-0.000 007 820 632 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 007 820 632(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
-0.000 007 820 632(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:

|-0.000 007 820 632| = 0.000 007 820 632


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: 0.000 007 820 632.

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) 0.000 007 820 632 × 2 = 0 + 0.000 015 641 264;
  • 2) 0.000 015 641 264 × 2 = 0 + 0.000 031 282 528;
  • 3) 0.000 031 282 528 × 2 = 0 + 0.000 062 565 056;
  • 4) 0.000 062 565 056 × 2 = 0 + 0.000 125 130 112;
  • 5) 0.000 125 130 112 × 2 = 0 + 0.000 250 260 224;
  • 6) 0.000 250 260 224 × 2 = 0 + 0.000 500 520 448;
  • 7) 0.000 500 520 448 × 2 = 0 + 0.001 001 040 896;
  • 8) 0.001 001 040 896 × 2 = 0 + 0.002 002 081 792;
  • 9) 0.002 002 081 792 × 2 = 0 + 0.004 004 163 584;
  • 10) 0.004 004 163 584 × 2 = 0 + 0.008 008 327 168;
  • 11) 0.008 008 327 168 × 2 = 0 + 0.016 016 654 336;
  • 12) 0.016 016 654 336 × 2 = 0 + 0.032 033 308 672;
  • 13) 0.032 033 308 672 × 2 = 0 + 0.064 066 617 344;
  • 14) 0.064 066 617 344 × 2 = 0 + 0.128 133 234 688;
  • 15) 0.128 133 234 688 × 2 = 0 + 0.256 266 469 376;
  • 16) 0.256 266 469 376 × 2 = 0 + 0.512 532 938 752;
  • 17) 0.512 532 938 752 × 2 = 1 + 0.025 065 877 504;
  • 18) 0.025 065 877 504 × 2 = 0 + 0.050 131 755 008;
  • 19) 0.050 131 755 008 × 2 = 0 + 0.100 263 510 016;
  • 20) 0.100 263 510 016 × 2 = 0 + 0.200 527 020 032;
  • 21) 0.200 527 020 032 × 2 = 0 + 0.401 054 040 064;
  • 22) 0.401 054 040 064 × 2 = 0 + 0.802 108 080 128;
  • 23) 0.802 108 080 128 × 2 = 1 + 0.604 216 160 256;
  • 24) 0.604 216 160 256 × 2 = 1 + 0.208 432 320 512;
  • 25) 0.208 432 320 512 × 2 = 0 + 0.416 864 641 024;
  • 26) 0.416 864 641 024 × 2 = 0 + 0.833 729 282 048;
  • 27) 0.833 729 282 048 × 2 = 1 + 0.667 458 564 096;
  • 28) 0.667 458 564 096 × 2 = 1 + 0.334 917 128 192;
  • 29) 0.334 917 128 192 × 2 = 0 + 0.669 834 256 384;
  • 30) 0.669 834 256 384 × 2 = 1 + 0.339 668 512 768;
  • 31) 0.339 668 512 768 × 2 = 0 + 0.679 337 025 536;
  • 32) 0.679 337 025 536 × 2 = 1 + 0.358 674 051 072;
  • 33) 0.358 674 051 072 × 2 = 0 + 0.717 348 102 144;
  • 34) 0.717 348 102 144 × 2 = 1 + 0.434 696 204 288;
  • 35) 0.434 696 204 288 × 2 = 0 + 0.869 392 408 576;
  • 36) 0.869 392 408 576 × 2 = 1 + 0.738 784 817 152;
  • 37) 0.738 784 817 152 × 2 = 1 + 0.477 569 634 304;
  • 38) 0.477 569 634 304 × 2 = 0 + 0.955 139 268 608;
  • 39) 0.955 139 268 608 × 2 = 1 + 0.910 278 537 216;
  • 40) 0.910 278 537 216 × 2 = 1 + 0.820 557 074 432;
  • 41) 0.820 557 074 432 × 2 = 1 + 0.641 114 148 864;
  • 42) 0.641 114 148 864 × 2 = 1 + 0.282 228 297 728;
  • 43) 0.282 228 297 728 × 2 = 0 + 0.564 456 595 456;
  • 44) 0.564 456 595 456 × 2 = 1 + 0.128 913 190 912;
  • 45) 0.128 913 190 912 × 2 = 0 + 0.257 826 381 824;
  • 46) 0.257 826 381 824 × 2 = 0 + 0.515 652 763 648;
  • 47) 0.515 652 763 648 × 2 = 1 + 0.031 305 527 296;
  • 48) 0.031 305 527 296 × 2 = 0 + 0.062 611 054 592;
  • 49) 0.062 611 054 592 × 2 = 0 + 0.125 222 109 184;
  • 50) 0.125 222 109 184 × 2 = 0 + 0.250 444 218 368;
  • 51) 0.250 444 218 368 × 2 = 0 + 0.500 888 436 736;
  • 52) 0.500 888 436 736 × 2 = 1 + 0.001 776 873 472;
  • 53) 0.001 776 873 472 × 2 = 0 + 0.003 553 746 944;
  • 54) 0.003 553 746 944 × 2 = 0 + 0.007 107 493 888;
  • 55) 0.007 107 493 888 × 2 = 0 + 0.014 214 987 776;
  • 56) 0.014 214 987 776 × 2 = 0 + 0.028 429 975 552;
  • 57) 0.028 429 975 552 × 2 = 0 + 0.056 859 951 104;
  • 58) 0.056 859 951 104 × 2 = 0 + 0.113 719 902 208;
  • 59) 0.113 719 902 208 × 2 = 0 + 0.227 439 804 416;
  • 60) 0.227 439 804 416 × 2 = 0 + 0.454 879 608 832;
  • 61) 0.454 879 608 832 × 2 = 0 + 0.909 759 217 664;
  • 62) 0.909 759 217 664 × 2 = 1 + 0.819 518 435 328;
  • 63) 0.819 518 435 328 × 2 = 1 + 0.639 036 870 656;
  • 64) 0.639 036 870 656 × 2 = 1 + 0.278 073 741 312;
  • 65) 0.278 073 741 312 × 2 = 0 + 0.556 147 482 624;
  • 66) 0.556 147 482 624 × 2 = 1 + 0.112 294 965 248;
  • 67) 0.112 294 965 248 × 2 = 0 + 0.224 589 930 496;
  • 68) 0.224 589 930 496 × 2 = 0 + 0.449 179 860 992;
  • 69) 0.449 179 860 992 × 2 = 0 + 0.898 359 721 984;

