Convert 1.570 796 326 794 896 619 231 321 691 636 to 64 Bit Double Precision IEEE 754 Binary Floating Point Standard, From a Number in Base 10 Decimal System

1.570 796 326 794 896 619 231 321 691 636(10) to 64 bit double precision IEEE 754 binary floating point (1 bit for sign, 11 bits for exponent, 52 bits for mantissa) = ?

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

2. 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.

1(10) =


1(2)


3. Convert to the binary (base 2) the fractional part: 0.570 796 326 794 896 619 231 321 691 636.

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) 0.570 796 326 794 896 619 231 321 691 636 × 2 = 1 + 0.141 592 653 589 793 238 462 643 383 272;
  • 2) 0.141 592 653 589 793 238 462 643 383 272 × 2 = 0 + 0.283 185 307 179 586 476 925 286 766 544;
  • 3) 0.283 185 307 179 586 476 925 286 766 544 × 2 = 0 + 0.566 370 614 359 172 953 850 573 533 088;
  • 4) 0.566 370 614 359 172 953 850 573 533 088 × 2 = 1 + 0.132 741 228 718 345 907 701 147 066 176;
  • 5) 0.132 741 228 718 345 907 701 147 066 176 × 2 = 0 + 0.265 482 457 436 691 815 402 294 132 352;
  • 6) 0.265 482 457 436 691 815 402 294 132 352 × 2 = 0 + 0.530 964 914 873 383 630 804 588 264 704;
  • 7) 0.530 964 914 873 383 630 804 588 264 704 × 2 = 1 + 0.061 929 829 746 767 261 609 176 529 408;
  • 8) 0.061 929 829 746 767 261 609 176 529 408 × 2 = 0 + 0.123 859 659 493 534 523 218 353 058 816;
  • 9) 0.123 859 659 493 534 523 218 353 058 816 × 2 = 0 + 0.247 719 318 987 069 046 436 706 117 632;
  • 10) 0.247 719 318 987 069 046 436 706 117 632 × 2 = 0 + 0.495 438 637 974 138 092 873 412 235 264;
  • 11) 0.495 438 637 974 138 092 873 412 235 264 × 2 = 0 + 0.990 877 275 948 276 185 746 824 470 528;
  • 12) 0.990 877 275 948 276 185 746 824 470 528 × 2 = 1 + 0.981 754 551 896 552 371 493 648 941 056;
  • 13) 0.981 754 551 896 552 371 493 648 941 056 × 2 = 1 + 0.963 509 103 793 104 742 987 297 882 112;
  • 14) 0.963 509 103 793 104 742 987 297 882 112 × 2 = 1 + 0.927 018 207 586 209 485 974 595 764 224;
  • 15) 0.927 018 207 586 209 485 974 595 764 224 × 2 = 1 + 0.854 036 415 172 418 971 949 191 528 448;
  • 16) 0.854 036 415 172 418 971 949 191 528 448 × 2 = 1 + 0.708 072 830 344 837 943 898 383 056 896;
  • 17) 0.708 072 830 344 837 943 898 383 056 896 × 2 = 1 + 0.416 145 660 689 675 887 796 766 113 792;
  • 18) 0.416 145 660 689 675 887 796 766 113 792 × 2 = 0 + 0.832 291 321 379 351 775 593 532 227 584;
  • 19) 0.832 291 321 379 351 775 593 532 227 584 × 2 = 1 + 0.664 582 642 758 703 551 187 064 455 168;
  • 20) 0.664 582 642 758 703 551 187 064 455 168 × 2 = 1 + 0.329 165 285 517 407 102 374 128 910 336;
  • 21) 0.329 165 285 517 407 102 374 128 910 336 × 2 = 0 + 0.658 330 571 034 814 204 748 257 820 672;
  • 22) 0.658 330 571 034 814 204 748 257 820 672 × 2 = 1 + 0.316 661 142 069 628 409 496 515 641 344;
  • 23) 0.316 661 142 069 628 409 496 515 641 344 × 2 = 0 + 0.633 322 284 139 256 818 993 031 282 688;
  • 24) 0.633 322 284 139 256 818 993 031 282 688 × 2 = 1 + 0.266 644 568 278 513 637 986 062 565 376;
