-0.000 000 000 079 354 13 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 079 354 13(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
-0.000 000 000 079 354 13(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:

|-0.000 000 000 079 354 13| = 0.000 000 000 079 354 13


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: 0.000 000 000 079 354 13.

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) 0.000 000 000 079 354 13 × 2 = 0 + 0.000 000 000 158 708 26;
  • 2) 0.000 000 000 158 708 26 × 2 = 0 + 0.000 000 000 317 416 52;
  • 3) 0.000 000 000 317 416 52 × 2 = 0 + 0.000 000 000 634 833 04;
  • 4) 0.000 000 000 634 833 04 × 2 = 0 + 0.000 000 001 269 666 08;
  • 5) 0.000 000 001 269 666 08 × 2 = 0 + 0.000 000 002 539 332 16;
  • 6) 0.000 000 002 539 332 16 × 2 = 0 + 0.000 000 005 078 664 32;
  • 7) 0.000 000 005 078 664 32 × 2 = 0 + 0.000 000 010 157 328 64;
  • 8) 0.000 000 010 157 328 64 × 2 = 0 + 0.000 000 020 314 657 28;
  • 9) 0.000 000 020 314 657 28 × 2 = 0 + 0.000 000 040 629 314 56;
  • 10) 0.000 000 040 629 314 56 × 2 = 0 + 0.000 000 081 258 629 12;
  • 11) 0.000 000 081 258 629 12 × 2 = 0 + 0.000 000 162 517 258 24;
  • 12) 0.000 000 162 517 258 24 × 2 = 0 + 0.000 000 325 034 516 48;
  • 13) 0.000 000 325 034 516 48 × 2 = 0 + 0.000 000 650 069 032 96;
  • 14) 0.000 000 650 069 032 96 × 2 = 0 + 0.000 001 300 138 065 92;
  • 15) 0.000 001 300 138 065 92 × 2 = 0 + 0.000 002 600 276 131 84;
  • 16) 0.000 002 600 276 131 84 × 2 = 0 + 0.000 005 200 552 263 68;
  • 17) 0.000 005 200 552 263 68 × 2 = 0 + 0.000 010 401 104 527 36;
  • 18) 0.000 010 401 104 527 36 × 2 = 0 + 0.000 020 802 209 054 72;
  • 19) 0.000 020 802 209 054 72 × 2 = 0 + 0.000 041 604 418 109 44;
  • 20) 0.000 041 604 418 109 44 × 2 = 0 + 0.000 083 208 836 218 88;
  • 21) 0.000 083 208 836 218 88 × 2 = 0 + 0.000 166 417 672 437 76;
  • 22) 0.000 166 417 672 437 76 × 2 = 0 + 0.000 332 835 344 875 52;
  • 23) 0.000 332 835 344 875 52 × 2 = 0 + 0.000 665 670 689 751 04;
  • 24) 0.000 665 670 689 751 04 × 2 = 0 + 0.001 331 341 379 502 08;
  • 25) 0.001 331 341 379 502 08 × 2 = 0 + 0.002 662 682 759 004 16;
  • 26) 0.002 662 682 759 004 16 × 2 = 0 + 0.005 325 365 518 008 32;
  • 27) 0.005 325 365 518 008 32 × 2 = 0 + 0.010 650 731 036 016 64;
  • 28) 0.010 650 731 036 016 64 × 2 = 0 + 0.021 301 462 072 033 28;
  • 29) 0.021 301 462 072 033 28 × 2 = 0 + 0.042 602 924 144 066 56;
  • 30) 0.042 602 924 144 066 56 × 2 = 0 + 0.085 205 848 288 133 12;
  • 31) 0.085 205 848 288 133 12 × 2 = 0 + 0.170 411 696 576 266 24;
  • 32) 0.170 411 696 576 266 24 × 2 = 0 + 0.340 823 393 152 532 48;
  • 33) 0.340 823 393 152 532 48 × 2 = 0 + 0.681 646 786 305 064 96;
  • 34) 0.681 646 786 305 064 96 × 2 = 1 + 0.363 293 572 610 129 92;
  • 35) 0.363 293 572 610 129 92 × 2 = 0 + 0.726 587 145 220 259 84;
  • 36) 0.726 587 145 220 259 84 × 2 = 1 + 0.453 174 290 440 519 68;
  • 37) 0.453 174 290 440 519 68 × 2 = 0 + 0.906 348 580 881 039 36;
  • 38) 0.906 348 580 881 039 36 × 2 = 1 + 0.812 697 161 762 078 72;
  • 39) 0.812 697 161 762 078 72 × 2 = 1 + 0.625 394 323 524 157 44;
  • 40) 0.625 394 323 524 157 44 × 2 = 1 + 0.250 788 647 048 314 88;
  • 41) 0.250 788 647 048 314 88 × 2 = 0 + 0.501 577 294 096 629 76;
  • 42) 0.501 577 294 096 629 76 × 2 = 1 + 0.003 154 588 193 259 52;
  • 43) 0.003 154 588 193 259 52 × 2 = 0 + 0.006 309 176 386 519 04;
  • 44) 0.006 309 176 386 519 04 × 2 = 0 + 0.012 618 352 773 038 08;
  • 45) 0.012 618 352 773 038 08 × 2 = 0 + 0.025 236 705 546 076 16;
  • 46) 0.025 236 705 546 076 16 × 2 = 0 + 0.050 473 411 092 152 32;
  • 47) 0.050 473 411 092 152 32 × 2 = 0 + 0.100 946 822 184 304 64;
  • 48) 0.100 946 822 184 304 64 × 2 = 0 + 0.201 893 644 368 609 28;
  • 49) 0.201 893 644 368 609 28 × 2 = 0 + 0.403 787 288 737 218 56;
  • 50) 0.403 787 288 737 218 56 × 2 = 0 + 0.807 574 577 474 437 12;
  • 51) 0.807 574 577 474 437 12 × 2 = 1 + 0.615 149 154 948 874 24;
  • 52) 0.615 149 154 948 874 24 × 2 = 1 + 0.230 298 309 897 748 48;
  • 53) 0.230 298 309 897 748 48 × 2 = 0 + 0.460 596 619 795 496 96;
  • 54) 0.460 596 619 795 496 96 × 2 = 0 + 0.921 193 239 590 993 92;
  • 55) 0.921 193 239 590 993 92 × 2 = 1 + 0.842 386 479 181 987 84;
  • 56) 0.842 386 479 181 987 84 × 2 = 1 + 0.684 772 958 363 975 68;
  • 57) 0.684 772 958 363 975 68 × 2 = 1 + 0.369 545 916 727 951 36;

