Convert 10 110 100 110 101 000 000 000 000 000 006 to 32 Bit Single Precision IEEE 754 Binary Floating Point Standard, From a Base 10 Decimal Number

10 110 100 110 101 000 000 000 000 000 006(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;
  • 10 110 100 110 101 000 000 000 000 000 006 ÷ 2 = 5 055 050 055 050 500 000 000 000 000 003 + 0;
  • 5 055 050 055 050 500 000 000 000 000 003 ÷ 2 = 2 527 525 027 525 250 000 000 000 000 001 + 1;
  • 2 527 525 027 525 250 000 000 000 000 001 ÷ 2 = 1 263 762 513 762 625 000 000 000 000 000 + 1;
  • 1 263 762 513 762 625 000 000 000 000 000 ÷ 2 = 631 881 256 881 312 500 000 000 000 000 + 0;
  • 631 881 256 881 312 500 000 000 000 000 ÷ 2 = 315 940 628 440 656 250 000 000 000 000 + 0;
  • 315 940 628 440 656 250 000 000 000 000 ÷ 2 = 157 970 314 220 328 125 000 000 000 000 + 0;
  • 157 970 314 220 328 125 000 000 000 000 ÷ 2 = 78 985 157 110 164 062 500 000 000 000 + 0;
  • 78 985 157 110 164 062 500 000 000 000 ÷ 2 = 39 492 578 555 082 031 250 000 000 000 + 0;
  • 39 492 578 555 082 031 250 000 000 000 ÷ 2 = 19 746 289 277 541 015 625 000 000 000 + 0;
  • 19 746 289 277 541 015 625 000 000 000 ÷ 2 = 9 873 144 638 770 507 812 500 000 000 + 0;
  • 9 873 144 638 770 507 812 500 000 000 ÷ 2 = 4 936 572 319 385 253 906 250 000 000 + 0;
  • 4 936 572 319 385 253 906 250 000 000 ÷ 2 = 2 468 286 159 692 626 953 125 000 000 + 0;
  • 2 468 286 159 692 626 953 125 000 000 ÷ 2 = 1 234 143 079 846 313 476 562 500 000 + 0;
  • 1 234 143 079 846 313 476 562 500 000 ÷ 2 = 617 071 539 923 156 738 281 250 000 + 0;
  • 617 071 539 923 156 738 281 250 000 ÷ 2 = 308 535 769 961 578 369 140 625 000 + 0;
  • 308 535 769 961 578 369 140 625 000 ÷ 2 = 154 267 884 980 789 184 570 312 500 + 0;
  • 154 267 884 980 789 184 570 312 500 ÷ 2 = 77 133 942 490 394 592 285 156 250 + 0;
  • 77 133 942 490 394 592 285 156 250 ÷ 2 = 38 566 971 245 197 296 142 578 125 + 0;
  • 38 566 971 245 197 296 142 578 125 ÷ 2 = 19 283 485 622 598 648 071 289 062 + 1;
  • 19 283 485 622 598 648 071 289 062 ÷ 2 = 9 641 742 811 299 324 035 644 531 + 0;
  • 9 641 742 811 299 324 035 644 531 ÷ 2 = 4 820 871 405 649 662 017 822 265 + 1;
  • 4 820 871 405 649 662 017 822 265 ÷ 2 = 2 410 435 702 824 831 008 911 132 + 1;
  • 2 410 435 702 824 831 008 911 132 ÷ 2 = 1 205 217 851 412 415 504 455 566 + 0;
  • 1 205 217 851 412 415 504 455 566 ÷ 2 = 602 608 925 706 207 752 227 783 + 0;
  • 602 608 925 706 207 752 227 783 ÷ 2 = 301 304 462 853 103 876 113 891 + 1;
  • 301 304 462 853 103 876 113 891 ÷ 2 = 150 652 231 426 551 938 056 945 + 1;
  • 150 652 231 426 551 938 056 945 ÷ 2 = 75 326 115 713 275 969 028 472 + 1;
  • 75 326 115 713 275 969 028 472 ÷ 2 = 37 663 057 856 637 984 514 236 + 0;
  • 37 663 057 856 637 984 514 236 ÷ 2 = 18 831 528 928 318 992 257 118 + 0;
  • 18 831 528 928 318 992 257 118 ÷ 2 = 9 415 764 464 159 496 128 559 + 0;
  • 9 415 764 464 159 496 128 559 ÷ 2 = 4 707 882 232 079 748 064 279 + 1;
  • 4 707 882 232 079 748 064 279 ÷ 2 = 2 353 941 116 039 874 032 139 + 1;
  • 2 353 941 116 039 874 032 139 ÷ 2 = 1 176 970 558 019 937 016 069 + 1;
