Convert 101 010 100 001 110 999 999 999 993 to 32 Bit Single Precision IEEE 754 Binary Floating Point Standard, From a Base 10 Decimal Number

101 010 100 001 110 999 999 999 993(10) to 32 bit single precision IEEE 754 binary floating point (1 bit for sign, 8 bits for exponent, 23 bits for mantissa) = ?

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;
  • 101 010 100 001 110 999 999 999 993 ÷ 2 = 50 505 050 000 555 499 999 999 996 + 1;
  • 50 505 050 000 555 499 999 999 996 ÷ 2 = 25 252 525 000 277 749 999 999 998 + 0;
  • 25 252 525 000 277 749 999 999 998 ÷ 2 = 12 626 262 500 138 874 999 999 999 + 0;
  • 12 626 262 500 138 874 999 999 999 ÷ 2 = 6 313 131 250 069 437 499 999 999 + 1;
  • 6 313 131 250 069 437 499 999 999 ÷ 2 = 3 156 565 625 034 718 749 999 999 + 1;
  • 3 156 565 625 034 718 749 999 999 ÷ 2 = 1 578 282 812 517 359 374 999 999 + 1;
  • 1 578 282 812 517 359 374 999 999 ÷ 2 = 789 141 406 258 679 687 499 999 + 1;
  • 789 141 406 258 679 687 499 999 ÷ 2 = 394 570 703 129 339 843 749 999 + 1;
  • 394 570 703 129 339 843 749 999 ÷ 2 = 197 285 351 564 669 921 874 999 + 1;
  • 197 285 351 564 669 921 874 999 ÷ 2 = 98 642 675 782 334 960 937 499 + 1;
  • 98 642 675 782 334 960 937 499 ÷ 2 = 49 321 337 891 167 480 468 749 + 1;
  • 49 321 337 891 167 480 468 749 ÷ 2 = 24 660 668 945 583 740 234 374 + 1;
  • 24 660 668 945 583 740 234 374 ÷ 2 = 12 330 334 472 791 870 117 187 + 0;
  • 12 330 334 472 791 870 117 187 ÷ 2 = 6 165 167 236 395 935 058 593 + 1;
  • 6 165 167 236 395 935 058 593 ÷ 2 = 3 082 583 618 197 967 529 296 + 1;
  • 3 082 583 618 197 967 529 296 ÷ 2 = 1 541 291 809 098 983 764 648 + 0;
  • 1 541 291 809 098 983 764 648 ÷ 2 = 770 645 904 549 491 882 324 + 0;
  • 770 645 904 549 491 882 324 ÷ 2 = 385 322 952 274 745 941 162 + 0;
  • 385 322 952 274 745 941 162 ÷ 2 = 192 661 476 137 372 970 581 + 0;
  • 192 661 476 137 372 970 581 ÷ 2 = 96 330 738 068 686 485 290 + 1;
  • 96 330 738 068 686 485 290 ÷ 2 = 48 165 369 034 343 242 645 + 0;
  • 48 165 369 034 343 242 645 ÷ 2 = 24 082 684 517 171 621 322 + 1;
  • 24 082 684 517 171 621 322 ÷ 2 = 12 041 342 258 585 810 661 + 0;
  • 12 041 342 258 585 810 661 ÷ 2 = 6 020 671 129 292 905 330 + 1;
  • 6 020 671 129 292 905 330 ÷ 2 = 3 010 335 564 646 452 665 + 0;
  • 3 010 335 564 646 452 665 ÷ 2 = 1 505 167 782 323 226 332 + 1;
  • 1 505 167 782 323 226 332 ÷ 2 = 752 583 891 161 613 166 + 0;
