1.745 459 324 169 999 826 281 659 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 1.745 459 324 169 999 826 281 659(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
1.745 459 324 169 999 826 281 659(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. First, convert to binary (in 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 binary (base 2) the fractional part: 0.745 459 324 169 999 826 281 659.

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) 0.745 459 324 169 999 826 281 659 × 2 = 1 + 0.490 918 648 339 999 652 563 318;
  • 2) 0.490 918 648 339 999 652 563 318 × 2 = 0 + 0.981 837 296 679 999 305 126 636;
  • 3) 0.981 837 296 679 999 305 126 636 × 2 = 1 + 0.963 674 593 359 998 610 253 272;
  • 4) 0.963 674 593 359 998 610 253 272 × 2 = 1 + 0.927 349 186 719 997 220 506 544;
  • 5) 0.927 349 186 719 997 220 506 544 × 2 = 1 + 0.854 698 373 439 994 441 013 088;
  • 6) 0.854 698 373 439 994 441 013 088 × 2 = 1 + 0.709 396 746 879 988 882 026 176;
  • 7) 0.709 396 746 879 988 882 026 176 × 2 = 1 + 0.418 793 493 759 977 764 052 352;
  • 8) 0.418 793 493 759 977 764 052 352 × 2 = 0 + 0.837 586 987 519 955 528 104 704;
  • 9) 0.837 586 987 519 955 528 104 704 × 2 = 1 + 0.675 173 975 039 911 056 209 408;
  • 10) 0.675 173 975 039 911 056 209 408 × 2 = 1 + 0.350 347 950 079 822 112 418 816;
  • 11) 0.350 347 950 079 822 112 418 816 × 2 = 0 + 0.700 695 900 159 644 224 837 632;
  • 12) 0.700 695 900 159 644 224 837 632 × 2 = 1 + 0.401 391 800 319 288 449 675 264;
  • 13) 0.401 391 800 319 288 449 675 264 × 2 = 0 + 0.802 783 600 638 576 899 350 528;
  • 14) 0.802 783 600 638 576 899 350 528 × 2 = 1 + 0.605 567 201 277 153 798 701 056;
  • 15) 0.605 567 201 277 153 798 701 056 × 2 = 1 + 0.211 134 402 554 307 597 402 112;
  • 16) 0.211 134 402 554 307 597 402 112 × 2 = 0 + 0.422 268 805 108 615 194 804 224;
  • 17) 0.422 268 805 108 615 194 804 224 × 2 = 0 + 0.844 537 610 217 230 389 608 448;
  • 18) 0.844 537 610 217 230 389 608 448 × 2 = 1 + 0.689 075 220 434 460 779 216 896;
  • 19) 0.689 075 220 434 460 779 216 896 × 2 = 1 + 0.378 150 440 868 921 558 433 792;
  • 20) 0.378 150 440 868 921 558 433 792 × 2 = 0 + 0.756 300 881 737 843 116 867 584;
  • 21) 0.756 300 881 737 843 116 867 584 × 2 = 1 + 0.512 601 763 475 686 233 735 168;
  • 22) 0.512 601 763 475 686 233 735 168 × 2 = 1 + 0.025 203 526 951 372 467 470 336;
  • 23) 0.025 203 526 951 372 467 470 336 × 2 = 0 + 0.050 407 053 902 744 934 940 672;
  • 24) 0.050 407 053 902 744 934 940 672 × 2 = 0 + 0.100 814 107 805 489 869 881 344;
  • 25) 0.100 814 107 805 489 869 881 344 × 2 = 0 + 0.201 628 215 610 979 739 762 688;
  • 26) 0.201 628 215 610 979 739 762 688 × 2 = 0 + 0.403 256 431 221 959 479 525 376;
  • 27) 0.403 256 431 221 959 479 525 376 × 2 = 0 + 0.806 512 862 443 918 959 050 752;
  • 28) 0.806 512 862 443 918 959 050 752 × 2 = 1 + 0.613 025 724 887 837 918 101 504;
  • 29) 0.613 025 724 887 837 918 101 504 × 2 = 1 + 0.226 051 449 775 675 836 203 008;
  • 30) 0.226 051 449 775 675 836 203 008 × 2 = 0 + 0.452 102 899 551 351 672 406 016;
  • 31) 0.452 102 899 551 351 672 406 016 × 2 = 0 + 0.904 205 799 102 703 344 812 032;
  • 32) 0.904 205 799 102 703 344 812 032 × 2 = 1 + 0.808 411 598 205 406 689 624 064;
  • 33) 0.808 411 598 205 406 689 624 064 × 2 = 1 + 0.616 823 196 410 813 379 248 128;
  • 34) 0.616 823 196 410 813 379 248 128 × 2 = 1 + 0.233 646 392 821 626 758 496 256;
  • 35) 0.233 646 392 821 626 758 496 256 × 2 = 0 + 0.467 292 785 643 253 516 992 512;
  • 36) 0.467 292 785 643 253 516 992 512 × 2 = 0 + 0.934 585 571 286 507 033 985 024;
  • 37) 0.934 585 571 286 507 033 985 024 × 2 = 1 + 0.869 171 142 573 014 067 970 048;
  • 38) 0.869 171 142 573 014 067 970 048 × 2 = 1 + 0.738 342 285 146 028 135 940 096;
  • 39) 0.738 342 285 146 028 135 940 096 × 2 = 1 + 0.476 684 570 292 056 271 880 192;
  • 40) 0.476 684 570 292 056 271 880 192 × 2 = 0 + 0.953 369 140 584 112 543 760 384;
  • 41) 0.953 369 140 584 112 543 760 384 × 2 = 1 + 0.906 738 281 168 225 087 520 768;
  • 42) 0.906 738 281 168 225 087 520 768 × 2 = 1 + 0.813 476 562 336 450 175 041 536;
  • 43) 0.813 476 562 336 450 175 041 536 × 2 = 1 + 0.626 953 124 672 900 350 083 072;
  • 44) 0.626 953 124 672 900 350 083 072 × 2 = 1 + 0.253 906 249 345 800 700 166 144;
  • 45) 0.253 906 249 345 800 700 166 144 × 2 = 0 + 0.507 812 498 691 601 400 332 288;
  • 46) 0.507 812 498 691 601 400 332 288 × 2 = 1 + 0.015 624 997 383 202 800 664 576;
  • 47) 0.015 624 997 383 202 800 664 576 × 2 = 0 + 0.031 249 994 766 405 601 329 152;
  • 48) 0.031 249 994 766 405 601 329 152 × 2 = 0 + 0.062 499 989 532 811 202 658 304;
  • 49) 0.062 499 989 532 811 202 658 304 × 2 = 0 + 0.124 999 979 065 622 405 316 608;
  • 50) 0.124 999 979 065 622 405 316 608 × 2 = 0 + 0.249 999 958 131 244 810 633 216;
  • 51) 0.249 999 958 131 244 810 633 216 × 2 = 0 + 0.499 999 916 262 489 621 266 432;
  • 52) 0.499 999 916 262 489 621 266 432 × 2 = 0 + 0.999 999 832 524 979 242 532 864;
  • 53) 0.999 999 832 524 979 242 532 864 × 2 = 1 + 0.999 999 665 049 958 485 065 728;

