-0.000 000 000 739 4 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 739 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
-0.000 000 000 739 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:

|-0.000 000 000 739 4| = 0.000 000 000 739 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: 0.000 000 000 739 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) 0.000 000 000 739 4 × 2 = 0 + 0.000 000 001 478 8;
  • 2) 0.000 000 001 478 8 × 2 = 0 + 0.000 000 002 957 6;
  • 3) 0.000 000 002 957 6 × 2 = 0 + 0.000 000 005 915 2;
  • 4) 0.000 000 005 915 2 × 2 = 0 + 0.000 000 011 830 4;
  • 5) 0.000 000 011 830 4 × 2 = 0 + 0.000 000 023 660 8;
  • 6) 0.000 000 023 660 8 × 2 = 0 + 0.000 000 047 321 6;
  • 7) 0.000 000 047 321 6 × 2 = 0 + 0.000 000 094 643 2;
  • 8) 0.000 000 094 643 2 × 2 = 0 + 0.000 000 189 286 4;
  • 9) 0.000 000 189 286 4 × 2 = 0 + 0.000 000 378 572 8;
  • 10) 0.000 000 378 572 8 × 2 = 0 + 0.000 000 757 145 6;
  • 11) 0.000 000 757 145 6 × 2 = 0 + 0.000 001 514 291 2;
  • 12) 0.000 001 514 291 2 × 2 = 0 + 0.000 003 028 582 4;
  • 13) 0.000 003 028 582 4 × 2 = 0 + 0.000 006 057 164 8;
  • 14) 0.000 006 057 164 8 × 2 = 0 + 0.000 012 114 329 6;
  • 15) 0.000 012 114 329 6 × 2 = 0 + 0.000 024 228 659 2;
  • 16) 0.000 024 228 659 2 × 2 = 0 + 0.000 048 457 318 4;
  • 17) 0.000 048 457 318 4 × 2 = 0 + 0.000 096 914 636 8;
  • 18) 0.000 096 914 636 8 × 2 = 0 + 0.000 193 829 273 6;
  • 19) 0.000 193 829 273 6 × 2 = 0 + 0.000 387 658 547 2;
  • 20) 0.000 387 658 547 2 × 2 = 0 + 0.000 775 317 094 4;
  • 21) 0.000 775 317 094 4 × 2 = 0 + 0.001 550 634 188 8;
  • 22) 0.001 550 634 188 8 × 2 = 0 + 0.003 101 268 377 6;
  • 23) 0.003 101 268 377 6 × 2 = 0 + 0.006 202 536 755 2;
  • 24) 0.006 202 536 755 2 × 2 = 0 + 0.012 405 073 510 4;
  • 25) 0.012 405 073 510 4 × 2 = 0 + 0.024 810 147 020 8;
  • 26) 0.024 810 147 020 8 × 2 = 0 + 0.049 620 294 041 6;
  • 27) 0.049 620 294 041 6 × 2 = 0 + 0.099 240 588 083 2;
  • 28) 0.099 240 588 083 2 × 2 = 0 + 0.198 481 176 166 4;
  • 29) 0.198 481 176 166 4 × 2 = 0 + 0.396 962 352 332 8;
  • 30) 0.396 962 352 332 8 × 2 = 0 + 0.793 924 704 665 6;
  • 31) 0.793 924 704 665 6 × 2 = 1 + 0.587 849 409 331 2;
  • 32) 0.587 849 409 331 2 × 2 = 1 + 0.175 698 818 662 4;
  • 33) 0.175 698 818 662 4 × 2 = 0 + 0.351 397 637 324 8;
  • 34) 0.351 397 637 324 8 × 2 = 0 + 0.702 795 274 649 6;
  • 35) 0.702 795 274 649 6 × 2 = 1 + 0.405 590 549 299 2;
  • 36) 0.405 590 549 299 2 × 2 = 0 + 0.811 181 098 598 4;
  • 37) 0.811 181 098 598 4 × 2 = 1 + 0.622 362 197 196 8;
  • 38) 0.622 362 197 196 8 × 2 = 1 + 0.244 724 394 393 6;
  • 39) 0.244 724 394 393 6 × 2 = 0 + 0.489 448 788 787 2;
  • 40) 0.489 448 788 787 2 × 2 = 0 + 0.978 897 577 574 4;
  • 41) 0.978 897 577 574 4 × 2 = 1 + 0.957 795 155 148 8;
  • 42) 0.957 795 155 148 8 × 2 = 1 + 0.915 590 310 297 6;
  • 43) 0.915 590 310 297 6 × 2 = 1 + 0.831 180 620 595 2;
  • 44) 0.831 180 620 595 2 × 2 = 1 + 0.662 361 241 190 4;
  • 45) 0.662 361 241 190 4 × 2 = 1 + 0.324 722 482 380 8;
  • 46) 0.324 722 482 380 8 × 2 = 0 + 0.649 444 964 761 6;
  • 47) 0.649 444 964 761 6 × 2 = 1 + 0.298 889 929 523 2;
  • 48) 0.298 889 929 523 2 × 2 = 0 + 0.597 779 859 046 4;
  • 49) 0.597 779 859 046 4 × 2 = 1 + 0.195 559 718 092 8;
  • 50) 0.195 559 718 092 8 × 2 = 0 + 0.391 119 436 185 6;
  • 51) 0.391 119 436 185 6 × 2 = 0 + 0.782 238 872 371 2;
  • 52) 0.782 238 872 371 2 × 2 = 1 + 0.564 477 744 742 4;
  • 53) 0.564 477 744 742 4 × 2 = 1 + 0.128 955 489 484 8;
  • 54) 0.128 955 489 484 8 × 2 = 0 + 0.257 910 978 969 6;

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:


0.000 000 000 739 4(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0010 1100 1111 1010 1001 10(2)

6. Positive number before normalization:

0.000 000 000 739 4(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0010 1100 1111 1010 1001 10(2)

7. Normalize the binary representation of the number.

Shift the decimal mark 31 positions to the right, so that only one non zero digit remains to the left of it:


0.000 000 000 739 4(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0010 1100 1111 1010 1001 10(2) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0010 1100 1111 1010 1001 10(2) × 20 =

1.1001 0110 0111 1101 0100 110(2) × 2-31


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): -31

Mantissa (not normalized):
1.1001 0110 0111 1101 0100 110


9. Adjust the exponent.

Use the 8 bit excess/bias notation:


Exponent (adjusted) =

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

-31 + 2(8-1) - 1 =

(-31 + 127)(10) =

96(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;
  • 96 ÷ 2 = 48 + 0;
  • 48 ÷ 2 = 24 + 0;
  • 24 ÷ 2 = 12 + 0;
  • 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) =

96(10) =

0110 0000(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 1011 0011 1110 1010 0110 =

100 1011 0011 1110 1010 0110


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 0000

Mantissa (23 bits) =
100 1011 0011 1110 1010 0110


Decimal number -0.000 000 000 739 4 converted to 32 bit single precision IEEE 754 binary floating point representation:

1 - 0110 0000 - 100 1011 0011 1110 1010 0110

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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