Convert 101 110 100 100 101 011 110 110 100 104 to 32 Bit Single Precision IEEE 754 Binary Floating Point Standard, From a Base 10 Decimal Number

101 110 100 100 101 011 110 110 100 104(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;
  • 101 110 100 100 101 011 110 110 100 104 ÷ 2 = 50 555 050 050 050 505 555 055 050 052 + 0;
  • 50 555 050 050 050 505 555 055 050 052 ÷ 2 = 25 277 525 025 025 252 777 527 525 026 + 0;
  • 25 277 525 025 025 252 777 527 525 026 ÷ 2 = 12 638 762 512 512 626 388 763 762 513 + 0;
  • 12 638 762 512 512 626 388 763 762 513 ÷ 2 = 6 319 381 256 256 313 194 381 881 256 + 1;
  • 6 319 381 256 256 313 194 381 881 256 ÷ 2 = 3 159 690 628 128 156 597 190 940 628 + 0;
  • 3 159 690 628 128 156 597 190 940 628 ÷ 2 = 1 579 845 314 064 078 298 595 470 314 + 0;
  • 1 579 845 314 064 078 298 595 470 314 ÷ 2 = 789 922 657 032 039 149 297 735 157 + 0;
  • 789 922 657 032 039 149 297 735 157 ÷ 2 = 394 961 328 516 019 574 648 867 578 + 1;
  • 394 961 328 516 019 574 648 867 578 ÷ 2 = 197 480 664 258 009 787 324 433 789 + 0;
  • 197 480 664 258 009 787 324 433 789 ÷ 2 = 98 740 332 129 004 893 662 216 894 + 1;
  • 98 740 332 129 004 893 662 216 894 ÷ 2 = 49 370 166 064 502 446 831 108 447 + 0;
  • 49 370 166 064 502 446 831 108 447 ÷ 2 = 24 685 083 032 251 223 415 554 223 + 1;
  • 24 685 083 032 251 223 415 554 223 ÷ 2 = 12 342 541 516 125 611 707 777 111 + 1;
  • 12 342 541 516 125 611 707 777 111 ÷ 2 = 6 171 270 758 062 805 853 888 555 + 1;
  • 6 171 270 758 062 805 853 888 555 ÷ 2 = 3 085 635 379 031 402 926 944 277 + 1;
  • 3 085 635 379 031 402 926 944 277 ÷ 2 = 1 542 817 689 515 701 463 472 138 + 1;
  • 1 542 817 689 515 701 463 472 138 ÷ 2 = 771 408 844 757 850 731 736 069 + 0;
  • 771 408 844 757 850 731 736 069 ÷ 2 = 385 704 422 378 925 365 868 034 + 1;
  • 385 704 422 378 925 365 868 034 ÷ 2 = 192 852 211 189 462 682 934 017 + 0;
  • 192 852 211 189 462 682 934 017 ÷ 2 = 96 426 105 594 731 341 467 008 + 1;
  • 96 426 105 594 731 341 467 008 ÷ 2 = 48 213 052 797 365 670 733 504 + 0;
  • 48 213 052 797 365 670 733 504 ÷ 2 = 24 106 526 398 682 835 366 752 + 0;
  • 24 106 526 398 682 835 366 752 ÷ 2 = 12 053 263 199 341 417 683 376 + 0;
  • 12 053 263 199 341 417 683 376 ÷ 2 = 6 026 631 599 670 708 841 688 + 0;
  • 6 026 631 599 670 708 841 688 ÷ 2 = 3 013 315 799 835 354 420 844 + 0;
  • 3 013 315 799 835 354 420 844 ÷ 2 = 1 506 657 899 917 677 210 422 + 0;
  • 1 506 657 899 917 677 210 422 ÷ 2 = 753 328 949 958 838 605 211 + 0;
  • 753 328 949 958 838 605 211 ÷ 2 = 376 664 474 979 419 302 605 + 1;
  • 376 664 474 979 419 302 605 ÷ 2 = 188 332 237 489 709 651 302 + 1;
  • 188 332 237 489 709 651 302 ÷ 2 = 94 166 118 744 854 825 651 + 0;
  • 94 166 118 744 854 825 651 ÷ 2 = 47 083 059 372 427 412 825 + 1;