  • 25) 0.266 644 568 278 513 637 986 062 565 376 × 2 = 0 + 0.533 289 136 557 027 275 972 125 130 752;
  • 26) 0.533 289 136 557 027 275 972 125 130 752 × 2 = 1 + 0.066 578 273 114 054 551 944 250 261 504;
  • 27) 0.066 578 273 114 054 551 944 250 261 504 × 2 = 0 + 0.133 156 546 228 109 103 888 500 523 008;
  • 28) 0.133 156 546 228 109 103 888 500 523 008 × 2 = 0 + 0.266 313 092 456 218 207 777 001 046 016;
  • 29) 0.266 313 092 456 218 207 777 001 046 016 × 2 = 0 + 0.532 626 184 912 436 415 554 002 092 032;
  • 30) 0.532 626 184 912 436 415 554 002 092 032 × 2 = 1 + 0.065 252 369 824 872 831 108 004 184 064;
  • 31) 0.065 252 369 824 872 831 108 004 184 064 × 2 = 0 + 0.130 504 739 649 745 662 216 008 368 128;
  • 32) 0.130 504 739 649 745 662 216 008 368 128 × 2 = 0 + 0.261 009 479 299 491 324 432 016 736 256;
  • 33) 0.261 009 479 299 491 324 432 016 736 256 × 2 = 0 + 0.522 018 958 598 982 648 864 033 472 512;
  • 34) 0.522 018 958 598 982 648 864 033 472 512 × 2 = 1 + 0.044 037 917 197 965 297 728 066 945 024;
  • 35) 0.044 037 917 197 965 297 728 066 945 024 × 2 = 0 + 0.088 075 834 395 930 595 456 133 890 048;
  • 36) 0.088 075 834 395 930 595 456 133 890 048 × 2 = 0 + 0.176 151 668 791 861 190 912 267 780 096;
  • 37) 0.176 151 668 791 861 190 912 267 780 096 × 2 = 0 + 0.352 303 337 583 722 381 824 535 560 192;
  • 38) 0.352 303 337 583 722 381 824 535 560 192 × 2 = 0 + 0.704 606 675 167 444 763 649 071 120 384;
  • 39) 0.704 606 675 167 444 763 649 071 120 384 × 2 = 1 + 0.409 213 350 334 889 527 298 142 240 768;
  • 40) 0.409 213 350 334 889 527 298 142 240 768 × 2 = 0 + 0.818 426 700 669 779 054 596 284 481 536;
  • 41) 0.818 426 700 669 779 054 596 284 481 536 × 2 = 1 + 0.636 853 401 339 558 109 192 568 963 072;
  • 42) 0.636 853 401 339 558 109 192 568 963 072 × 2 = 1 + 0.273 706 802 679 116 218 385 137 926 144;
  • 43) 0.273 706 802 679 116 218 385 137 926 144 × 2 = 0 + 0.547 413 605 358 232 436 770 275 852 288;
  • 44) 0.547 413 605 358 232 436 770 275 852 288 × 2 = 1 + 0.094 827 210 716 464 873 540 551 704 576;
  • 45) 0.094 827 210 716 464 873 540 551 704 576 × 2 = 0 + 0.189 654 421 432 929 747 081 103 409 152;
  • 46) 0.189 654 421 432 929 747 081 103 409 152 × 2 = 0 + 0.379 308 842 865 859 494 162 206 818 304;
  • 47) 0.379 308 842 865 859 494 162 206 818 304 × 2 = 0 + 0.758 617 685 731 718 988 324 413 636 608;
  • 48) 0.758 617 685 731 718 988 324 413 636 608 × 2 = 1 + 0.517 235 371 463 437 976 648 827 273 216;
  • 49) 0.517 235 371 463 437 976 648 827 273 216 × 2 = 1 + 0.034 470 742 926 875 953 297 654 546 432;
  • 50) 0.034 470 742 926 875 953 297 654 546 432 × 2 = 0 + 0.068 941 485 853 751 906 595 309 092 864;
  • 51) 0.068 941 485 853 751 906 595 309 092 864 × 2 = 0 + 0.137 882 971 707 503 813 190 618 185 728;
  • 52) 0.137 882 971 707 503 813 190 618 185 728 × 2 = 0 + 0.275 765 943 415 007 626 381 236 371 456;
  • 53) 0.275 765 943 415 007 626 381 236 371 456 × 2 = 0 + 0.551 531 886 830 015 252 762 472 742 912;

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...)


