1.761 828 659 535 915 416 672 685 000 071 263 416 998 590 400 908 142 328 267 63 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 1.761 828 659 535 915 416 672 685 000 071 263 416 998 590 400 908 142 328 267 63(10) to 32 bit single precision IEEE 754 binary floating point representation standard (1 bit for sign, 8 bits for exponent, 23 bits for mantissa)

What are the steps to convert decimal number
1.761 828 659 535 915 416 672 685 000 071 263 416 998 590 400 908 142 328 267 63(10) to 32 bit single precision IEEE 754 binary floating point representation (1 bit for sign, 8 bits for exponent, 23 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: 0.761 828 659 535 915 416 672 685 000 071 263 416 998 590 400 908 142 328 267 63.

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) 0.761 828 659 535 915 416 672 685 000 071 263 416 998 590 400 908 142 328 267 63 × 2 = 1 + 0.523 657 319 071 830 833 345 370 000 142 526 833 997 180 801 816 284 656 535 26;
  • 2) 0.523 657 319 071 830 833 345 370 000 142 526 833 997 180 801 816 284 656 535 26 × 2 = 1 + 0.047 314 638 143 661 666 690 740 000 285 053 667 994 361 603 632 569 313 070 52;
  • 3) 0.047 314 638 143 661 666 690 740 000 285 053 667 994 361 603 632 569 313 070 52 × 2 = 0 + 0.094 629 276 287 323 333 381 480 000 570 107 335 988 723 207 265 138 626 141 04;
  • 4) 0.094 629 276 287 323 333 381 480 000 570 107 335 988 723 207 265 138 626 141 04 × 2 = 0 + 0.189 258 552 574 646 666 762 960 001 140 214 671 977 446 414 530 277 252 282 08;
  • 5) 0.189 258 552 574 646 666 762 960 001 140 214 671 977 446 414 530 277 252 282 08 × 2 = 0 + 0.378 517 105 149 293 333 525 920 002 280 429 343 954 892 829 060 554 504 564 16;
  • 6) 0.378 517 105 149 293 333 525 920 002 280 429 343 954 892 829 060 554 504 564 16 × 2 = 0 + 0.757 034 210 298 586 667 051 840 004 560 858 687 909 785 658 121 109 009 128 32;
  • 7) 0.757 034 210 298 586 667 051 840 004 560 858 687 909 785 658 121 109 009 128 32 × 2 = 1 + 0.514 068 420 597 173 334 103 680 009 121 717 375 819 571 316 242 218 018 256 64;
  • 8) 0.514 068 420 597 173 334 103 680 009 121 717 375 819 571 316 242 218 018 256 64 × 2 = 1 + 0.028 136 841 194 346 668 207 360 018 243 434 751 639 142 632 484 436 036 513 28;
  • 9) 0.028 136 841 194 346 668 207 360 018 243 434 751 639 142 632 484 436 036 513 28 × 2 = 0 + 0.056 273 682 388 693 336 414 720 036 486 869 503 278 285 264 968 872 073 026 56;
  • 10) 0.056 273 682 388 693 336 414 720 036 486 869 503 278 285 264 968 872 073 026 56 × 2 = 0 + 0.112 547 364 777 386 672 829 440 072 973 739 006 556 570 529 937 744 146 053 12;
  • 11) 0.112 547 364 777 386 672 829 440 072 973 739 006 556 570 529 937 744 146 053 12 × 2 = 0 + 0.225 094 729 554 773 345 658 880 145 947 478 013 113 141 059 875 488 292 106 24;
  • 12) 0.225 094 729 554 773 345 658 880 145 947 478 013 113 141 059 875 488 292 106 24 × 2 = 0 + 0.450 189 459 109 546 691 317 760 291 894 956 026 226 282 119 750 976 584 212 48;
  • 13) 0.450 189 459 109 546 691 317 760 291 894 956 026 226 282 119 750 976 584 212 48 × 2 = 0 + 0.900 378 918 219 093 382 635 520 583 789 912 052 452 564 239 501 953 168 424 96;
  • 14) 0.900 378 918 219 093 382 635 520 583 789 912 052 452 564 239 501 953 168 424 96 × 2 = 1 + 0.800 757 836 438 186 765 271 041 167 579 824 104 905 128 479 003 906 336 849 92;
  • 15) 0.800 757 836 438 186 765 271 041 167 579 824 104 905 128 479 003 906 336 849 92 × 2 = 1 + 0.601 515 672 876 373 530 542 082 335 159 648 209 810 256 958 007 812 673 699 84;
  • 16) 0.601 515 672 876 373 530 542 082 335 159 648 209 810 256 958 007 812 673 699 84 × 2 = 1 + 0.203 031 345 752 747 061 084 164 670 319 296 419 620 513 916 015 625 347 399 68;
  • 17) 0.203 031 345 752 747 061 084 164 670 319 296 419 620 513 916 015 625 347 399 68 × 2 = 0 + 0.406 062 691 505 494 122 168 329 340 638 592 839 241 027 832 031 250 694 799 36;
  • 18) 0.406 062 691 505 494 122 168 329 340 638 592 839 241 027 832 031 250 694 799 36 × 2 = 0 + 0.812 125 383 010 988 244 336 658 681 277 185 678 482 055 664 062 501 389 598 72;
  • 19) 0.812 125 383 010 988 244 336 658 681 277 185 678 482 055 664 062 501 389 598 72 × 2 = 1 + 0.624 250 766 021 976 488 673 317 362 554 371 356 964 111 328 125 002 779 197 44;
  • 20) 0.624 250 766 021 976 488 673 317 362 554 371 356 964 111 328 125 002 779 197 44 × 2 = 1 + 0.248 501 532 043 952 977 346 634 725 108 742 713 928 222 656 250 005 558 394 88;
  • 21) 0.248 501 532 043 952 977 346 634 725 108 742 713 928 222 656 250 005 558 394 88 × 2 = 0 + 0.497 003 064 087 905 954 693 269 450 217 485 427 856 445 312 500 011 116 789 76;
  • 22) 0.497 003 064 087 905 954 693 269 450 217 485 427 856 445 312 500 011 116 789 76 × 2 = 0 + 0.994 006 128 175 811 909 386 538 900 434 970 855 712 890 625 000 022 233 579 52;
  • 23) 0.994 006 128 175 811 909 386 538 900 434 970 855 712 890 625 000 022 233 579 52 × 2 = 1 + 0.988 012 256 351 623 818 773 077 800 869 941 711 425 781 250 000 044 467 159 04;
  • 24) 0.988 012 256 351 623 818 773 077 800 869 941 711 425 781 250 000 044 467 159 04 × 2 = 1 + 0.976 024 512 703 247 637 546 155 601 739 883 422 851 562 500 000 088 934 318 08;

