1.745 459 324 169 999 826 281 608 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 1.745 459 324 169 999 826 281 608(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
1.745 459 324 169 999 826 281 608(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: 0.745 459 324 169 999 826 281 608.

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) 0.745 459 324 169 999 826 281 608 × 2 = 1 + 0.490 918 648 339 999 652 563 216;
  • 2) 0.490 918 648 339 999 652 563 216 × 2 = 0 + 0.981 837 296 679 999 305 126 432;
  • 3) 0.981 837 296 679 999 305 126 432 × 2 = 1 + 0.963 674 593 359 998 610 252 864;
  • 4) 0.963 674 593 359 998 610 252 864 × 2 = 1 + 0.927 349 186 719 997 220 505 728;
  • 5) 0.927 349 186 719 997 220 505 728 × 2 = 1 + 0.854 698 373 439 994 441 011 456;
  • 6) 0.854 698 373 439 994 441 011 456 × 2 = 1 + 0.709 396 746 879 988 882 022 912;
  • 7) 0.709 396 746 879 988 882 022 912 × 2 = 1 + 0.418 793 493 759 977 764 045 824;
  • 8) 0.418 793 493 759 977 764 045 824 × 2 = 0 + 0.837 586 987 519 955 528 091 648;
  • 9) 0.837 586 987 519 955 528 091 648 × 2 = 1 + 0.675 173 975 039 911 056 183 296;
  • 10) 0.675 173 975 039 911 056 183 296 × 2 = 1 + 0.350 347 950 079 822 112 366 592;
  • 11) 0.350 347 950 079 822 112 366 592 × 2 = 0 + 0.700 695 900 159 644 224 733 184;
  • 12) 0.700 695 900 159 644 224 733 184 × 2 = 1 + 0.401 391 800 319 288 449 466 368;
  • 13) 0.401 391 800 319 288 449 466 368 × 2 = 0 + 0.802 783 600 638 576 898 932 736;
  • 14) 0.802 783 600 638 576 898 932 736 × 2 = 1 + 0.605 567 201 277 153 797 865 472;
  • 15) 0.605 567 201 277 153 797 865 472 × 2 = 1 + 0.211 134 402 554 307 595 730 944;
  • 16) 0.211 134 402 554 307 595 730 944 × 2 = 0 + 0.422 268 805 108 615 191 461 888;
  • 17) 0.422 268 805 108 615 191 461 888 × 2 = 0 + 0.844 537 610 217 230 382 923 776;
  • 18) 0.844 537 610 217 230 382 923 776 × 2 = 1 + 0.689 075 220 434 460 765 847 552;
  • 19) 0.689 075 220 434 460 765 847 552 × 2 = 1 + 0.378 150 440 868 921 531 695 104;
  • 20) 0.378 150 440 868 921 531 695 104 × 2 = 0 + 0.756 300 881 737 843 063 390 208;
  • 21) 0.756 300 881 737 843 063 390 208 × 2 = 1 + 0.512 601 763 475 686 126 780 416;
  • 22) 0.512 601 763 475 686 126 780 416 × 2 = 1 + 0.025 203 526 951 372 253 560 832;
  • 23) 0.025 203 526 951 372 253 560 832 × 2 = 0 + 0.050 407 053 902 744 507 121 664;
  • 24) 0.050 407 053 902 744 507 121 664 × 2 = 0 + 0.100 814 107 805 489 014 243 328;
  • 25) 0.100 814 107 805 489 014 243 328 × 2 = 0 + 0.201 628 215 610 978 028 486 656;
  • 26) 0.201 628 215 610 978 028 486 656 × 2 = 0 + 0.403 256 431 221 956 056 973 312;
  • 27) 0.403 256 431 221 956 056 973 312 × 2 = 0 + 0.806 512 862 443 912 113 946 624;
  • 28) 0.806 512 862 443 912 113 946 624 × 2 = 1 + 0.613 025 724 887 824 227 893 248;
  • 29) 0.613 025 724 887 824 227 893 248 × 2 = 1 + 0.226 051 449 775 648 455 786 496;
  • 30) 0.226 051 449 775 648 455 786 496 × 2 = 0 + 0.452 102 899 551 296 911 572 992;
  • 31) 0.452 102 899 551 296 911 572 992 × 2 = 0 + 0.904 205 799 102 593 823 145 984;
  • 32) 0.904 205 799 102 593 823 145 984 × 2 = 1 + 0.808 411 598 205 187 646 291 968;
  • 33) 0.808 411 598 205 187 646 291 968 × 2 = 1 + 0.616 823 196 410 375 292 583 936;
  • 34) 0.616 823 196 410 375 292 583 936 × 2 = 1 + 0.233 646 392 820 750 585 167 872;
  • 35) 0.233 646 392 820 750 585 167 872 × 2 = 0 + 0.467 292 785 641 501 170 335 744;
  • 36) 0.467 292 785 641 501 170 335 744 × 2 = 0 + 0.934 585 571 283 002 340 671 488;
  • 37) 0.934 585 571 283 002 340 671 488 × 2 = 1 + 0.869 171 142 566 004 681 342 976;
  • 38) 0.869 171 142 566 004 681 342 976 × 2 = 1 + 0.738 342 285 132 009 362 685 952;
  • 39) 0.738 342 285 132 009 362 685 952 × 2 = 1 + 0.476 684 570 264 018 725 371 904;
  • 40) 0.476 684 570 264 018 725 371 904 × 2 = 0 + 0.953 369 140 528 037 450 743 808;
  • 41) 0.953 369 140 528 037 450 743 808 × 2 = 1 + 0.906 738 281 056 074 901 487 616;
  • 42) 0.906 738 281 056 074 901 487 616 × 2 = 1 + 0.813 476 562 112 149 802 975 232;
  • 43) 0.813 476 562 112 149 802 975 232 × 2 = 1 + 0.626 953 124 224 299 605 950 464;
  • 44) 0.626 953 124 224 299 605 950 464 × 2 = 1 + 0.253 906 248 448 599 211 900 928;
  • 45) 0.253 906 248 448 599 211 900 928 × 2 = 0 + 0.507 812 496 897 198 423 801 856;
  • 46) 0.507 812 496 897 198 423 801 856 × 2 = 1 + 0.015 624 993 794 396 847 603 712;
  • 47) 0.015 624 993 794 396 847 603 712 × 2 = 0 + 0.031 249 987 588 793 695 207 424;
  • 48) 0.031 249 987 588 793 695 207 424 × 2 = 0 + 0.062 499 975 177 587 390 414 848;
  • 49) 0.062 499 975 177 587 390 414 848 × 2 = 0 + 0.124 999 950 355 174 780 829 696;
  • 50) 0.124 999 950 355 174 780 829 696 × 2 = 0 + 0.249 999 900 710 349 561 659 392;
  • 51) 0.249 999 900 710 349 561 659 392 × 2 = 0 + 0.499 999 801 420 699 123 318 784;
  • 52) 0.499 999 801 420 699 123 318 784 × 2 = 0 + 0.999 999 602 841 398 246 637 568;
  • 53) 0.999 999 602 841 398 246 637 568 × 2 = 1 + 0.999 999 205 682 796 493 275 136;

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:


0.745 459 324 169 999 826 281 608(10) =

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

5. Positive number before normalization:

1.745 459 324 169 999 826 281 608(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:


1.745 459 324 169 999 826 281 608(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 1.745 459 324 169 999 826 281 608 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 1.745 459 324 169 999 826 281 683 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 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