Convert Decimal -3.402 725 076 131 482 to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -3.402 725 076 131 482(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
-3.402 725 076 131 482(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:

|-3.402 725 076 131 482| = 3.402 725 076 131 482


2. First, convert to binary (in base 2) the integer part: 3.
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;
  • 3 ÷ 2 = 1 + 1;
  • 1 ÷ 2 = 0 + 1;

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.

3(10) =

11(2)


4. Convert to binary (base 2) the fractional part: 0.402 725 076 131 482.

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) 0.402 725 076 131 482 × 2 = 0 + 0.805 450 152 262 964;
  • 2) 0.805 450 152 262 964 × 2 = 1 + 0.610 900 304 525 928;
  • 3) 0.610 900 304 525 928 × 2 = 1 + 0.221 800 609 051 856;
  • 4) 0.221 800 609 051 856 × 2 = 0 + 0.443 601 218 103 712;
  • 5) 0.443 601 218 103 712 × 2 = 0 + 0.887 202 436 207 424;
  • 6) 0.887 202 436 207 424 × 2 = 1 + 0.774 404 872 414 848;
  • 7) 0.774 404 872 414 848 × 2 = 1 + 0.548 809 744 829 696;
  • 8) 0.548 809 744 829 696 × 2 = 1 + 0.097 619 489 659 392;
  • 9) 0.097 619 489 659 392 × 2 = 0 + 0.195 238 979 318 784;
  • 10) 0.195 238 979 318 784 × 2 = 0 + 0.390 477 958 637 568;
  • 11) 0.390 477 958 637 568 × 2 = 0 + 0.780 955 917 275 136;
  • 12) 0.780 955 917 275 136 × 2 = 1 + 0.561 911 834 550 272;
  • 13) 0.561 911 834 550 272 × 2 = 1 + 0.123 823 669 100 544;
  • 14) 0.123 823 669 100 544 × 2 = 0 + 0.247 647 338 201 088;
  • 15) 0.247 647 338 201 088 × 2 = 0 + 0.495 294 676 402 176;
  • 16) 0.495 294 676 402 176 × 2 = 0 + 0.990 589 352 804 352;
  • 17) 0.990 589 352 804 352 × 2 = 1 + 0.981 178 705 608 704;
  • 18) 0.981 178 705 608 704 × 2 = 1 + 0.962 357 411 217 408;
  • 19) 0.962 357 411 217 408 × 2 = 1 + 0.924 714 822 434 816;
  • 20) 0.924 714 822 434 816 × 2 = 1 + 0.849 429 644 869 632;
  • 21) 0.849 429 644 869 632 × 2 = 1 + 0.698 859 289 739 264;
  • 22) 0.698 859 289 739 264 × 2 = 1 + 0.397 718 579 478 528;
  • 23) 0.397 718 579 478 528 × 2 = 0 + 0.795 437 158 957 056;
  • 24) 0.795 437 158 957 056 × 2 = 1 + 0.590 874 317 914 112;
  • 25) 0.590 874 317 914 112 × 2 = 1 + 0.181 748 635 828 224;
  • 26) 0.181 748 635 828 224 × 2 = 0 + 0.363 497 271 656 448;
  • 27) 0.363 497 271 656 448 × 2 = 0 + 0.726 994 543 312 896;
  • 28) 0.726 994 543 312 896 × 2 = 1 + 0.453 989 086 625 792;
  • 29) 0.453 989 086 625 792 × 2 = 0 + 0.907 978 173 251 584;
  • 30) 0.907 978 173 251 584 × 2 = 1 + 0.815 956 346 503 168;
  • 31) 0.815 956 346 503 168 × 2 = 1 + 0.631 912 693 006 336;
  • 32) 0.631 912 693 006 336 × 2 = 1 + 0.263 825 386 012 672;
  • 33) 0.263 825 386 012 672 × 2 = 0 + 0.527 650 772 025 344;
  • 34) 0.527 650 772 025 344 × 2 = 1 + 0.055 301 544 050 688;
  • 35) 0.055 301 544 050 688 × 2 = 0 + 0.110 603 088 101 376;
  • 36) 0.110 603 088 101 376 × 2 = 0 + 0.221 206 176 202 752;
  • 37) 0.221 206 176 202 752 × 2 = 0 + 0.442 412 352 405 504;
  • 38) 0.442 412 352 405 504 × 2 = 0 + 0.884 824 704 811 008;
  • 39) 0.884 824 704 811 008 × 2 = 1 + 0.769 649 409 622 016;
  • 40) 0.769 649 409 622 016 × 2 = 1 + 0.539 298 819 244 032;
  • 41) 0.539 298 819 244 032 × 2 = 1 + 0.078 597 638 488 064;
  • 42) 0.078 597 638 488 064 × 2 = 0 + 0.157 195 276 976 128;
  • 43) 0.157 195 276 976 128 × 2 = 0 + 0.314 390 553 952 256;
  • 44) 0.314 390 553 952 256 × 2 = 0 + 0.628 781 107 904 512;
  • 45) 0.628 781 107 904 512 × 2 = 1 + 0.257 562 215 809 024;
  • 46) 0.257 562 215 809 024 × 2 = 0 + 0.515 124 431 618 048;
  • 47) 0.515 124 431 618 048 × 2 = 1 + 0.030 248 863 236 096;
  • 48) 0.030 248 863 236 096 × 2 = 0 + 0.060 497 726 472 192;
  • 49) 0.060 497 726 472 192 × 2 = 0 + 0.120 995 452 944 384;
  • 50) 0.120 995 452 944 384 × 2 = 0 + 0.241 990 905 888 768;
  • 51) 0.241 990 905 888 768 × 2 = 0 + 0.483 981 811 777 536;
  • 52) 0.483 981 811 777 536 × 2 = 0 + 0.967 963 623 555 072;
  • 53) 0.967 963 623 555 072 × 2 = 1 + 0.935 927 247 110 144;

