-0.000 000 000 000 176 426 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 000 176 426(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
-0.000 000 000 000 176 426(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:

|-0.000 000 000 000 176 426| = 0.000 000 000 000 176 426


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: 0.000 000 000 000 176 426.

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) 0.000 000 000 000 176 426 × 2 = 0 + 0.000 000 000 000 352 852;
  • 2) 0.000 000 000 000 352 852 × 2 = 0 + 0.000 000 000 000 705 704;
  • 3) 0.000 000 000 000 705 704 × 2 = 0 + 0.000 000 000 001 411 408;
  • 4) 0.000 000 000 001 411 408 × 2 = 0 + 0.000 000 000 002 822 816;
  • 5) 0.000 000 000 002 822 816 × 2 = 0 + 0.000 000 000 005 645 632;
  • 6) 0.000 000 000 005 645 632 × 2 = 0 + 0.000 000 000 011 291 264;
  • 7) 0.000 000 000 011 291 264 × 2 = 0 + 0.000 000 000 022 582 528;
  • 8) 0.000 000 000 022 582 528 × 2 = 0 + 0.000 000 000 045 165 056;
  • 9) 0.000 000 000 045 165 056 × 2 = 0 + 0.000 000 000 090 330 112;
  • 10) 0.000 000 000 090 330 112 × 2 = 0 + 0.000 000 000 180 660 224;
  • 11) 0.000 000 000 180 660 224 × 2 = 0 + 0.000 000 000 361 320 448;
  • 12) 0.000 000 000 361 320 448 × 2 = 0 + 0.000 000 000 722 640 896;
  • 13) 0.000 000 000 722 640 896 × 2 = 0 + 0.000 000 001 445 281 792;
  • 14) 0.000 000 001 445 281 792 × 2 = 0 + 0.000 000 002 890 563 584;
  • 15) 0.000 000 002 890 563 584 × 2 = 0 + 0.000 000 005 781 127 168;
  • 16) 0.000 000 005 781 127 168 × 2 = 0 + 0.000 000 011 562 254 336;
  • 17) 0.000 000 011 562 254 336 × 2 = 0 + 0.000 000 023 124 508 672;
  • 18) 0.000 000 023 124 508 672 × 2 = 0 + 0.000 000 046 249 017 344;
  • 19) 0.000 000 046 249 017 344 × 2 = 0 + 0.000 000 092 498 034 688;
  • 20) 0.000 000 092 498 034 688 × 2 = 0 + 0.000 000 184 996 069 376;
  • 21) 0.000 000 184 996 069 376 × 2 = 0 + 0.000 000 369 992 138 752;
  • 22) 0.000 000 369 992 138 752 × 2 = 0 + 0.000 000 739 984 277 504;
  • 23) 0.000 000 739 984 277 504 × 2 = 0 + 0.000 001 479 968 555 008;
  • 24) 0.000 001 479 968 555 008 × 2 = 0 + 0.000 002 959 937 110 016;
  • 25) 0.000 002 959 937 110 016 × 2 = 0 + 0.000 005 919 874 220 032;
  • 26) 0.000 005 919 874 220 032 × 2 = 0 + 0.000 011 839 748 440 064;
  • 27) 0.000 011 839 748 440 064 × 2 = 0 + 0.000 023 679 496 880 128;
  • 28) 0.000 023 679 496 880 128 × 2 = 0 + 0.000 047 358 993 760 256;
  • 29) 0.000 047 358 993 760 256 × 2 = 0 + 0.000 094 717 987 520 512;
  • 30) 0.000 094 717 987 520 512 × 2 = 0 + 0.000 189 435 975 041 024;
  • 31) 0.000 189 435 975 041 024 × 2 = 0 + 0.000 378 871 950 082 048;
  • 32) 0.000 378 871 950 082 048 × 2 = 0 + 0.000 757 743 900 164 096;
  • 33) 0.000 757 743 900 164 096 × 2 = 0 + 0.001 515 487 800 328 192;
  • 34) 0.001 515 487 800 328 192 × 2 = 0 + 0.003 030 975 600 656 384;
  • 35) 0.003 030 975 600 656 384 × 2 = 0 + 0.006 061 951 201 312 768;
  • 36) 0.006 061 951 201 312 768 × 2 = 0 + 0.012 123 902 402 625 536;
  • 37) 0.012 123 902 402 625 536 × 2 = 0 + 0.024 247 804 805 251 072;
  • 38) 0.024 247 804 805 251 072 × 2 = 0 + 0.048 495 609 610 502 144;
  • 39) 0.048 495 609 610 502 144 × 2 = 0 + 0.096 991 219 221 004 288;
  • 40) 0.096 991 219 221 004 288 × 2 = 0 + 0.193 982 438 442 008 576;
  • 41) 0.193 982 438 442 008 576 × 2 = 0 + 0.387 964 876 884 017 152;
  • 42) 0.387 964 876 884 017 152 × 2 = 0 + 0.775 929 753 768 034 304;
  • 43) 0.775 929 753 768 034 304 × 2 = 1 + 0.551 859 507 536 068 608;
  • 44) 0.551 859 507 536 068 608 × 2 = 1 + 0.103 719 015 072 137 216;
  • 45) 0.103 719 015 072 137 216 × 2 = 0 + 0.207 438 030 144 274 432;
