Convert -0.000 352 859 497 070 312 7 to 32 Bit Single Precision IEEE 754 Binary Floating Point Standard, From a Base 10 Decimal Number

-0.000 352 859 497 070 312 7(10) to 32 bit single precision IEEE 754 binary floating point (1 bit for sign, 8 bits for exponent, 23 bits for mantissa) = ?

1. Start with the positive version of the number:

|-0.000 352 859 497 070 312 7| = 0.000 352 859 497 070 312 7

2. First, convert to the binary (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 the binary (base 2) the fractional part: 0.000 352 859 497 070 312 7.

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) 0.000 352 859 497 070 312 7 × 2 = 0 + 0.000 705 718 994 140 625 4;
  • 2) 0.000 705 718 994 140 625 4 × 2 = 0 + 0.001 411 437 988 281 250 8;
  • 3) 0.001 411 437 988 281 250 8 × 2 = 0 + 0.002 822 875 976 562 501 6;
  • 4) 0.002 822 875 976 562 501 6 × 2 = 0 + 0.005 645 751 953 125 003 2;
  • 5) 0.005 645 751 953 125 003 2 × 2 = 0 + 0.011 291 503 906 250 006 4;
  • 6) 0.011 291 503 906 250 006 4 × 2 = 0 + 0.022 583 007 812 500 012 8;
  • 7) 0.022 583 007 812 500 012 8 × 2 = 0 + 0.045 166 015 625 000 025 6;
  • 8) 0.045 166 015 625 000 025 6 × 2 = 0 + 0.090 332 031 250 000 051 2;
  • 9) 0.090 332 031 250 000 051 2 × 2 = 0 + 0.180 664 062 500 000 102 4;
  • 10) 0.180 664 062 500 000 102 4 × 2 = 0 + 0.361 328 125 000 000 204 8;
  • 11) 0.361 328 125 000 000 204 8 × 2 = 0 + 0.722 656 250 000 000 409 6;
  • 12) 0.722 656 250 000 000 409 6 × 2 = 1 + 0.445 312 500 000 000 819 2;
  • 13) 0.445 312 500 000 000 819 2 × 2 = 0 + 0.890 625 000 000 001 638 4;
  • 14) 0.890 625 000 000 001 638 4 × 2 = 1 + 0.781 250 000 000 003 276 8;
  • 15) 0.781 250 000 000 003 276 8 × 2 = 1 + 0.562 500 000 000 006 553 6;
  • 16) 0.562 500 000 000 006 553 6 × 2 = 1 + 0.125 000 000 000 013 107 2;
  • 17) 0.125 000 000 000 013 107 2 × 2 = 0 + 0.250 000 000 000 026 214 4;
  • 18) 0.250 000 000 000 026 214 4 × 2 = 0 + 0.500 000 000 000 052 428 8;
  • 19) 0.500 000 000 000 052 428 8 × 2 = 1 + 0.000 000 000 000 104 857 6;
  • 20) 0.000 000 000 000 104 857 6 × 2 = 0 + 0.000 000 000 000 209 715 2;
  • 21) 0.000 000 000 000 209 715 2 × 2 = 0 + 0.000 000 000 000 419 430 4;
  • 22) 0.000 000 000 000 419 430 4 × 2 = 0 + 0.000 000 000 000 838 860 8;
  • 23) 0.000 000 000 000 838 860 8 × 2 = 0 + 0.000 000 000 001 677 721 6;
  • 24) 0.000 000 000 001 677 721 6 × 2 = 0 + 0.000 000 000 003 355 443 2;
  • 25) 0.000 000 000 003 355 443 2 × 2 = 0 + 0.000 000 000 006 710 886 4;
  • 26) 0.000 000 000 006 710 886 4 × 2 = 0 + 0.000 000 000 013 421 772 8;
  • 27) 0.000 000 000 013 421 772 8 × 2 = 0 + 0.000 000 000 026 843 545 6;
  • 28) 0.000 000 000 026 843 545 6 × 2 = 0 + 0.000 000 000 053 687 091 2;
  • 29) 0.000 000 000 053 687 091 2 × 2 = 0 + 0.000 000 000 107 374 182 4;
  • 30) 0.000 000 000 107 374 182 4 × 2 = 0 + 0.000 000 000 214 748 364 8;
  • 31) 0.000 000 000 214 748 364 8 × 2 = 0 + 0.000 000 000 429 496 729 6;
  • 32) 0.000 000 000 429 496 729 6 × 2 = 0 + 0.000 000 000 858 993 459 2;
  • 33) 0.000 000 000 858 993 459 2 × 2 = 0 + 0.000 000 001 717 986 918 4;
  • 34) 0.000 000 001 717 986 918 4 × 2 = 0 + 0.000 000 003 435 973 836 8;
  • 35) 0.000 000 003 435 973 836 8 × 2 = 0 + 0.000 000 006 871 947 673 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...)


