3.402 823 466 385 288 598 117 041 834 857 04 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 3.402 823 466 385 288 598 117 041 834 857 04(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
3.402 823 466 385 288 598 117 041 834 857 04(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: 3.
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;
  • 3 ÷ 2 = 1 + 1;
  • 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.

3(10) =

11(2)


3. Convert to binary (base 2) the fractional part: 0.402 823 466 385 288 598 117 041 834 857 04.

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) 0.402 823 466 385 288 598 117 041 834 857 04 × 2 = 0 + 0.805 646 932 770 577 196 234 083 669 714 08;
  • 2) 0.805 646 932 770 577 196 234 083 669 714 08 × 2 = 1 + 0.611 293 865 541 154 392 468 167 339 428 16;
  • 3) 0.611 293 865 541 154 392 468 167 339 428 16 × 2 = 1 + 0.222 587 731 082 308 784 936 334 678 856 32;
  • 4) 0.222 587 731 082 308 784 936 334 678 856 32 × 2 = 0 + 0.445 175 462 164 617 569 872 669 357 712 64;
  • 5) 0.445 175 462 164 617 569 872 669 357 712 64 × 2 = 0 + 0.890 350 924 329 235 139 745 338 715 425 28;
  • 6) 0.890 350 924 329 235 139 745 338 715 425 28 × 2 = 1 + 0.780 701 848 658 470 279 490 677 430 850 56;
  • 7) 0.780 701 848 658 470 279 490 677 430 850 56 × 2 = 1 + 0.561 403 697 316 940 558 981 354 861 701 12;
  • 8) 0.561 403 697 316 940 558 981 354 861 701 12 × 2 = 1 + 0.122 807 394 633 881 117 962 709 723 402 24;
  • 9) 0.122 807 394 633 881 117 962 709 723 402 24 × 2 = 0 + 0.245 614 789 267 762 235 925 419 446 804 48;
  • 10) 0.245 614 789 267 762 235 925 419 446 804 48 × 2 = 0 + 0.491 229 578 535 524 471 850 838 893 608 96;
  • 11) 0.491 229 578 535 524 471 850 838 893 608 96 × 2 = 0 + 0.982 459 157 071 048 943 701 677 787 217 92;
  • 12) 0.982 459 157 071 048 943 701 677 787 217 92 × 2 = 1 + 0.964 918 314 142 097 887 403 355 574 435 84;
  • 13) 0.964 918 314 142 097 887 403 355 574 435 84 × 2 = 1 + 0.929 836 628 284 195 774 806 711 148 871 68;
  • 14) 0.929 836 628 284 195 774 806 711 148 871 68 × 2 = 1 + 0.859 673 256 568 391 549 613 422 297 743 36;
  • 15) 0.859 673 256 568 391 549 613 422 297 743 36 × 2 = 1 + 0.719 346 513 136 783 099 226 844 595 486 72;
  • 16) 0.719 346 513 136 783 099 226 844 595 486 72 × 2 = 1 + 0.438 693 026 273 566 198 453 689 190 973 44;
  • 17) 0.438 693 026 273 566 198 453 689 190 973 44 × 2 = 0 + 0.877 386 052 547 132 396 907 378 381 946 88;
  • 18) 0.877 386 052 547 132 396 907 378 381 946 88 × 2 = 1 + 0.754 772 105 094 264 793 814 756 763 893 76;
  • 19) 0.754 772 105 094 264 793 814 756 763 893 76 × 2 = 1 + 0.509 544 210 188 529 587 629 513 527 787 52;
  • 20) 0.509 544 210 188 529 587 629 513 527 787 52 × 2 = 1 + 0.019 088 420 377 059 175 259 027 055 575 04;
  • 21) 0.019 088 420 377 059 175 259 027 055 575 04 × 2 = 0 + 0.038 176 840 754 118 350 518 054 111 150 08;
  • 22) 0.038 176 840 754 118 350 518 054 111 150 08 × 2 = 0 + 0.076 353 681 508 236 701 036 108 222 300 16;
  • 23) 0.076 353 681 508 236 701 036 108 222 300 16 × 2 = 0 + 0.152 707 363 016 473 402 072 216 444 600 32;
  • 24) 0.152 707 363 016 473 402 072 216 444 600 32 × 2 = 0 + 0.305 414 726 032 946 804 144 432 889 200 64;

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:


0.402 823 466 385 288 598 117 041 834 857 04(10) =

0.0110 0111 0001 1111 0111 0000(2)

5. Positive number before normalization:

3.402 823 466 385 288 598 117 041 834 857 04(10) =

11.0110 0111 0001 1111 0111 0000(2)

6. Normalize the binary representation of the number.

Shift the decimal mark 1 positions to the left, so that only one non zero digit remains to the left of it:


3.402 823 466 385 288 598 117 041 834 857 04(10) =

11.0110 0111 0001 1111 0111 0000(2) =

11.0110 0111 0001 1111 0111 0000(2) × 20 =

1.1011 0011 1000 1111 1011 1000 0(2) × 21


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): 1

Mantissa (not normalized):
1.1011 0011 1000 1111 1011 1000 0


8. Adjust the exponent.

Use the 8 bit excess/bias notation:


Exponent (adjusted) =

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

1 + 2(8-1) - 1 =

(1 + 127)(10) =

128(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;
  • 128 ÷ 2 = 64 + 0;
  • 64 ÷ 2 = 32 + 0;
  • 32 ÷ 2 = 16 + 0;
  • 16 ÷ 2 = 8 + 0;
  • 8 ÷ 2 = 4 + 0;
  • 4 ÷ 2 = 2 + 0;
  • 2 ÷ 2 = 1 + 0;
  • 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) =

128(10) =

1000 0000(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. 101 1001 1100 0111 1101 1100 00 =

101 1001 1100 0111 1101 1100


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) =
1000 0000

Mantissa (23 bits) =
101 1001 1100 0111 1101 1100


Decimal number 3.402 823 466 385 288 598 117 041 834 857 04 converted to 32 bit single precision IEEE 754 binary floating point representation:

0 - 1000 0000 - 101 1001 1100 0111 1101 1100

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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