-0.016 738 891 601 562 496 74 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.016 738 891 601 562 496 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
-0.016 738 891 601 562 496 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:

|-0.016 738 891 601 562 496 74| = 0.016 738 891 601 562 496 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: 0.016 738 891 601 562 496 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) 0.016 738 891 601 562 496 74 × 2 = 0 + 0.033 477 783 203 124 993 48;
  • 2) 0.033 477 783 203 124 993 48 × 2 = 0 + 0.066 955 566 406 249 986 96;
  • 3) 0.066 955 566 406 249 986 96 × 2 = 0 + 0.133 911 132 812 499 973 92;
  • 4) 0.133 911 132 812 499 973 92 × 2 = 0 + 0.267 822 265 624 999 947 84;
  • 5) 0.267 822 265 624 999 947 84 × 2 = 0 + 0.535 644 531 249 999 895 68;
  • 6) 0.535 644 531 249 999 895 68 × 2 = 1 + 0.071 289 062 499 999 791 36;
  • 7) 0.071 289 062 499 999 791 36 × 2 = 0 + 0.142 578 124 999 999 582 72;
  • 8) 0.142 578 124 999 999 582 72 × 2 = 0 + 0.285 156 249 999 999 165 44;
  • 9) 0.285 156 249 999 999 165 44 × 2 = 0 + 0.570 312 499 999 998 330 88;
  • 10) 0.570 312 499 999 998 330 88 × 2 = 1 + 0.140 624 999 999 996 661 76;
  • 11) 0.140 624 999 999 996 661 76 × 2 = 0 + 0.281 249 999 999 993 323 52;
  • 12) 0.281 249 999 999 993 323 52 × 2 = 0 + 0.562 499 999 999 986 647 04;
  • 13) 0.562 499 999 999 986 647 04 × 2 = 1 + 0.124 999 999 999 973 294 08;
  • 14) 0.124 999 999 999 973 294 08 × 2 = 0 + 0.249 999 999 999 946 588 16;
  • 15) 0.249 999 999 999 946 588 16 × 2 = 0 + 0.499 999 999 999 893 176 32;
  • 16) 0.499 999 999 999 893 176 32 × 2 = 0 + 0.999 999 999 999 786 352 64;
  • 17) 0.999 999 999 999 786 352 64 × 2 = 1 + 0.999 999 999 999 572 705 28;
  • 18) 0.999 999 999 999 572 705 28 × 2 = 1 + 0.999 999 999 999 145 410 56;
  • 19) 0.999 999 999 999 145 410 56 × 2 = 1 + 0.999 999 999 998 290 821 12;
  • 20) 0.999 999 999 998 290 821 12 × 2 = 1 + 0.999 999 999 996 581 642 24;
  • 21) 0.999 999 999 996 581 642 24 × 2 = 1 + 0.999 999 999 993 163 284 48;
  • 22) 0.999 999 999 993 163 284 48 × 2 = 1 + 0.999 999 999 986 326 568 96;
  • 23) 0.999 999 999 986 326 568 96 × 2 = 1 + 0.999 999 999 972 653 137 92;
  • 24) 0.999 999 999 972 653 137 92 × 2 = 1 + 0.999 999 999 945 306 275 84;
  • 25) 0.999 999 999 945 306 275 84 × 2 = 1 + 0.999 999 999 890 612 551 68;
  • 26) 0.999 999 999 890 612 551 68 × 2 = 1 + 0.999 999 999 781 225 103 36;
  • 27) 0.999 999 999 781 225 103 36 × 2 = 1 + 0.999 999 999 562 450 206 72;
  • 28) 0.999 999 999 562 450 206 72 × 2 = 1 + 0.999 999 999 124 900 413 44;
  • 29) 0.999 999 999 124 900 413 44 × 2 = 1 + 0.999 999 998 249 800 826 88;
  • 30) 0.999 999 998 249 800 826 88 × 2 = 1 + 0.999 999 996 499 601 653 76;
  • 31) 0.999 999 996 499 601 653 76 × 2 = 1 + 0.999 999 992 999 203 307 52;
  • 32) 0.999 999 992 999 203 307 52 × 2 = 1 + 0.999 999 985 998 406 615 04;
  • 33) 0.999 999 985 998 406 615 04 × 2 = 1 + 0.999 999 971 996 813 230 08;
  • 34) 0.999 999 971 996 813 230 08 × 2 = 1 + 0.999 999 943 993 626 460 16;
  • 35) 0.999 999 943 993 626 460 16 × 2 = 1 + 0.999 999 887 987 252 920 32;
  • 36) 0.999 999 887 987 252 920 32 × 2 = 1 + 0.999 999 775 974 505 840 64;
  • 37) 0.999 999 775 974 505 840 64 × 2 = 1 + 0.999 999 551 949 011 681 28;
  • 38) 0.999 999 551 949 011 681 28 × 2 = 1 + 0.999 999 103 898 023 362 56;
  • 39) 0.999 999 103 898 023 362 56 × 2 = 1 + 0.999 998 207 796 046 725 12;
  • 40) 0.999 998 207 796 046 725 12 × 2 = 1 + 0.999 996 415 592 093 450 24;
  • 41) 0.999 996 415 592 093 450 24 × 2 = 1 + 0.999 992 831 184 186 900 48;
  • 42) 0.999 992 831 184 186 900 48 × 2 = 1 + 0.999 985 662 368 373 800 96;
  • 43) 0.999 985 662 368 373 800 96 × 2 = 1 + 0.999 971 324 736 747 601 92;
  • 44) 0.999 971 324 736 747 601 92 × 2 = 1 + 0.999 942 649 473 495 203 84;
  • 45) 0.999 942 649 473 495 203 84 × 2 = 1 + 0.999 885 298 946 990 407 68;
  • 46) 0.999 885 298 946 990 407 68 × 2 = 1 + 0.999 770 597 893 980 815 36;
  • 47) 0.999 770 597 893 980 815 36 × 2 = 1 + 0.999 541 195 787 961 630 72;
  • 48) 0.999 541 195 787 961 630 72 × 2 = 1 + 0.999 082 391 575 923 261 44;
  • 49) 0.999 082 391 575 923 261 44 × 2 = 1 + 0.998 164 783 151 846 522 88;
  • 50) 0.998 164 783 151 846 522 88 × 2 = 1 + 0.996 329 566 303 693 045 76;
  • 51) 0.996 329 566 303 693 045 76 × 2 = 1 + 0.992 659 132 607 386 091 52;
  • 52) 0.992 659 132 607 386 091 52 × 2 = 1 + 0.985 318 265 214 772 183 04;
  • 53) 0.985 318 265 214 772 183 04 × 2 = 1 + 0.970 636 530 429 544 366 08;
  • 54) 0.970 636 530 429 544 366 08 × 2 = 1 + 0.941 273 060 859 088 732 16;
  • 55) 0.941 273 060 859 088 732 16 × 2 = 1 + 0.882 546 121 718 177 464 32;
  • 56) 0.882 546 121 718 177 464 32 × 2 = 1 + 0.765 092 243 436 354 928 64;
  • 57) 0.765 092 243 436 354 928 64 × 2 = 1 + 0.530 184 486 872 709 857 28;
  • 58) 0.530 184 486 872 709 857 28 × 2 = 1 + 0.060 368 973 745 419 714 56;