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:


0.000 007 820 632(10) =

0.0000 0000 0000 0000 1000 0011 0011 0101 0101 1011 1101 0010 0001 0000 0000 0111 0100 0(2)

6. Positive number before normalization:

0.000 007 820 632(10) =

0.0000 0000 0000 0000 1000 0011 0011 0101 0101 1011 1101 0010 0001 0000 0000 0111 0100 0(2)

7. Normalize the binary representation of the number.

Shift the decimal mark 17 positions to the right, so that only one non zero digit remains to the left of it:


0.000 007 820 632(10) =

0.0000 0000 0000 0000 1000 0011 0011 0101 0101 1011 1101 0010 0001 0000 0000 0111 0100 0(2) =

0.0000 0000 0000 0000 1000 0011 0011 0101 0101 1011 1101 0010 0001 0000 0000 0111 0100 0(2) × 20 =

1.0000 0110 0110 1010 1011 0111 1010 0100 0010 0000 0000 1110 1000(2) × 2-17


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): -17

Mantissa (not normalized):
1.0000 0110 0110 1010 1011 0111 1010 0100 0010 0000 0000 1110 1000


9. Adjust the exponent.

Use the 11 bit excess/bias notation:


Exponent (adjusted) =

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

-17 + 2(11-1) - 1 =

(-17 + 1 023)(10) =

1 006(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;
  • 1 006 ÷ 2 = 503 + 0;
  • 503 ÷ 2 = 251 + 1;
  • 251 ÷ 2 = 125 + 1;
  • 125 ÷ 2 = 62 + 1;
  • 62 ÷ 2 = 31 + 0;
  • 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) =

1006(10) =

011 1110 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. 0000 0110 0110 1010 1011 0111 1010 0100 0010 0000 0000 1110 1000 =

0000 0110 0110 1010 1011 0111 1010 0100 0010 0000 0000 1110 1000


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 1110 1110

Mantissa (52 bits) =
0000 0110 0110 1010 1011 0111 1010 0100 0010 0000 0000 1110 1000


Decimal number -0.000 007 820 632 converted to 64 bit double precision IEEE 754 binary floating point representation:

1 - 011 1110 1110 - 0000 0110 0110 1010 1011 0111 1010 0100 0010 0000 0000 1110 1000

Share

» Convert decimal -0.000 007 820 538 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 09, 2026 [September]: Decimal Numbers Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard. Calculations performed during the month of: September - by our visitors

» Improve your social skills: Reputation: Bridge Between Company's Actions & Market Value. Gemini Notebook YouTube Video of 6.55 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