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:


0.761 828 659 535 915 416 672 685 000 071 263 416 998 590 400 908 142 328 267 63(10) =

0.1100 0011 0000 0111 0011 0011(2)

5. Positive number before normalization:

1.761 828 659 535 915 416 672 685 000 071 263 416 998 590 400 908 142 328 267 63(10) =

1.1100 0011 0000 0111 0011 0011(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:


1.761 828 659 535 915 416 672 685 000 071 263 416 998 590 400 908 142 328 267 63(10) =

1.1100 0011 0000 0111 0011 0011(2) =

1.1100 0011 0000 0111 0011 0011(2) × 20


7. Up to this moment, there are the following elements that would feed into the 32 bit single precision IEEE 754 binary floating point representation:

Sign 0 (a positive number)

Exponent (unadjusted): 0

Mantissa (not normalized):
1.1100 0011 0000 0111 0011 0011


8. Adjust the exponent.

Use the 8 bit excess/bias notation:


Exponent (adjusted) =

Exponent (unadjusted) + 2(8-1) - 1 =

0 + 2(8-1) - 1 =

(0 + 127)(10) =

127(10)


9. Convert the adjusted exponent from the decimal (base 10) to 8 bit binary.

Use the same technique of repeatedly dividing by 2:


  • division = quotient + remainder;
  • 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) =

127(10) =

0111 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 23 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. 110 0001 1000 0011 1001 1001 1 =

110 0001 1000 0011 1001 1001


12. The three elements that make up the number's 32 bit single precision IEEE 754 binary floating point representation:

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

Exponent (8 bits) =
0111 1111

Mantissa (23 bits) =
110 0001 1000 0011 1001 1001


Decimal number 1.761 828 659 535 915 416 672 685 000 071 263 416 998 590 400 908 142 328 267 63 converted to 32 bit single precision IEEE 754 binary floating point representation:

0 - 0111 1111 - 110 0001 1000 0011 1001 1001

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How to convert decimal numbers from base ten to 32 bit single precision IEEE 754 binary floating point standard