  • 752 583 891 161 613 166 ÷ 2 = 376 291 945 580 806 583 + 0;
  • 376 291 945 580 806 583 ÷ 2 = 188 145 972 790 403 291 + 1;
  • 188 145 972 790 403 291 ÷ 2 = 94 072 986 395 201 645 + 1;
  • 94 072 986 395 201 645 ÷ 2 = 47 036 493 197 600 822 + 1;
  • 47 036 493 197 600 822 ÷ 2 = 23 518 246 598 800 411 + 0;
  • 23 518 246 598 800 411 ÷ 2 = 11 759 123 299 400 205 + 1;
  • 11 759 123 299 400 205 ÷ 2 = 5 879 561 649 700 102 + 1;
  • 5 879 561 649 700 102 ÷ 2 = 2 939 780 824 850 051 + 0;
  • 2 939 780 824 850 051 ÷ 2 = 1 469 890 412 425 025 + 1;
  • 1 469 890 412 425 025 ÷ 2 = 734 945 206 212 512 + 1;
  • 734 945 206 212 512 ÷ 2 = 367 472 603 106 256 + 0;
  • 367 472 603 106 256 ÷ 2 = 183 736 301 553 128 + 0;
  • 183 736 301 553 128 ÷ 2 = 91 868 150 776 564 + 0;
  • 91 868 150 776 564 ÷ 2 = 45 934 075 388 282 + 0;
  • 45 934 075 388 282 ÷ 2 = 22 967 037 694 141 + 0;
  • 22 967 037 694 141 ÷ 2 = 11 483 518 847 070 + 1;
  • 11 483 518 847 070 ÷ 2 = 5 741 759 423 535 + 0;
  • 5 741 759 423 535 ÷ 2 = 2 870 879 711 767 + 1;
  • 2 870 879 711 767 ÷ 2 = 1 435 439 855 883 + 1;
  • 1 435 439 855 883 ÷ 2 = 717 719 927 941 + 1;
  • 717 719 927 941 ÷ 2 = 358 859 963 970 + 1;
  • 358 859 963 970 ÷ 2 = 179 429 981 985 + 0;
  • 179 429 981 985 ÷ 2 = 89 714 990 992 + 1;
  • 89 714 990 992 ÷ 2 = 44 857 495 496 + 0;
  • 44 857 495 496 ÷ 2 = 22 428 747 748 + 0;
  • 22 428 747 748 ÷ 2 = 11 214 373 874 + 0;
  • 11 214 373 874 ÷ 2 = 5 607 186 937 + 0;
  • 5 607 186 937 ÷ 2 = 2 803 593 468 + 1;
  • 2 803 593 468 ÷ 2 = 1 401 796 734 + 0;
  • 1 401 796 734 ÷ 2 = 700 898 367 + 0;
  • 700 898 367 ÷ 2 = 350 449 183 + 1;
  • 350 449 183 ÷ 2 = 175 224 591 + 1;
  • 175 224 591 ÷ 2 = 87 612 295 + 1;
  • 87 612 295 ÷ 2 = 43 806 147 + 1;
  • 43 806 147 ÷ 2 = 21 903 073 + 1;
  • 21 903 073 ÷ 2 = 10 951 536 + 1;
  • 10 951 536 ÷ 2 = 5 475 768 + 0;
  • 5 475 768 ÷ 2 = 2 737 884 + 0;
  • 2 737 884 ÷ 2 = 1 368 942 + 0;
  • 1 368 942 ÷ 2 = 684 471 + 0;
  • 684 471 ÷ 2 = 342 235 + 1;
  • 342 235 ÷ 2 = 171 117 + 1;
  • 171 117 ÷ 2 = 85 558 + 1;
  • 85 558 ÷ 2 = 42 779 + 0;
  • 42 779 ÷ 2 = 21 389 + 1;
  • 21 389 ÷ 2 = 10 694 + 1;
  • 10 694 ÷ 2 = 5 347 + 0;
  • 5 347 ÷ 2 = 2 673 + 1;
  • 2 673 ÷ 2 = 1 336 + 1;
  • 1 336 ÷ 2 = 668 + 0;
  • 668 ÷ 2 = 334 + 0;
  • 334 ÷ 2 = 167 + 0;
  • 167 ÷ 2 = 83 + 1;
  • 83 ÷ 2 = 41 + 1;
  • 41 ÷ 2 = 20 + 1;
  • 20 ÷ 2 = 10 + 0;
  • 10 ÷ 2 = 5 + 0;
  • 5 ÷ 2 = 2 + 1;
  • 2 ÷ 2 = 1 + 0;
  • 1 ÷ 2 = 0 + 1;