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).


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:


0.745 459 324 169 999 826 281 659(10) =

0.1011 1110 1101 0110 0110 1100 0001 1001 1100 1110 1111 0100 0000 1(2)

5. Positive number before normalization:

1.745 459 324 169 999 826 281 659(10) =

1.1011 1110 1101 0110 0110 1100 0001 1001 1100 1110 1111 0100 0000 1(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:


1.745 459 324 169 999 826 281 659(10) =

1.1011 1110 1101 0110 0110 1100 0001 1001 1100 1110 1111 0100 0000 1(2) =

1.1011 1110 1101 0110 0110 1100 0001 1001 1100 1110 1111 0100 0000 1(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):
1.1011 1110 1101 0110 0110 1100 0001 1001 1100 1110 1111 0100 0000 1


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. 1011 1110 1101 0110 0110 1100 0001 1001 1100 1110 1111 0100 0000 1 =

1011 1110 1101 0110 0110 1100 0001 1001 1100 1110 1111 0100 0000


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) =
1011 1110 1101 0110 0110 1100 0001 1001 1100 1110 1111 0100 0000


Decimal number 1.745 459 324 169 999 826 281 659 converted to 64 bit double precision IEEE 754 binary floating point representation:

0 - 011 1111 1111 - 1011 1110 1101 0110 0110 1100 0001 1001 1100 1110 1111 0100 0000

Share

» Convert decimal 1.745 459 324 169 999 826 281 572 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 07, 2026 [July]: Decimal Numbers Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard. Calculations performed during the month of: July - by our visitors

» Improve your social skills: The Hidden Requirement in Every Single Job Description. NotebookLM YouTube Video of 8.15 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