Follow the steps below to convert a base 10 decimal number to 32 bit single 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 base ten 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 of the previous dividing 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 previous multiplying operations, starting from the top of the constructed list 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, by shifting the decimal point (or if you prefer, the decimal mark) "n" positions either to the left or to the right, so that only one non zero digit remains to the left of the decimal point.
  • 7. Adjust the exponent in 8 bit excess/bias notation and then convert it from decimal (base 10) to 8 bit binary, by using the same technique of repeatedly dividing by 2, as shown above:
    Exponent (adjusted) = Exponent (unadjusted) + 2(8-1) - 1
  • 8. Normalize mantissa, remove the leading (leftmost) bit, since it's allways '1' (and the decimal sign if the case) and adjust its length to 23 bits, either by removing the excess bits from the right (losing precision...) or by adding extra '0' bits 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 -25.347 from decimal system (base ten) to 32 bit single precision IEEE 754 binary floating point:

  • 1. Start with the positive version of the number:

    |-25.347| = 25.347

  • 2. First convert the integer part, 25. Divide it repeatedly by 2, keeping track of each remainder, until we get a quotient that is equal to zero:
    • division = quotient + remainder;
    • 25 ÷ 2 = 12 + 1;
    • 12 ÷ 2 = 6 + 0;
    • 6 ÷ 2 = 3 + 0;
    • 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:

    25(10) = 1 1001(2)

  • 4. Then convert the fractional part, 0.347. 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) 0.347 × 2 = 0 + 0.694;
    • 2) 0.694 × 2 = 1 + 0.388;
    • 3) 0.388 × 2 = 0 + 0.776;
    • 4) 0.776 × 2 = 1 + 0.552;
    • 5) 0.552 × 2 = 1 + 0.104;
    • 6) 0.104 × 2 = 0 + 0.208;
    • 7) 0.208 × 2 = 0 + 0.416;
    • 8) 0.416 × 2 = 0 + 0.832;
    • 9) 0.832 × 2 = 1 + 0.664;
    • 10) 0.664 × 2 = 1 + 0.328;
    • 11) 0.328 × 2 = 0 + 0.656;
    • 12) 0.656 × 2 = 1 + 0.312;
    • 13) 0.312 × 2 = 0 + 0.624;
    • 14) 0.624 × 2 = 1 + 0.248;
    • 15) 0.248 × 2 = 0 + 0.496;
    • 16) 0.496 × 2 = 0 + 0.992;
    • 17) 0.992 × 2 = 1 + 0.984;
    • 18) 0.984 × 2 = 1 + 0.968;
    • 19) 0.968 × 2 = 1 + 0.936;
    • 20) 0.936 × 2 = 1 + 0.872;
    • 21) 0.872 × 2 = 1 + 0.744;
    • 22) 0.744 × 2 = 1 + 0.488;
    • 23) 0.488 × 2 = 0 + 0.976;
    • 24) 0.976 × 2 = 1 + 0.952;
    • We didn't get any fractional part that was equal to zero. But we had enough iterations (over Mantissa limit = 23) 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:

    0.347(10) = 0.0101 1000 1101 0100 1111 1101(2)

  • 6. Summarizing - the positive number before normalization:

    25.347(10) = 1 1001.0101 1000 1101 0100 1111 1101(2)

  • 7. Normalize the binary representation of the number, shifting the decimal point 4 positions to the left so that only one non-zero digit stays to the left of the decimal point:

    25.347(10) =
    1 1001.0101 1000 1101 0100 1111 1101(2) =
    1 1001.0101 1000 1101 0100 1111 1101(2) × 20 =
    1.1001 0101 1000 1101 0100 1111 1101(2) × 24

  • 8. Up to this moment, there are the following elements that would feed into the 32 bit single precision IEEE 754 binary floating point:

    Sign: 1 (a negative number)

    Exponent (unadjusted): 4

    Mantissa (not-normalized): 1.1001 0101 1000 1101 0100 1111 1101

  • 9. Adjust the exponent in 8 bit excess/bias notation and then convert it from decimal (base 10) to 8 bit binary (base 2), by using the same technique of repeatedly dividing it by 2, as already demonstrated above:

    Exponent (adjusted) = Exponent (unadjusted) + 2(8-1) - 1 = (4 + 127)(10) = 131(10) =
    1000 0011(2)

  • 10. Normalize the mantissa, remove the leading (leftmost) bit, since it's allways '1' (and the decimal point) and adjust its length to 23 bits, by removing the excess bits from the right (losing precision...):

    Mantissa (not-normalized): 1.1001 0101 1000 1101 0100 1111 1101

    Mantissa (normalized): 100 1010 1100 0110 1010 0111

  • Conclusion:

    Sign (1 bit) = 1 (a negative number)

    Exponent (8 bits) = 1000 0011

    Mantissa (23 bits) = 100 1010 1100 0110 1010 0111

  • Number -25.347, converted from the decimal system (base 10) to 32 bit single precision IEEE 754 binary floating point =
    1 - 1000 0011 - 100 1010 1100 0110 1010 0111