2. Construct the base 2 representation of the positive number.

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

101 010 100 001 110 999 999 999 993(10) =


101 0011 1000 1101 1011 1000 0111 1110 0100 0010 1111 0100 0001 1011 0111 0010 1010 1000 0110 1111 1111 1001(2)


3. Normalize the binary representation of the number.

Shift the decimal mark 86 positions to the left so that only one non zero digit remains to the left of it:

101 010 100 001 110 999 999 999 993(10) =


101 0011 1000 1101 1011 1000 0111 1110 0100 0010 1111 0100 0001 1011 0111 0010 1010 1000 0110 1111 1111 1001(2) =


101 0011 1000 1101 1011 1000 0111 1110 0100 0010 1111 0100 0001 1011 0111 0010 1010 1000 0110 1111 1111 1001(2) × 20 =


1.0100 1110 0011 0110 1110 0001 1111 1001 0000 1011 1101 0000 0110 1101 1100 1010 1010 0001 1011 1111 1110 01(2) × 286


4. 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): 86


Mantissa (not normalized):
1.0100 1110 0011 0110 1110 0001 1111 1001 0000 1011 1101 0000 0110 1101 1100 1010 1010 0001 1011 1111 1110 01


5. Adjust the exponent.

Use the 8 bit excess/bias notation:

Exponent (adjusted) =


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


86 + 2(8-1) - 1 =


(86 + 127)(10) =


213(10)


6. 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;
  • 213 ÷ 2 = 106 + 1;
  • 106 ÷ 2 = 53 + 0;
  • 53 ÷ 2 = 26 + 1;
  • 26 ÷ 2 = 13 + 0;
  • 13 ÷ 2 = 6 + 1;
  • 6 ÷ 2 = 3 + 0;
  • 3 ÷ 2 = 1 + 1;
  • 1 ÷ 2 = 0 + 1;

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

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

Exponent (adjusted) =


213(10) =


1101 0101(2)


8. 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. 010 0111 0001 1011 0111 0000 111 1110 0100 0010 1111 0100 0001 1011 0111 0010 1010 1000 0110 1111 1111 1001 =


010 0111 0001 1011 0111 0000


9. 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) =
1101 0101


Mantissa (23 bits) =
010 0111 0001 1011 0111 0000


Number 101 010 100 001 110 999 999 999 993 converted from decimal system (base 10) to 32 bit single precision IEEE 754 binary floating point:
0 - 1101 0101 - 010 0111 0001 1011 0111 0000

(32 bits IEEE 754)
  • Sign (1 bit):

    • 0

      31
  • Exponent (8 bits):

    • 1

      30
    • 1

      29
    • 0

      28
    • 1

      27
    • 0

      26
    • 1

      25
    • 0

      24
    • 1

      23
  • Mantissa (23 bits):

    • 0

      22
    • 1

      21
    • 0

      20
    • 0

      19
    • 1

      18
    • 1

      17
    • 1

      16
    • 0

      15
    • 0

      14
    • 0

      13
    • 1

      12
    • 1

      11
    • 0

      10
    • 1

      9
    • 1

      8
    • 0

      7
    • 1

      6
    • 1

      5
    • 1

      4
    • 0

      3
    • 0

      2
    • 0

      1
    • 0

      0

More operations of this kind:

101 010 100 001 110 999 999 999 992 = ? ... 101 010 100 001 110 999 999 999 994 = ?


Convert to 32 bit single precision IEEE 754 binary floating point standard

A number in 32 bit single precision IEEE 754 binary floating point standard representation requires three building elements: sign (it takes 1 bit and it's either 0 for positive or 1 for negative numbers), exponent (8 bits) and mantissa (23 bits)

Latest decimal numbers converted from base ten to 32 bit single precision IEEE 754 floating point binary standard representation

101 010 100 001 110 999 999 999 993 to 32 bit single precision IEEE 754 binary floating point = ? Oct 28 12:09 UTC (GMT)
389 711 to 32 bit single precision IEEE 754 binary floating point = ? Oct 28 12:09 UTC (GMT)
-389 715 to 32 bit single precision IEEE 754 binary floating point = ? Oct 28 12:09 UTC (GMT)
-123.674 to 32 bit single precision IEEE 754 binary floating point = ? Oct 28 12:09 UTC (GMT)
-515.38 to 32 bit single precision IEEE 754 binary floating point = ? Oct 28 12:09 UTC (GMT)
-27 479 to 32 bit single precision IEEE 754 binary floating point = ? Oct 28 12:09 UTC (GMT)
-6.800 000 190 734 863 287 to 32 bit single precision IEEE 754 binary floating point = ? Oct 28 12:09 UTC (GMT)
100 100 001 011 000 000 000 000 000 000 000 000 016 to 32 bit single precision IEEE 754 binary floating point = ? Oct 28 12:09 UTC (GMT)
-1 148 846 084 to 32 bit single precision IEEE 754 binary floating point = ? Oct 28 12:09 UTC (GMT)
3.806 3 to 32 bit single precision IEEE 754 binary floating point = ? Oct 28 12:08 UTC (GMT)
114.937 3 to 32 bit single precision IEEE 754 binary floating point = ? Oct 28 12:08 UTC (GMT)
262.74 to 32 bit single precision IEEE 754 binary floating point = ? Oct 28 12:08 UTC (GMT)
93 449 692 270 000 000.000 000 000 106 51 to 32 bit single precision IEEE 754 binary floating point = ? Oct 28 12:08 UTC (GMT)
All base ten decimal numbers converted to 32 bit single precision IEEE 754 binary floating point

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