  • 47 083 059 372 427 412 825 ÷ 2 = 23 541 529 686 213 706 412 + 1;
  • 23 541 529 686 213 706 412 ÷ 2 = 11 770 764 843 106 853 206 + 0;
  • 11 770 764 843 106 853 206 ÷ 2 = 5 885 382 421 553 426 603 + 0;
  • 5 885 382 421 553 426 603 ÷ 2 = 2 942 691 210 776 713 301 + 1;
  • 2 942 691 210 776 713 301 ÷ 2 = 1 471 345 605 388 356 650 + 1;
  • 1 471 345 605 388 356 650 ÷ 2 = 735 672 802 694 178 325 + 0;
  • 735 672 802 694 178 325 ÷ 2 = 367 836 401 347 089 162 + 1;
  • 367 836 401 347 089 162 ÷ 2 = 183 918 200 673 544 581 + 0;
  • 183 918 200 673 544 581 ÷ 2 = 91 959 100 336 772 290 + 1;
  • 91 959 100 336 772 290 ÷ 2 = 45 979 550 168 386 145 + 0;
  • 45 979 550 168 386 145 ÷ 2 = 22 989 775 084 193 072 + 1;
  • 22 989 775 084 193 072 ÷ 2 = 11 494 887 542 096 536 + 0;
  • 11 494 887 542 096 536 ÷ 2 = 5 747 443 771 048 268 + 0;
  • 5 747 443 771 048 268 ÷ 2 = 2 873 721 885 524 134 + 0;
  • 2 873 721 885 524 134 ÷ 2 = 1 436 860 942 762 067 + 0;
  • 1 436 860 942 762 067 ÷ 2 = 718 430 471 381 033 + 1;
  • 718 430 471 381 033 ÷ 2 = 359 215 235 690 516 + 1;
  • 359 215 235 690 516 ÷ 2 = 179 607 617 845 258 + 0;
  • 179 607 617 845 258 ÷ 2 = 89 803 808 922 629 + 0;
  • 89 803 808 922 629 ÷ 2 = 44 901 904 461 314 + 1;
  • 44 901 904 461 314 ÷ 2 = 22 450 952 230 657 + 0;
  • 22 450 952 230 657 ÷ 2 = 11 225 476 115 328 + 1;
  • 11 225 476 115 328 ÷ 2 = 5 612 738 057 664 + 0;
  • 5 612 738 057 664 ÷ 2 = 2 806 369 028 832 + 0;
  • 2 806 369 028 832 ÷ 2 = 1 403 184 514 416 + 0;
  • 1 403 184 514 416 ÷ 2 = 701 592 257 208 + 0;
  • 701 592 257 208 ÷ 2 = 350 796 128 604 + 0;
  • 350 796 128 604 ÷ 2 = 175 398 064 302 + 0;
  • 175 398 064 302 ÷ 2 = 87 699 032 151 + 0;
  • 87 699 032 151 ÷ 2 = 43 849 516 075 + 1;
  • 43 849 516 075 ÷ 2 = 21 924 758 037 + 1;
  • 21 924 758 037 ÷ 2 = 10 962 379 018 + 1;
  • 10 962 379 018 ÷ 2 = 5 481 189 509 + 0;
  • 5 481 189 509 ÷ 2 = 2 740 594 754 + 1;
  • 2 740 594 754 ÷ 2 = 1 370 297 377 + 0;
  • 1 370 297 377 ÷ 2 = 685 148 688 + 1;
  • 685 148 688 ÷ 2 = 342 574 344 + 0;
  • 342 574 344 ÷ 2 = 171 287 172 + 0;
  • 171 287 172 ÷ 2 = 85 643 586 + 0;
  • 85 643 586 ÷ 2 = 42 821 793 + 0;
  • 42 821 793 ÷ 2 = 21 410 896 + 1;
  • 21 410 896 ÷ 2 = 10 705 448 + 0;
  • 10 705 448 ÷ 2 = 5 352 724 + 0;
  • 5 352 724 ÷ 2 = 2 676 362 + 0;
  • 2 676 362 ÷ 2 = 1 338 181 + 0;
  • 1 338 181 ÷ 2 = 669 090 + 1;
  • 669 090 ÷ 2 = 334 545 + 0;
  • 334 545 ÷ 2 = 167 272 + 1;
  • 167 272 ÷ 2 = 83 636 + 0;
  • 83 636 ÷ 2 = 41 818 + 0;
  • 41 818 ÷ 2 = 20 909 + 0;
  • 20 909 ÷ 2 = 10 454 + 1;
  • 10 454 ÷ 2 = 5 227 + 0;
  • 5 227 ÷ 2 = 2 613 + 1;
  • 2 613 ÷ 2 = 1 306 + 1;
  • 1 306 ÷ 2 = 653 + 0;
  • 653 ÷ 2 = 326 + 1;
  • 326 ÷ 2 = 163 + 0;
  • 163 ÷ 2 = 81 + 1;
  • 81 ÷ 2 = 40 + 1;
  • 40 ÷ 2 = 20 + 0;
  • 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.

