-0.016 738 891 601 562 496 530 506 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.016 738 891 601 562 496 530 506(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
-0.016 738 891 601 562 496 530 506(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:

|-0.016 738 891 601 562 496 530 506| = 0.016 738 891 601 562 496 530 506


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: 0.016 738 891 601 562 496 530 506.

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) 0.016 738 891 601 562 496 530 506 × 2 = 0 + 0.033 477 783 203 124 993 061 012;
  • 2) 0.033 477 783 203 124 993 061 012 × 2 = 0 + 0.066 955 566 406 249 986 122 024;
  • 3) 0.066 955 566 406 249 986 122 024 × 2 = 0 + 0.133 911 132 812 499 972 244 048;
  • 4) 0.133 911 132 812 499 972 244 048 × 2 = 0 + 0.267 822 265 624 999 944 488 096;
  • 5) 0.267 822 265 624 999 944 488 096 × 2 = 0 + 0.535 644 531 249 999 888 976 192;
  • 6) 0.535 644 531 249 999 888 976 192 × 2 = 1 + 0.071 289 062 499 999 777 952 384;
  • 7) 0.071 289 062 499 999 777 952 384 × 2 = 0 + 0.142 578 124 999 999 555 904 768;
  • 8) 0.142 578 124 999 999 555 904 768 × 2 = 0 + 0.285 156 249 999 999 111 809 536;
  • 9) 0.285 156 249 999 999 111 809 536 × 2 = 0 + 0.570 312 499 999 998 223 619 072;
  • 10) 0.570 312 499 999 998 223 619 072 × 2 = 1 + 0.140 624 999 999 996 447 238 144;
  • 11) 0.140 624 999 999 996 447 238 144 × 2 = 0 + 0.281 249 999 999 992 894 476 288;
  • 12) 0.281 249 999 999 992 894 476 288 × 2 = 0 + 0.562 499 999 999 985 788 952 576;
  • 13) 0.562 499 999 999 985 788 952 576 × 2 = 1 + 0.124 999 999 999 971 577 905 152;
  • 14) 0.124 999 999 999 971 577 905 152 × 2 = 0 + 0.249 999 999 999 943 155 810 304;
  • 15) 0.249 999 999 999 943 155 810 304 × 2 = 0 + 0.499 999 999 999 886 311 620 608;
  • 16) 0.499 999 999 999 886 311 620 608 × 2 = 0 + 0.999 999 999 999 772 623 241 216;
  • 17) 0.999 999 999 999 772 623 241 216 × 2 = 1 + 0.999 999 999 999 545 246 482 432;
  • 18) 0.999 999 999 999 545 246 482 432 × 2 = 1 + 0.999 999 999 999 090 492 964 864;
  • 19) 0.999 999 999 999 090 492 964 864 × 2 = 1 + 0.999 999 999 998 180 985 929 728;
  • 20) 0.999 999 999 998 180 985 929 728 × 2 = 1 + 0.999 999 999 996 361 971 859 456;
  • 21) 0.999 999 999 996 361 971 859 456 × 2 = 1 + 0.999 999 999 992 723 943 718 912;
  • 22) 0.999 999 999 992 723 943 718 912 × 2 = 1 + 0.999 999 999 985 447 887 437 824;
  • 23) 0.999 999 999 985 447 887 437 824 × 2 = 1 + 0.999 999 999 970 895 774 875 648;
  • 24) 0.999 999 999 970 895 774 875 648 × 2 = 1 + 0.999 999 999 941 791 549 751 296;
  • 25) 0.999 999 999 941 791 549 751 296 × 2 = 1 + 0.999 999 999 883 583 099 502 592;
  • 26) 0.999 999 999 883 583 099 502 592 × 2 = 1 + 0.999 999 999 767 166 199 005 184;
  • 27) 0.999 999 999 767 166 199 005 184 × 2 = 1 + 0.999 999 999 534 332 398 010 368;
  • 28) 0.999 999 999 534 332 398 010 368 × 2 = 1 + 0.999 999 999 068 664 796 020 736;
  • 29) 0.999 999 999 068 664 796 020 736 × 2 = 1 + 0.999 999 998 137 329 592 041 472;
  • 30) 0.999 999 998 137 329 592 041 472 × 2 = 1 + 0.999 999 996 274 659 184 082 944;
  • 31) 0.999 999 996 274 659 184 082 944 × 2 = 1 + 0.999 999 992 549 318 368 165 888;
  • 32) 0.999 999 992 549 318 368 165 888 × 2 = 1 + 0.999 999 985 098 636 736 331 776;
  • 33) 0.999 999 985 098 636 736 331 776 × 2 = 1 + 0.999 999 970 197 273 472 663 552;
  • 34) 0.999 999 970 197 273 472 663 552 × 2 = 1 + 0.999 999 940 394 546 945 327 104;
