-0.000 000 047 684 613 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 047 684 613(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
-0.000 000 047 684 613(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. Start with the positive version of the number:

|-0.000 000 047 684 613| = 0.000 000 047 684 613


2. First, convert to binary (in base 2) the integer part: 0.
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;
  • 0 ÷ 2 = 0 + 0;

3. 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.

0(10) =

0(2)


4. Convert to binary (base 2) the fractional part: 0.000 000 047 684 613.

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) 0.000 000 047 684 613 × 2 = 0 + 0.000 000 095 369 226;
  • 2) 0.000 000 095 369 226 × 2 = 0 + 0.000 000 190 738 452;
  • 3) 0.000 000 190 738 452 × 2 = 0 + 0.000 000 381 476 904;
  • 4) 0.000 000 381 476 904 × 2 = 0 + 0.000 000 762 953 808;
  • 5) 0.000 000 762 953 808 × 2 = 0 + 0.000 001 525 907 616;
  • 6) 0.000 001 525 907 616 × 2 = 0 + 0.000 003 051 815 232;
  • 7) 0.000 003 051 815 232 × 2 = 0 + 0.000 006 103 630 464;
  • 8) 0.000 006 103 630 464 × 2 = 0 + 0.000 012 207 260 928;
  • 9) 0.000 012 207 260 928 × 2 = 0 + 0.000 024 414 521 856;
  • 10) 0.000 024 414 521 856 × 2 = 0 + 0.000 048 829 043 712;
  • 11) 0.000 048 829 043 712 × 2 = 0 + 0.000 097 658 087 424;
  • 12) 0.000 097 658 087 424 × 2 = 0 + 0.000 195 316 174 848;
  • 13) 0.000 195 316 174 848 × 2 = 0 + 0.000 390 632 349 696;
  • 14) 0.000 390 632 349 696 × 2 = 0 + 0.000 781 264 699 392;
  • 15) 0.000 781 264 699 392 × 2 = 0 + 0.001 562 529 398 784;
  • 16) 0.001 562 529 398 784 × 2 = 0 + 0.003 125 058 797 568;
  • 17) 0.003 125 058 797 568 × 2 = 0 + 0.006 250 117 595 136;
  • 18) 0.006 250 117 595 136 × 2 = 0 + 0.012 500 235 190 272;
  • 19) 0.012 500 235 190 272 × 2 = 0 + 0.025 000 470 380 544;
  • 20) 0.025 000 470 380 544 × 2 = 0 + 0.050 000 940 761 088;
  • 21) 0.050 000 940 761 088 × 2 = 0 + 0.100 001 881 522 176;
  • 22) 0.100 001 881 522 176 × 2 = 0 + 0.200 003 763 044 352;
  • 23) 0.200 003 763 044 352 × 2 = 0 + 0.400 007 526 088 704;
  • 24) 0.400 007 526 088 704 × 2 = 0 + 0.800 015 052 177 408;
  • 25) 0.800 015 052 177 408 × 2 = 1 + 0.600 030 104 354 816;
  • 26) 0.600 030 104 354 816 × 2 = 1 + 0.200 060 208 709 632;
  • 27) 0.200 060 208 709 632 × 2 = 0 + 0.400 120 417 419 264;
  • 28) 0.400 120 417 419 264 × 2 = 0 + 0.800 240 834 838 528;
  • 29) 0.800 240 834 838 528 × 2 = 1 + 0.600 481 669 677 056;
  • 30) 0.600 481 669 677 056 × 2 = 1 + 0.200 963 339 354 112;
  • 31) 0.200 963 339 354 112 × 2 = 0 + 0.401 926 678 708 224;
  • 32) 0.401 926 678 708 224 × 2 = 0 + 0.803 853 357 416 448;
  • 33) 0.803 853 357 416 448 × 2 = 1 + 0.607 706 714 832 896;
  • 34) 0.607 706 714 832 896 × 2 = 1 + 0.215 413 429 665 792;
  • 35) 0.215 413 429 665 792 × 2 = 0 + 0.430 826 859 331 584;
  • 36) 0.430 826 859 331 584 × 2 = 0 + 0.861 653 718 663 168;
  • 37) 0.861 653 718 663 168 × 2 = 1 + 0.723 307 437 326 336;
  • 38) 0.723 307 437 326 336 × 2 = 1 + 0.446 614 874 652 672;
  • 39) 0.446 614 874 652 672 × 2 = 0 + 0.893 229 749 305 344;
  • 40) 0.893 229 749 305 344 × 2 = 1 + 0.786 459 498 610 688;
  • 41) 0.786 459 498 610 688 × 2 = 1 + 0.572 918 997 221 376;
  • 42) 0.572 918 997 221 376 × 2 = 1 + 0.145 837 994 442 752;
  • 43) 0.145 837 994 442 752 × 2 = 0 + 0.291 675 988 885 504;
  • 44) 0.291 675 988 885 504 × 2 = 0 + 0.583 351 977 771 008;
  • 45) 0.583 351 977 771 008 × 2 = 1 + 0.166 703 955 542 016;
  • 46) 0.166 703 955 542 016 × 2 = 0 + 0.333 407 911 084 032;
  • 47) 0.333 407 911 084 032 × 2 = 0 + 0.666 815 822 168 064;
  • 48) 0.666 815 822 168 064 × 2 = 1 + 0.333 631 644 336 128;

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).


5. 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:


0.000 000 047 684 613(10) =

0.0000 0000 0000 0000 0000 0000 1100 1100 1100 1101 1100 1001(2)

6. Positive number before normalization:

0.000 000 047 684 613(10) =

0.0000 0000 0000 0000 0000 0000 1100 1100 1100 1101 1100 1001(2)

7. Normalize the binary representation of the number.

Shift the decimal mark 25 positions to the right, so that only one non zero digit remains to the left of it:


0.000 000 047 684 613(10) =

0.0000 0000 0000 0000 0000 0000 1100 1100 1100 1101 1100 1001(2) =

0.0000 0000 0000 0000 0000 0000 1100 1100 1100 1101 1100 1001(2) × 20 =

1.1001 1001 1001 1011 1001 001(2) × 2-25


8. 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 1 (a negative number)

Exponent (unadjusted): -25

Mantissa (not normalized):
1.1001 1001 1001 1011 1001 001


9. Adjust the exponent.

Use the 8 bit excess/bias notation:


Exponent (adjusted) =

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

-25 + 2(8-1) - 1 =

(-25 + 127)(10) =

102(10)


10. 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;
  • 102 ÷ 2 = 51 + 0;
  • 51 ÷ 2 = 25 + 1;
  • 25 ÷ 2 = 12 + 1;
  • 12 ÷ 2 = 6 + 0;
  • 6 ÷ 2 = 3 + 0;
  • 3 ÷ 2 = 1 + 1;
  • 1 ÷ 2 = 0 + 1;

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

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


Exponent (adjusted) =

102(10) =

0110 0110(2)


12. 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, only if necessary (not the case here).


Mantissa (normalized) =

1. 100 1100 1100 1101 1100 1001 =

100 1100 1100 1101 1100 1001


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

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

Exponent (8 bits) =
0110 0110

Mantissa (23 bits) =
100 1100 1100 1101 1100 1001


Decimal number -0.000 000 047 684 613 converted to 32 bit single precision IEEE 754 binary floating point representation:

1 - 0110 0110 - 100 1100 1100 1101 1100 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