  • 1 176 970 558 019 937 016 069 ÷ 2 = 588 485 279 009 968 508 034 + 1;
  • 588 485 279 009 968 508 034 ÷ 2 = 294 242 639 504 984 254 017 + 0;
  • 294 242 639 504 984 254 017 ÷ 2 = 147 121 319 752 492 127 008 + 1;
  • 147 121 319 752 492 127 008 ÷ 2 = 73 560 659 876 246 063 504 + 0;
  • 73 560 659 876 246 063 504 ÷ 2 = 36 780 329 938 123 031 752 + 0;
  • 36 780 329 938 123 031 752 ÷ 2 = 18 390 164 969 061 515 876 + 0;
  • 18 390 164 969 061 515 876 ÷ 2 = 9 195 082 484 530 757 938 + 0;
  • 9 195 082 484 530 757 938 ÷ 2 = 4 597 541 242 265 378 969 + 0;
  • 4 597 541 242 265 378 969 ÷ 2 = 2 298 770 621 132 689 484 + 1;
  • 2 298 770 621 132 689 484 ÷ 2 = 1 149 385 310 566 344 742 + 0;
  • 1 149 385 310 566 344 742 ÷ 2 = 574 692 655 283 172 371 + 0;
  • 574 692 655 283 172 371 ÷ 2 = 287 346 327 641 586 185 + 1;
  • 287 346 327 641 586 185 ÷ 2 = 143 673 163 820 793 092 + 1;
  • 143 673 163 820 793 092 ÷ 2 = 71 836 581 910 396 546 + 0;
  • 71 836 581 910 396 546 ÷ 2 = 35 918 290 955 198 273 + 0;
  • 35 918 290 955 198 273 ÷ 2 = 17 959 145 477 599 136 + 1;
  • 17 959 145 477 599 136 ÷ 2 = 8 979 572 738 799 568 + 0;
  • 8 979 572 738 799 568 ÷ 2 = 4 489 786 369 399 784 + 0;
  • 4 489 786 369 399 784 ÷ 2 = 2 244 893 184 699 892 + 0;
  • 2 244 893 184 699 892 ÷ 2 = 1 122 446 592 349 946 + 0;
  • 1 122 446 592 349 946 ÷ 2 = 561 223 296 174 973 + 0;
  • 561 223 296 174 973 ÷ 2 = 280 611 648 087 486 + 1;
  • 280 611 648 087 486 ÷ 2 = 140 305 824 043 743 + 0;
  • 140 305 824 043 743 ÷ 2 = 70 152 912 021 871 + 1;
  • 70 152 912 021 871 ÷ 2 = 35 076 456 010 935 + 1;
  • 35 076 456 010 935 ÷ 2 = 17 538 228 005 467 + 1;
  • 17 538 228 005 467 ÷ 2 = 8 769 114 002 733 + 1;
  • 8 769 114 002 733 ÷ 2 = 4 384 557 001 366 + 1;
  • 4 384 557 001 366 ÷ 2 = 2 192 278 500 683 + 0;
  • 2 192 278 500 683 ÷ 2 = 1 096 139 250 341 + 1;
  • 1 096 139 250 341 ÷ 2 = 548 069 625 170 + 1;
  • 548 069 625 170 ÷ 2 = 274 034 812 585 + 0;
  • 274 034 812 585 ÷ 2 = 137 017 406 292 + 1;
  • 137 017 406 292 ÷ 2 = 68 508 703 146 + 0;
  • 68 508 703 146 ÷ 2 = 34 254 351 573 + 0;
  • 34 254 351 573 ÷ 2 = 17 127 175 786 + 1;
  • 17 127 175 786 ÷ 2 = 8 563 587 893 + 0;
  • 8 563 587 893 ÷ 2 = 4 281 793 946 + 1;
  • 4 281 793 946 ÷ 2 = 2 140 896 973 + 0;
  • 2 140 896 973 ÷ 2 = 1 070 448 486 + 1;
  • 1 070 448 486 ÷ 2 = 535 224 243 + 0;
  • 535 224 243 ÷ 2 = 267 612 121 + 1;
  • 267 612 121 ÷ 2 = 133 806 060 + 1;
  • 133 806 060 ÷ 2 = 66 903 030 + 0;
  • 66 903 030 ÷ 2 = 33 451 515 + 0;
  • 33 451 515 ÷ 2 = 16 725 757 + 1;
  • 16 725 757 ÷ 2 = 8 362 878 + 1;
  • 8 362 878 ÷ 2 = 4 181 439 + 0;
  • 4 181 439 ÷ 2 = 2 090 719 + 1;
  • 2 090 719 ÷ 2 = 1 045 359 + 1;
  • 1 045 359 ÷ 2 = 522 679 + 1;
  • 522 679 ÷ 2 = 261 339 + 1;
  • 261 339 ÷ 2 = 130 669 + 1;
  • 130 669 ÷ 2 = 65 334 + 1;
  • 65 334 ÷ 2 = 32 667 + 0;
  • 32 667 ÷ 2 = 16 333 + 1;
  • 16 333 ÷ 2 = 8 166 + 1;
  • 8 166 ÷ 2 = 4 083 + 0;
  • 4 083 ÷ 2 = 2 041 + 1;
  • 2 041 ÷ 2 = 1 020 + 1;
  • 1 020 ÷ 2 = 510 + 0;
  • 510 ÷ 2 = 255 + 0;
  • 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;

