0.974 013 318 541 668 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 0.974 013 318 541 668(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
0.974 013 318 541 668(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: 0.974 013 318 541 668.

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) 0.974 013 318 541 668 × 2 = 1 + 0.948 026 637 083 336;
  • 2) 0.948 026 637 083 336 × 2 = 1 + 0.896 053 274 166 672;
  • 3) 0.896 053 274 166 672 × 2 = 1 + 0.792 106 548 333 344;
  • 4) 0.792 106 548 333 344 × 2 = 1 + 0.584 213 096 666 688;
  • 5) 0.584 213 096 666 688 × 2 = 1 + 0.168 426 193 333 376;
  • 6) 0.168 426 193 333 376 × 2 = 0 + 0.336 852 386 666 752;
  • 7) 0.336 852 386 666 752 × 2 = 0 + 0.673 704 773 333 504;
  • 8) 0.673 704 773 333 504 × 2 = 1 + 0.347 409 546 667 008;
  • 9) 0.347 409 546 667 008 × 2 = 0 + 0.694 819 093 334 016;
  • 10) 0.694 819 093 334 016 × 2 = 1 + 0.389 638 186 668 032;
  • 11) 0.389 638 186 668 032 × 2 = 0 + 0.779 276 373 336 064;
  • 12) 0.779 276 373 336 064 × 2 = 1 + 0.558 552 746 672 128;
  • 13) 0.558 552 746 672 128 × 2 = 1 + 0.117 105 493 344 256;
  • 14) 0.117 105 493 344 256 × 2 = 0 + 0.234 210 986 688 512;
  • 15) 0.234 210 986 688 512 × 2 = 0 + 0.468 421 973 377 024;
  • 16) 0.468 421 973 377 024 × 2 = 0 + 0.936 843 946 754 048;
  • 17) 0.936 843 946 754 048 × 2 = 1 + 0.873 687 893 508 096;
  • 18) 0.873 687 893 508 096 × 2 = 1 + 0.747 375 787 016 192;
  • 19) 0.747 375 787 016 192 × 2 = 1 + 0.494 751 574 032 384;
  • 20) 0.494 751 574 032 384 × 2 = 0 + 0.989 503 148 064 768;
  • 21) 0.989 503 148 064 768 × 2 = 1 + 0.979 006 296 129 536;
  • 22) 0.979 006 296 129 536 × 2 = 1 + 0.958 012 592 259 072;
  • 23) 0.958 012 592 259 072 × 2 = 1 + 0.916 025 184 518 144;
  • 24) 0.916 025 184 518 144 × 2 = 1 + 0.832 050 369 036 288;
  • 25) 0.832 050 369 036 288 × 2 = 1 + 0.664 100 738 072 576;
  • 26) 0.664 100 738 072 576 × 2 = 1 + 0.328 201 476 145 152;
  • 27) 0.328 201 476 145 152 × 2 = 0 + 0.656 402 952 290 304;
  • 28) 0.656 402 952 290 304 × 2 = 1 + 0.312 805 904 580 608;
  • 29) 0.312 805 904 580 608 × 2 = 0 + 0.625 611 809 161 216;
  • 30) 0.625 611 809 161 216 × 2 = 1 + 0.251 223 618 322 432;
  • 31) 0.251 223 618 322 432 × 2 = 0 + 0.502 447 236 644 864;
  • 32) 0.502 447 236 644 864 × 2 = 1 + 0.004 894 473 289 728;
  • 33) 0.004 894 473 289 728 × 2 = 0 + 0.009 788 946 579 456;
  • 34) 0.009 788 946 579 456 × 2 = 0 + 0.019 577 893 158 912;
  • 35) 0.019 577 893 158 912 × 2 = 0 + 0.039 155 786 317 824;
  • 36) 0.039 155 786 317 824 × 2 = 0 + 0.078 311 572 635 648;
  • 37) 0.078 311 572 635 648 × 2 = 0 + 0.156 623 145 271 296;
  • 38) 0.156 623 145 271 296 × 2 = 0 + 0.313 246 290 542 592;
  • 39) 0.313 246 290 542 592 × 2 = 0 + 0.626 492 581 085 184;
  • 40) 0.626 492 581 085 184 × 2 = 1 + 0.252 985 162 170 368;
  • 41) 0.252 985 162 170 368 × 2 = 0 + 0.505 970 324 340 736;
  • 42) 0.505 970 324 340 736 × 2 = 1 + 0.011 940 648 681 472;
  • 43) 0.011 940 648 681 472 × 2 = 0 + 0.023 881 297 362 944;
  • 44) 0.023 881 297 362 944 × 2 = 0 + 0.047 762 594 725 888;
  • 45) 0.047 762 594 725 888 × 2 = 0 + 0.095 525 189 451 776;
  • 46) 0.095 525 189 451 776 × 2 = 0 + 0.191 050 378 903 552;
  • 47) 0.191 050 378 903 552 × 2 = 0 + 0.382 100 757 807 104;
  • 48) 0.382 100 757 807 104 × 2 = 0 + 0.764 201 515 614 208;
  • 49) 0.764 201 515 614 208 × 2 = 1 + 0.528 403 031 228 416;
  • 50) 0.528 403 031 228 416 × 2 = 1 + 0.056 806 062 456 832;
  • 51) 0.056 806 062 456 832 × 2 = 0 + 0.113 612 124 913 664;
  • 52) 0.113 612 124 913 664 × 2 = 0 + 0.227 224 249 827 328;
  • 53) 0.227 224 249 827 328 × 2 = 0 + 0.454 448 499 654 656;

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:


0.974 013 318 541 668(10) =

0.1111 1001 0101 1000 1110 1111 1101 0101 0000 0001 0100 0000 1100 0(2)

5. Positive number before normalization:

0.974 013 318 541 668(10) =

0.1111 1001 0101 1000 1110 1111 1101 0101 0000 0001 0100 0000 1100 0(2)

6. 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:


0.974 013 318 541 668(10) =

0.1111 1001 0101 1000 1110 1111 1101 0101 0000 0001 0100 0000 1100 0(2) =

0.1111 1001 0101 1000 1110 1111 1101 0101 0000 0001 0100 0000 1100 0(2) × 20 =

1.1111 0010 1011 0001 1101 1111 1010 1010 0000 0010 1000 0001 1000(2) × 2-1


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): -1

Mantissa (not normalized):
1.1111 0010 1011 0001 1101 1111 1010 1010 0000 0010 1000 0001 1000


8. 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)


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;
  • 1 022 ÷ 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;

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) =

1022(10) =

011 1111 1110(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. 1111 0010 1011 0001 1101 1111 1010 1010 0000 0010 1000 0001 1000 =

1111 0010 1011 0001 1101 1111 1010 1010 0000 0010 1000 0001 1000


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 1110

Mantissa (52 bits) =
1111 0010 1011 0001 1101 1111 1010 1010 0000 0010 1000 0001 1000


Decimal number 0.974 013 318 541 668 converted to 64 bit double precision IEEE 754 binary floating point representation:

0 - 011 1111 1110 - 1111 0010 1011 0001 1101 1111 1010 1010 0000 0010 1000 0001 1000

Share

» Convert decimal 0.974 013 318 541 577 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 05, 2026 [May]: Decimal Numbers Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard. Calculations performed during the month of: May - by our visitors

» Improve your social skills: Are you using communication by design, or by default? NotebookLM YouTube Video of 6.33 minutes

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