  • 35) 0.999 999 940 394 546 945 327 104 × 2 = 1 + 0.999 999 880 789 093 890 654 208;
  • 36) 0.999 999 880 789 093 890 654 208 × 2 = 1 + 0.999 999 761 578 187 781 308 416;
  • 37) 0.999 999 761 578 187 781 308 416 × 2 = 1 + 0.999 999 523 156 375 562 616 832;
  • 38) 0.999 999 523 156 375 562 616 832 × 2 = 1 + 0.999 999 046 312 751 125 233 664;
  • 39) 0.999 999 046 312 751 125 233 664 × 2 = 1 + 0.999 998 092 625 502 250 467 328;
  • 40) 0.999 998 092 625 502 250 467 328 × 2 = 1 + 0.999 996 185 251 004 500 934 656;
  • 41) 0.999 996 185 251 004 500 934 656 × 2 = 1 + 0.999 992 370 502 009 001 869 312;
  • 42) 0.999 992 370 502 009 001 869 312 × 2 = 1 + 0.999 984 741 004 018 003 738 624;
  • 43) 0.999 984 741 004 018 003 738 624 × 2 = 1 + 0.999 969 482 008 036 007 477 248;
  • 44) 0.999 969 482 008 036 007 477 248 × 2 = 1 + 0.999 938 964 016 072 014 954 496;
  • 45) 0.999 938 964 016 072 014 954 496 × 2 = 1 + 0.999 877 928 032 144 029 908 992;
  • 46) 0.999 877 928 032 144 029 908 992 × 2 = 1 + 0.999 755 856 064 288 059 817 984;
  • 47) 0.999 755 856 064 288 059 817 984 × 2 = 1 + 0.999 511 712 128 576 119 635 968;
  • 48) 0.999 511 712 128 576 119 635 968 × 2 = 1 + 0.999 023 424 257 152 239 271 936;
  • 49) 0.999 023 424 257 152 239 271 936 × 2 = 1 + 0.998 046 848 514 304 478 543 872;
  • 50) 0.998 046 848 514 304 478 543 872 × 2 = 1 + 0.996 093 697 028 608 957 087 744;
  • 51) 0.996 093 697 028 608 957 087 744 × 2 = 1 + 0.992 187 394 057 217 914 175 488;
  • 52) 0.992 187 394 057 217 914 175 488 × 2 = 1 + 0.984 374 788 114 435 828 350 976;
  • 53) 0.984 374 788 114 435 828 350 976 × 2 = 1 + 0.968 749 576 228 871 656 701 952;
  • 54) 0.968 749 576 228 871 656 701 952 × 2 = 1 + 0.937 499 152 457 743 313 403 904;
  • 55) 0.937 499 152 457 743 313 403 904 × 2 = 1 + 0.874 998 304 915 486 626 807 808;
  • 56) 0.874 998 304 915 486 626 807 808 × 2 = 1 + 0.749 996 609 830 973 253 615 616;
  • 57) 0.749 996 609 830 973 253 615 616 × 2 = 1 + 0.499 993 219 661 946 507 231 232;
  • 58) 0.499 993 219 661 946 507 231 232 × 2 = 0 + 0.999 986 439 323 893 014 462 464;

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:


0.016 738 891 601 562 496 530 506(10) =

0.0000 0100 0100 1000 1111 1111 1111 1111 1111 1111 1111 1111 1111 1111 10(2)

6. Positive number before normalization:

0.016 738 891 601 562 496 530 506(10) =

0.0000 0100 0100 1000 1111 1111 1111 1111 1111 1111 1111 1111 1111 1111 10(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:


0.016 738 891 601 562 496 530 506(10) =

0.0000 0100 0100 1000 1111 1111 1111 1111 1111 1111 1111 1111 1111 1111 10(2) =

0.0000 0100 0100 1000 1111 1111 1111 1111 1111 1111 1111 1111 1111 1111 10(2) × 20 =

1.0001 0010 0011 1111 1111 1111 1111 1111 1111 1111 1111 1111 1110(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 1110


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 1110 =

0001 0010 0011 1111 1111 1111 1111 1111 1111 1111 1111 1111 1110


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 1110


Decimal number -0.016 738 891 601 562 496 530 506 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 1110

Share

» Convert decimal -0.016 738 891 601 562 496 530 513 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: Magnetic Leadership: The Invisible Power of the Pull. NotebookLM YouTube Video of 11.31 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