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:

0.000 352 859 497 070 312 7(10) =


0.0000 0000 0001 0111 0010 0000 0000 0000 000(2)


6. Positive number before normalization:

0.000 352 859 497 070 312 7(10) =


0.0000 0000 0001 0111 0010 0000 0000 0000 000(2)


7. Normalize the binary representation of the number.

Shift the decimal mark 12 positions to the right so that only one non zero digit remains to the left of it:

0.000 352 859 497 070 312 7(10) =


0.0000 0000 0001 0111 0010 0000 0000 0000 000(2) =


0.0000 0000 0001 0111 0010 0000 0000 0000 000(2) × 20 =


1.0111 0010 0000 0000 0000 000(2) × 2-12


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): -12


Mantissa (not normalized):
1.0111 0010 0000 0000 0000 000


9. Adjust the exponent.

Use the 8 bit excess/bias notation:

Exponent (adjusted) =


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


-12 + 2(8-1) - 1 =


(-12 + 127)(10) =


115(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;
  • 115 ÷ 2 = 57 + 1;
  • 57 ÷ 2 = 28 + 1;
  • 28 ÷ 2 = 14 + 0;
  • 14 ÷ 2 = 7 + 0;
  • 7 ÷ 2 = 3 + 1;
  • 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) =


115(10) =


0111 0011(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. 011 1001 0000 0000 0000 0000 =


011 1001 0000 0000 0000 0000


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) =
0111 0011


Mantissa (23 bits) =
011 1001 0000 0000 0000 0000


Number -0.000 352 859 497 070 312 7 converted from decimal system (base 10) to 32 bit single precision IEEE 754 binary floating point:
1 - 0111 0011 - 011 1001 0000 0000 0000 0000

(32 bits IEEE 754)
  • Sign (1 bit):

    • 1

      31
  • Exponent (8 bits):

    • 0

      30
    • 1

      29
    • 1

      28
    • 1

      27
    • 0

      26
    • 0

      25
    • 1

      24
    • 1

      23
  • Mantissa (23 bits):

    • 0

      22
    • 1

      21
    • 1

      20
    • 1

      19
    • 0

      18
    • 0

      17
    • 1

      16
    • 0

      15
    • 0

      14
    • 0

      13
    • 0

      12
    • 0

      11
    • 0

      10
    • 0

      9
    • 0

      8
    • 0

      7
    • 0

      6
    • 0

      5
    • 0

      4
    • 0

      3
    • 0

      2
    • 0

      1
    • 0

      0

More operations of this kind:

-0.000 352 859 497 070 312 8 = ? ... -0.000 352 859 497 070 312 6 = ?


Convert to 32 bit single precision IEEE 754 binary floating point standard

A number in 32 bit single precision IEEE 754 binary floating point standard representation requires three building elements: sign (it takes 1 bit and it's either 0 for positive or 1 for negative numbers), exponent (8 bits) and mantissa (23 bits)

Latest decimal numbers converted from base ten to 32 bit single precision IEEE 754 floating point binary standard representation

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All base ten decimal numbers converted to 32 bit single precision IEEE 754 binary floating point

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