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:


0.016 738 891 601 562 496 74(10) =

0.0000 0100 0100 1000 1111 1111 1111 1111 1111 1111 1111 1111 1111 1111 11(2)

6. Positive number before normalization:

0.016 738 891 601 562 496 74(10) =

0.0000 0100 0100 1000 1111 1111 1111 1111 1111 1111 1111 1111 1111 1111 11(2)

7. Normalize the binary representation of the number.

Shift the decimal mark 6 positions to the right, so that only one non zero digit remains to the left of it:


0.016 738 891 601 562 496 74(10) =

0.0000 0100 0100 1000 1111 1111 1111 1111 1111 1111 1111 1111 1111 1111 11(2) =

0.0000 0100 0100 1000 1111 1111 1111 1111 1111 1111 1111 1111 1111 1111 11(2) × 20 =

1.0001 0010 0011 1111 1111 1111 1111 1111 1111 1111 1111 1111 1111(2) × 2-6


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): -6

Mantissa (not normalized):
1.0001 0010 0011 1111 1111 1111 1111 1111 1111 1111 1111 1111 1111


9. Adjust the exponent.

Use the 11 bit excess/bias notation:


Exponent (adjusted) =

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

-6 + 2(11-1) - 1 =

(-6 + 1 023)(10) =

1 017(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;
  • 1 017 ÷ 2 = 508 + 1;
  • 508 ÷ 2 = 254 + 0;
  • 254 ÷ 2 = 127 + 0;
  • 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) =

1017(10) =

011 1111 1001(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. 0001 0010 0011 1111 1111 1111 1111 1111 1111 1111 1111 1111 1111 =

0001 0010 0011 1111 1111 1111 1111 1111 1111 1111 1111 1111 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 1001

Mantissa (52 bits) =
0001 0010 0011 1111 1111 1111 1111 1111 1111 1111 1111 1111 1111


Decimal number -0.016 738 891 601 562 496 74 converted to 64 bit double precision IEEE 754 binary floating point representation:

1 - 011 1111 1001 - 0001 0010 0011 1111 1111 1111 1111 1111 1111 1111 1111 1111 1111

Share

» Convert decimal -0.016 738 891 601 562 497 28 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 06, 2026 [June]: Decimal Numbers Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard. Calculations performed during the month of: June - by our visitors

» Improve your social skills: The Art of Strategic Ambiguity and Concealed Intentions. NotebookLM YouTube Video of 6.59 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