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:


0.402 725 076 131 482(10) =

0.0110 0111 0001 1000 1111 1101 1001 0111 0100 0011 1000 1010 0000 1(2)

6. Positive number before normalization:

3.402 725 076 131 482(10) =

11.0110 0111 0001 1000 1111 1101 1001 0111 0100 0011 1000 1010 0000 1(2)

7. Normalize the binary representation of the number.

Shift the decimal mark 1 positions to the left, so that only one non zero digit remains to the left of it:


3.402 725 076 131 482(10) =

11.0110 0111 0001 1000 1111 1101 1001 0111 0100 0011 1000 1010 0000 1(2) =

11.0110 0111 0001 1000 1111 1101 1001 0111 0100 0011 1000 1010 0000 1(2) × 20 =

1.1011 0011 1000 1100 0111 1110 1100 1011 1010 0001 1100 0101 0000 01(2) × 21


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):
1.1011 0011 1000 1100 0111 1110 1100 1011 1010 0001 1100 0101 0000 01


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) =

1 024(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;
  • 1 024 ÷ 2 = 512 + 0;
  • 512 ÷ 2 = 256 + 0;
  • 256 ÷ 2 = 128 + 0;
  • 128 ÷ 2 = 64 + 0;
  • 64 ÷ 2 = 32 + 0;
  • 32 ÷ 2 = 16 + 0;
  • 16 ÷ 2 = 8 + 0;
  • 8 ÷ 2 = 4 + 0;
  • 4 ÷ 2 = 2 + 0;
  • 2 ÷ 2 = 1 + 0;
  • 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) =

1024(10) =

100 0000 0000(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, by removing the excess bits, from the right (if any of the excess bits is set on 1, we are losing precision...).


Mantissa (normalized) =

1. 1011 0011 1000 1100 0111 1110 1100 1011 1010 0001 1100 0101 0000 01 =

1011 0011 1000 1100 0111 1110 1100 1011 1010 0001 1100 0101 0000


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) =
100 0000 0000

Mantissa (52 bits) =
1011 0011 1000 1100 0111 1110 1100 1011 1010 0001 1100 0101 0000


Decimal number -3.402 725 076 131 482 converted to 64 bit double precision IEEE 754 binary floating point representation:

1 - 100 0000 0000 - 1011 0011 1000 1100 0111 1110 1100 1011 1010 0001 1100 0101 0000

Share

» Convert decimal -3.402 725 076 131 414 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, 2025 [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

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