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

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

10 110 100 110 101 000 000 000 000 000 006(10) =


111 1111 1001 1011 0111 1110 1100 1101 0101 0010 1101 1111 0100 0001 0011 0010 0000 1011 1100 0111 0011 0100 0000 0000 0000 0110(2)


3. Normalize the binary representation of the number.

Shift the decimal mark 102 positions to the left so that only one non zero digit remains to the left of it:

10 110 100 110 101 000 000 000 000 000 006(10) =


111 1111 1001 1011 0111 1110 1100 1101 0101 0010 1101 1111 0100 0001 0011 0010 0000 1011 1100 0111 0011 0100 0000 0000 0000 0110(2) =


111 1111 1001 1011 0111 1110 1100 1101 0101 0010 1101 1111 0100 0001 0011 0010 0000 1011 1100 0111 0011 0100 0000 0000 0000 0110(2) × 20 =


1.1111 1110 0110 1101 1111 1011 0011 0101 0100 1011 0111 1101 0000 0100 1100 1000 0010 1111 0001 1100 1101 0000 0000 0000 0001 10(2) × 2102


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): 102


Mantissa (not normalized):
1.1111 1110 0110 1101 1111 1011 0011 0101 0100 1011 0111 1101 0000 0100 1100 1000 0010 1111 0001 1100 1101 0000 0000 0000 0001 10


5. Adjust the exponent.

Use the 8 bit excess/bias notation:

Exponent (adjusted) =


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


102 + 2(8-1) - 1 =


(102 + 127)(10) =


229(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;
  • 229 ÷ 2 = 114 + 1;
  • 114 ÷ 2 = 57 + 0;
  • 57 ÷ 2 = 28 + 1;
  • 28 ÷ 2 = 14 + 0;
  • 14 ÷ 2 = 7 + 0;
  • 7 ÷ 2 = 3 + 1;
  • 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) =


229(10) =


1110 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. 111 1111 0011 0110 1111 1101 100 1101 0101 0010 1101 1111 0100 0001 0011 0010 0000 1011 1100 0111 0011 0100 0000 0000 0000 0110 =


111 1111 0011 0110 1111 1101


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) =
1110 0101


Mantissa (23 bits) =
111 1111 0011 0110 1111 1101


Number 10 110 100 110 101 000 000 000 000 000 006 converted from decimal system (base 10) to 32 bit single precision IEEE 754 binary floating point:
0 - 1110 0101 - 111 1111 0011 0110 1111 1101

(32 bits IEEE 754)
  • Sign (1 bit):

    • 0

      31
  • Exponent (8 bits):

    • 1

      30
    • 1

      29
    • 1

      28
    • 0

      27
    • 0

      26
    • 1

      25
    • 0

      24
    • 1

      23
  • Mantissa (23 bits):

    • 1

      22
    • 1

      21
    • 1

      20
    • 1

      19
    • 1

      18
    • 1

      17
    • 1

      16
    • 0

      15
    • 0

      14
    • 1

      13
    • 1

      12
    • 0

      11
    • 1

      10
    • 1

      9
    • 0

      8
    • 1

      7
    • 1

      6
    • 1

      5
    • 1

      4
    • 1

      3
    • 1

      2
    • 0

      1
    • 1

      0

More operations of this kind:

10 110 100 110 101 000 000 000 000 000 005 = ? ... 10 110 100 110 101 000 000 000 000 000 007 = ?


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

10 110 100 110 101 000 000 000 000 000 006 to 32 bit single precision IEEE 754 binary floating point = ? Oct 28 10:55 UTC (GMT)
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0.000 015 258 789 7 to 32 bit single precision IEEE 754 binary floating point = ? Oct 28 10:55 UTC (GMT)
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3.941 05 to 32 bit single precision IEEE 754 binary floating point = ? Oct 28 10:55 UTC (GMT)
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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