4. 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:

0.570 796 326 794 896 619 231 321 691 636(10) =


0.1001 0010 0001 1111 1011 0101 0100 0100 0100 0010 1101 0001 1000 0(2)


5. Positive number before normalization:

1.570 796 326 794 896 619 231 321 691 636(10) =


1.1001 0010 0001 1111 1011 0101 0100 0100 0100 0010 1101 0001 1000 0(2)


6. Normalize the binary representation of the number.

Shift the decimal mark 0 positions to the left so that only one non zero digit remains to the left of it:

1.570 796 326 794 896 619 231 321 691 636(10) =


1.1001 0010 0001 1111 1011 0101 0100 0100 0100 0010 1101 0001 1000 0(2) =


1.1001 0010 0001 1111 1011 0101 0100 0100 0100 0010 1101 0001 1000 0(2) × 20


7. 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: 0 (a positive number)


Exponent (unadjusted): 0


Mantissa (not normalized):
1.1001 0010 0001 1111 1011 0101 0100 0100 0100 0010 1101 0001 1000 0


8. Adjust the exponent.

Use the 11 bit excess/bias notation:

Exponent (adjusted) =


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


0 + 2(11-1) - 1 =


(0 + 1 023)(10) =


1 023(10)


9. 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;
  • 1 023 ÷ 2 = 511 + 1;
  • 511 ÷ 2 = 255 + 1;
  • 255 ÷ 2 = 127 + 1;
  • 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;

10. Construct the base 2 representation of the adjusted exponent.

Take all the remainders starting from the bottom of the list constructed above:

Exponent (adjusted) =


1023(10) =


011 1111 1111(2)


11. 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. 1001 0010 0001 1111 1011 0101 0100 0100 0100 0010 1101 0001 1000 0 =


1001 0010 0001 1111 1011 0101 0100 0100 0100 0010 1101 0001 1000


12. The three elements that make up the number's 64 bit double precision IEEE 754 binary floating point representation:

Sign (1 bit) =
0 (a positive number)


Exponent (11 bits) =
011 1111 1111


Mantissa (52 bits) =
1001 0010 0001 1111 1011 0101 0100 0100 0100 0010 1101 0001 1000


Number 1.570 796 326 794 896 619 231 321 691 636 converted from decimal system (base 10) to 64 bit double precision IEEE 754 binary floating point:
0 - 011 1111 1111 - 1001 0010 0001 1111 1011 0101 0100 0100 0100 0010 1101 0001 1000

(64 bits IEEE 754)
  • Sign (1 bit):

    • 0

      63
  • Exponent (11 bits):

    • 0

      62
    • 1

      61
    • 1

      60
    • 1

      59
    • 1

      58
    • 1

      57
    • 1

      56
    • 1

      55
    • 1

      54
    • 1

      53
    • 1

      52
  • Mantissa (52 bits):

    • 1

      51
    • 0

      50
    • 0

      49
    • 1

      48
    • 0

      47
    • 0

      46
    • 1

      45
    • 0

      44
    • 0

      43
    • 0

      42
    • 0

      41
    • 1

      40
    • 1

      39
    • 1

      38
    • 1

      37
    • 1

      36
    • 1

      35
    • 0

      34
    • 1

      33
    • 1

      32
    • 0

      31
    • 1

      30
    • 0

      29
    • 1

      28
    • 0

      27
    • 1

      26
    • 0

      25
    • 0

      24
    • 0

      23
    • 1

      22
    • 0

      21
    • 0

      20
    • 0

      19
    • 1

      18
    • 0

      17
    • 0

      16
    • 0

      15
    • 0

      14
    • 1

      13
    • 0

      12
    • 1

      11
    • 1

      10
    • 0

      9
    • 1

      8
    • 0

      7
    • 0

      6
    • 0

      5
    • 1

      4
    • 1

      3
    • 0

      2
    • 0

      1
    • 0

      0

More operations of this kind:

1.570 796 326 794 896 619 231 321 691 635 = ? ... 1.570 796 326 794 896 619 231 321 691 637 = ?


Convert to 64 bit double precision IEEE 754 binary floating point standard

A number in 64 bit double precision IEEE 754 binary floating point standard representation requires three building elements: sign (it takes one bit and it's either 0 for positive or 1 for negative numbers), exponent (11 bits), mantissa (52 bits)

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All base ten decimal numbers converted to 64 bit double precision IEEE 754 binary floating point

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