  • 46) 0.207 438 030 144 274 432 × 2 = 0 + 0.414 876 060 288 548 864;
  • 47) 0.414 876 060 288 548 864 × 2 = 0 + 0.829 752 120 577 097 728;
  • 48) 0.829 752 120 577 097 728 × 2 = 1 + 0.659 504 241 154 195 456;
  • 49) 0.659 504 241 154 195 456 × 2 = 1 + 0.319 008 482 308 390 912;
  • 50) 0.319 008 482 308 390 912 × 2 = 0 + 0.638 016 964 616 781 824;
  • 51) 0.638 016 964 616 781 824 × 2 = 1 + 0.276 033 929 233 563 648;
  • 52) 0.276 033 929 233 563 648 × 2 = 0 + 0.552 067 858 467 127 296;
  • 53) 0.552 067 858 467 127 296 × 2 = 1 + 0.104 135 716 934 254 592;
  • 54) 0.104 135 716 934 254 592 × 2 = 0 + 0.208 271 433 868 509 184;
  • 55) 0.208 271 433 868 509 184 × 2 = 0 + 0.416 542 867 737 018 368;
  • 56) 0.416 542 867 737 018 368 × 2 = 0 + 0.833 085 735 474 036 736;
  • 57) 0.833 085 735 474 036 736 × 2 = 1 + 0.666 171 470 948 073 472;
  • 58) 0.666 171 470 948 073 472 × 2 = 1 + 0.332 342 941 896 146 944;
  • 59) 0.332 342 941 896 146 944 × 2 = 0 + 0.664 685 883 792 293 888;
  • 60) 0.664 685 883 792 293 888 × 2 = 1 + 0.329 371 767 584 587 776;
  • 61) 0.329 371 767 584 587 776 × 2 = 0 + 0.658 743 535 169 175 552;
  • 62) 0.658 743 535 169 175 552 × 2 = 1 + 0.317 487 070 338 351 104;
  • 63) 0.317 487 070 338 351 104 × 2 = 0 + 0.634 974 140 676 702 208;
  • 64) 0.634 974 140 676 702 208 × 2 = 1 + 0.269 948 281 353 404 416;
  • 65) 0.269 948 281 353 404 416 × 2 = 0 + 0.539 896 562 706 808 832;
  • 66) 0.539 896 562 706 808 832 × 2 = 1 + 0.079 793 125 413 617 664;
  • 67) 0.079 793 125 413 617 664 × 2 = 0 + 0.159 586 250 827 235 328;
  • 68) 0.159 586 250 827 235 328 × 2 = 0 + 0.319 172 501 654 470 656;
  • 69) 0.319 172 501 654 470 656 × 2 = 0 + 0.638 345 003 308 941 312;
  • 70) 0.638 345 003 308 941 312 × 2 = 1 + 0.276 690 006 617 882 624;
  • 71) 0.276 690 006 617 882 624 × 2 = 0 + 0.553 380 013 235 765 248;
  • 72) 0.553 380 013 235 765 248 × 2 = 1 + 0.106 760 026 471 530 496;
  • 73) 0.106 760 026 471 530 496 × 2 = 0 + 0.213 520 052 943 060 992;
  • 74) 0.213 520 052 943 060 992 × 2 = 0 + 0.427 040 105 886 121 984;
  • 75) 0.427 040 105 886 121 984 × 2 = 0 + 0.854 080 211 772 243 968;
  • 76) 0.854 080 211 772 243 968 × 2 = 1 + 0.708 160 423 544 487 936;
  • 77) 0.708 160 423 544 487 936 × 2 = 1 + 0.416 320 847 088 975 872;
  • 78) 0.416 320 847 088 975 872 × 2 = 0 + 0.832 641 694 177 951 744;
  • 79) 0.832 641 694 177 951 744 × 2 = 1 + 0.665 283 388 355 903 488;
  • 80) 0.665 283 388 355 903 488 × 2 = 1 + 0.330 566 776 711 806 976;
  • 81) 0.330 566 776 711 806 976 × 2 = 0 + 0.661 133 553 423 613 952;
  • 82) 0.661 133 553 423 613 952 × 2 = 1 + 0.322 267 106 847 227 904;
  • 83) 0.322 267 106 847 227 904 × 2 = 0 + 0.644 534 213 694 455 808;
  • 84) 0.644 534 213 694 455 808 × 2 = 1 + 0.289 068 427 388 911 616;
  • 85) 0.289 068 427 388 911 616 × 2 = 0 + 0.578 136 854 777 823 232;
  • 86) 0.578 136 854 777 823 232 × 2 = 1 + 0.156 273 709 555 646 464;
  • 87) 0.156 273 709 555 646 464 × 2 = 0 + 0.312 547 419 111 292 928;
  • 88) 0.312 547 419 111 292 928 × 2 = 0 + 0.625 094 838 222 585 856;
  • 89) 0.625 094 838 222 585 856 × 2 = 1 + 0.250 189 676 445 171 712;
  • 90) 0.250 189 676 445 171 712 × 2 = 0 + 0.500 379 352 890 343 424;
  • 91) 0.500 379 352 890 343 424 × 2 = 1 + 0.000 758 705 780 686 848;
  • 92) 0.000 758 705 780 686 848 × 2 = 0 + 0.001 517 411 561 373 696;
  • 93) 0.001 517 411 561 373 696 × 2 = 0 + 0.003 034 823 122 747 392;
  • 94) 0.003 034 823 122 747 392 × 2 = 0 + 0.006 069 646 245 494 784;
  • 95) 0.006 069 646 245 494 784 × 2 = 0 + 0.012 139 292 490 989 568;