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:


0.000 000 000 079 354 13(10) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0101 0111 0100 0000 0011 0011 1(2)

6. Positive number before normalization:

0.000 000 000 079 354 13(10) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0101 0111 0100 0000 0011 0011 1(2)

7. Normalize the binary representation of the number.

Shift the decimal mark 34 positions to the right, so that only one non zero digit remains to the left of it:


0.000 000 000 079 354 13(10) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0101 0111 0100 0000 0011 0011 1(2) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0101 0111 0100 0000 0011 0011 1(2) × 20 =

1.0101 1101 0000 0000 1100 111(2) × 2-34


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): -34

Mantissa (not normalized):
1.0101 1101 0000 0000 1100 111


9. Adjust the exponent.

Use the 8 bit excess/bias notation:


Exponent (adjusted) =

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

-34 + 2(8-1) - 1 =

(-34 + 127)(10) =

93(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;
  • 93 ÷ 2 = 46 + 1;
  • 46 ÷ 2 = 23 + 0;
  • 23 ÷ 2 = 11 + 1;
  • 11 ÷ 2 = 5 + 1;
  • 5 ÷ 2 = 2 + 1;
  • 2 ÷ 2 = 1 + 0;
  • 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) =

93(10) =

0101 1101(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. 010 1110 1000 0000 0110 0111 =

010 1110 1000 0000 0110 0111


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) =
0101 1101

Mantissa (23 bits) =
010 1110 1000 0000 0110 0111


Decimal number -0.000 000 000 079 354 13 converted to 32 bit single precision IEEE 754 binary floating point representation:

1 - 0101 1101 - 010 1110 1000 0000 0110 0111

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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