101 110 100 100 101 011 110 110 100 104(10) =


1 0100 0110 1011 0100 0101 0000 1000 0101 0111 0000 0001 0100 1100 0010 1010 1100 1101 1000 0000 1010 1111 1010 1000 1000(2)


3. Normalize the binary representation of the number.

Shift the decimal mark 96 positions to the left so that only one non zero digit remains to the left of it:

101 110 100 100 101 011 110 110 100 104(10) =


1 0100 0110 1011 0100 0101 0000 1000 0101 0111 0000 0001 0100 1100 0010 1010 1100 1101 1000 0000 1010 1111 1010 1000 1000(2) =


1 0100 0110 1011 0100 0101 0000 1000 0101 0111 0000 0001 0100 1100 0010 1010 1100 1101 1000 0000 1010 1111 1010 1000 1000(2) × 20 =


1.0100 0110 1011 0100 0101 0000 1000 0101 0111 0000 0001 0100 1100 0010 1010 1100 1101 1000 0000 1010 1111 1010 1000 1000(2) × 296


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): 96


Mantissa (not normalized):
1.0100 0110 1011 0100 0101 0000 1000 0101 0111 0000 0001 0100 1100 0010 1010 1100 1101 1000 0000 1010 1111 1010 1000 1000


5. Adjust the exponent.

Use the 8 bit excess/bias notation:

Exponent (adjusted) =


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


96 + 2(8-1) - 1 =


(96 + 127)(10) =


223(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;
  • 223 ÷ 2 = 111 + 1;
  • 111 ÷ 2 = 55 + 1;
  • 55 ÷ 2 = 27 + 1;
  • 27 ÷ 2 = 13 + 1;
  • 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) =


223(10) =


1101 1111(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 0011 0101 1010 0010 1000 0 1000 0101 0111 0000 0001 0100 1100 0010 1010 1100 1101 1000 0000 1010 1111 1010 1000 1000 =


010 0011 0101 1010 0010 1000


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 1111


Mantissa (23 bits) =
010 0011 0101 1010 0010 1000


Number 101 110 100 100 101 011 110 110 100 104 converted from decimal system (base 10) to 32 bit single precision IEEE 754 binary floating point:
0 - 1101 1111 - 010 0011 0101 1010 0010 1000

(32 bits IEEE 754)
  • Sign (1 bit):

    • 0

      31
  • Exponent (8 bits):

    • 1

      30
    • 1

      29
    • 0

      28
    • 1

      27
    • 1

      26
    • 1

      25
    • 1

      24
    • 1

      23
  • Mantissa (23 bits):

    • 0

      22
    • 1

      21
    • 0

      20
    • 0

      19
    • 0

      18
    • 1

      17
    • 1

      16
    • 0

      15
    • 1

      14
    • 0

      13
    • 1

      12
    • 1

      11
    • 0

      10
    • 1

      9
    • 0

      8
    • 0

      7
    • 0

      6
    • 1

      5
    • 0

      4
    • 1

      3
    • 0

      2
    • 0

      1
    • 0

      0

More operations of this kind:

101 110 100 100 101 011 110 110 100 103 = ? ... 101 110 100 100 101 011 110 110 100 105 = ?


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

101 110 100 100 101 011 110 110 100 104 to 32 bit single precision IEEE 754 binary floating point = ? Nov 30 09:55 UTC (GMT)
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17 054 to 32 bit single precision IEEE 754 binary floating point = ? Nov 30 09:54 UTC (GMT)
34.891 to 32 bit single precision IEEE 754 binary floating point = ? Nov 30 09:54 UTC (GMT)
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-0.5 to 32 bit single precision IEEE 754 binary floating point = ? Nov 30 09:54 UTC (GMT)
-132.142 to 32 bit single precision IEEE 754 binary floating point = ? Nov 30 09:54 UTC (GMT)
53.57 to 32 bit single precision IEEE 754 binary floating point = ? Nov 30 09:54 UTC (GMT)
-23 518 to 32 bit single precision IEEE 754 binary floating point = ? Nov 30 09:54 UTC (GMT)
55.659 to 32 bit single precision IEEE 754 binary floating point = ? Nov 30 09:54 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