939 287 04 × 2 = 1 + 0.277 403 853 383 878 574 08;

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 74(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 74(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 74(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 74 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

» Convert decimal -0.738 560 627 359 831 221 25 to 64 bit double precision IEEE 754 binary floating point representation standard

» Calculations Performed by Our Visitors: Decimal Numbers Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard. Data organized on a Monthly Basis

» Month 10, 2026 [October]: Decimal Numbers Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard. Calculations performed during the month of: October - by our visitors

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