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:


0.000 000 000 000 176 426(10) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0011 0001 1010 1000 1101 0101 0100 0101 0001 1011 0101 0100 1010 000(2)

6. Positive number before normalization:

0.000 000 000 000 176 426(10) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0011 0001 1010 1000 1101 0101 0100 0101 0001 1011 0101 0100 1010 000(2)

7. Normalize the binary representation of the number.

Shift the decimal mark 43 positions to the right, so that only one non zero digit remains to the left of it:


0.000 000 000 000 176 426(10) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0011 0001 1010 1000 1101 0101 0100 0101 0001 1011 0101 0100 1010 000(2) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0011 0001 1010 1000 1101 0101 0100 0101 0001 1011 0101 0100 1010 000(2) × 20 =

1.1000 1101 0100 0110 1010 1010 0010 1000 1101 1010 1010 0101 0000(2) × 2-43


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): -43

Mantissa (not normalized):
1.1000 1101 0100 0110 1010 1010 0010 1000 1101 1010 1010 0101 0000


9. Adjust the exponent.

Use the 11 bit excess/bias notation:


Exponent (adjusted) =

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

-43 + 2(11-1) - 1 =

(-43 + 1 023)(10) =

980(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;
  • 980 ÷ 2 = 490 + 0;
  • 490 ÷ 2 = 245 + 0;
  • 245 ÷ 2 = 122 + 1;
  • 122 ÷ 2 = 61 + 0;
  • 61 ÷ 2 = 30 + 1;
  • 30 ÷ 2 = 15 + 0;
  • 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) =

980(10) =

011 1101 0100(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. 1000 1101 0100 0110 1010 1010 0010 1000 1101 1010 1010 0101 0000 =

1000 1101 0100 0110 1010 1010 0010 1000 1101 1010 1010 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) =
011 1101 0100

Mantissa (52 bits) =
1000 1101 0100 0110 1010 1010 0010 1000 1101 1010 1010 0101 0000


Decimal number -0.000 000 000 000 176 426 converted to 64 bit double precision IEEE 754 binary floating point representation:

1 - 011 1101 0100 - 1000 1101 0100 0110 1010 1010 0010 1000 1101 1010 1010 0101 0000

Share

» Convert decimal -0.000 000 000 000 176 334 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 Attention Economy - The Battlefield For Your Focus. NotebookLM YouTube Video of 7.50 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