-0.000 001 457 896 927 4 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 001 457 896 927 4(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
-0.000 001 457 896 927 4(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:

|-0.000 001 457 896 927 4| = 0.000 001 457 896 927 4


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: 0.000 001 457 896 927 4.

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) 0.000 001 457 896 927 4 × 2 = 0 + 0.000 002 915 793 854 8;
  • 2) 0.000 002 915 793 854 8 × 2 = 0 + 0.000 005 831 587 709 6;
  • 3) 0.000 005 831 587 709 6 × 2 = 0 + 0.000 011 663 175 419 2;
  • 4) 0.000 011 663 175 419 2 × 2 = 0 + 0.000 023 326 350 838 4;
  • 5) 0.000 023 326 350 838 4 × 2 = 0 + 0.000 046 652 701 676 8;
  • 6) 0.000 046 652 701 676 8 × 2 = 0 + 0.000 093 305 403 353 6;
  • 7) 0.000 093 305 403 353 6 × 2 = 0 + 0.000 186 610 806 707 2;
  • 8) 0.000 186 610 806 707 2 × 2 = 0 + 0.000 373 221 613 414 4;
  • 9) 0.000 373 221 613 414 4 × 2 = 0 + 0.000 746 443 226 828 8;
  • 10) 0.000 746 443 226 828 8 × 2 = 0 + 0.001 492 886 453 657 6;
  • 11) 0.001 492 886 453 657 6 × 2 = 0 + 0.002 985 772 907 315 2;
  • 12) 0.002 985 772 907 315 2 × 2 = 0 + 0.005 971 545 814 630 4;
  • 13) 0.005 971 545 814 630 4 × 2 = 0 + 0.011 943 091 629 260 8;
  • 14) 0.011 943 091 629 260 8 × 2 = 0 + 0.023 886 183 258 521 6;
  • 15) 0.023 886 183 258 521 6 × 2 = 0 + 0.047 772 366 517 043 2;
  • 16) 0.047 772 366 517 043 2 × 2 = 0 + 0.095 544 733 034 086 4;
  • 17) 0.095 544 733 034 086 4 × 2 = 0 + 0.191 089 466 068 172 8;
  • 18) 0.191 089 466 068 172 8 × 2 = 0 + 0.382 178 932 136 345 6;
  • 19) 0.382 178 932 136 345 6 × 2 = 0 + 0.764 357 864 272 691 2;
  • 20) 0.764 357 864 272 691 2 × 2 = 1 + 0.528 715 728 545 382 4;
  • 21) 0.528 715 728 545 382 4 × 2 = 1 + 0.057 431 457 090 764 8;
  • 22) 0.057 431 457 090 764 8 × 2 = 0 + 0.114 862 914 181 529 6;
  • 23) 0.114 862 914 181 529 6 × 2 = 0 + 0.229 725 828 363 059 2;
  • 24) 0.229 725 828 363 059 2 × 2 = 0 + 0.459 451 656 726 118 4;
  • 25) 0.459 451 656 726 118 4 × 2 = 0 + 0.918 903 313 452 236 8;
  • 26) 0.918 903 313 452 236 8 × 2 = 1 + 0.837 806 626 904 473 6;
  • 27) 0.837 806 626 904 473 6 × 2 = 1 + 0.675 613 253 808 947 2;
  • 28) 0.675 613 253 808 947 2 × 2 = 1 + 0.351 226 507 617 894 4;
  • 29) 0.351 226 507 617 894 4 × 2 = 0 + 0.702 453 015 235 788 8;
  • 30) 0.702 453 015 235 788 8 × 2 = 1 + 0.404 906 030 471 577 6;
  • 31) 0.404 906 030 471 577 6 × 2 = 0 + 0.809 812 060 943 155 2;
  • 32) 0.809 812 060 943 155 2 × 2 = 1 + 0.619 624 121 886 310 4;
  • 33) 0.619 624 121 886 310 4 × 2 = 1 + 0.239 248 243 772 620 8;
  • 34) 0.239 248 243 772 620 8 × 2 = 0 + 0.478 496 487 545 241 6;
  • 35) 0.478 496 487 545 241 6 × 2 = 0 + 0.956 992 975 090 483 2;
  • 36) 0.956 992 975 090 483 2 × 2 = 1 + 0.913 985 950 180 966 4;
  • 37) 0.913 985 950 180 966 4 × 2 = 1 + 0.827 971 900 361 932 8;
  • 38) 0.827 971 900 361 932 8 × 2 = 1 + 0.655 943 800 723 865 6;
  • 39) 0.655 943 800 723 865 6 × 2 = 1 + 0.311 887 601 447 731 2;
  • 40) 0.311 887 601 447 731 2 × 2 = 0 + 0.623 775 202 895 462 4;
  • 41) 0.623 775 202 895 462 4 × 2 = 1 + 0.247 550 405 790 924 8;
  • 42) 0.247 550 405 790 924 8 × 2 = 0 + 0.495 100 811 581 849 6;
  • 43) 0.495 100 811 581 849 6 × 2 = 0 + 0.990 201 623 163 699 2;

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:


0.000 001 457 896 927 4(10) =

0.0000 0000 0000 0000 0001 1000 0111 0101 1001 1110 100(2)

6. Positive number before normalization:

0.000 001 457 896 927 4(10) =

0.0000 0000 0000 0000 0001 1000 0111 0101 1001 1110 100(2)

7. Normalize the binary representation of the number.

Shift the decimal mark 20 positions to the right, so that only one non zero digit remains to the left of it:


0.000 001 457 896 927 4(10) =

0.0000 0000 0000 0000 0001 1000 0111 0101 1001 1110 100(2) =

0.0000 0000 0000 0000 0001 1000 0111 0101 1001 1110 100(2) × 20 =

1.1000 0111 0101 1001 1110 100(2) × 2-20


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): -20

Mantissa (not normalized):
1.1000 0111 0101 1001 1110 100


9. Adjust the exponent.

Use the 8 bit excess/bias notation:


Exponent (adjusted) =

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

-20 + 2(8-1) - 1 =

(-20 + 127)(10) =

107(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;
  • 107 ÷ 2 = 53 + 1;
  • 53 ÷ 2 = 26 + 1;
  • 26 ÷ 2 = 13 + 0;
  • 13 ÷ 2 = 6 + 1;
  • 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) =

107(10) =

0110 1011(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 0011 1010 1100 1111 0100 =

100 0011 1010 1100 1111 0100


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 1011

Mantissa (23 bits) =
100 0011 1010 1100 1111 0100


Decimal number -0.000 001 457 896 927 4 converted to 32 bit single precision IEEE 754 binary floating point representation:

1 - 0110 1011 - 100 0011 1010 1100 1111 0100

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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