-0.000 000 000 839 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 839(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
-0.000 000 000 839(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:

|-0.000 000 000 839| = 0.000 000 000 839


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: 0.000 000 000 839.

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) 0.000 000 000 839 × 2 = 0 + 0.000 000 001 678;
  • 2) 0.000 000 001 678 × 2 = 0 + 0.000 000 003 356;
  • 3) 0.000 000 003 356 × 2 = 0 + 0.000 000 006 712;
  • 4) 0.000 000 006 712 × 2 = 0 + 0.000 000 013 424;
  • 5) 0.000 000 013 424 × 2 = 0 + 0.000 000 026 848;
  • 6) 0.000 000 026 848 × 2 = 0 + 0.000 000 053 696;
  • 7) 0.000 000 053 696 × 2 = 0 + 0.000 000 107 392;
  • 8) 0.000 000 107 392 × 2 = 0 + 0.000 000 214 784;
  • 9) 0.000 000 214 784 × 2 = 0 + 0.000 000 429 568;
  • 10) 0.000 000 429 568 × 2 = 0 + 0.000 000 859 136;
  • 11) 0.000 000 859 136 × 2 = 0 + 0.000 001 718 272;
  • 12) 0.000 001 718 272 × 2 = 0 + 0.000 003 436 544;
  • 13) 0.000 003 436 544 × 2 = 0 + 0.000 006 873 088;
  • 14) 0.000 006 873 088 × 2 = 0 + 0.000 013 746 176;
  • 15) 0.000 013 746 176 × 2 = 0 + 0.000 027 492 352;
  • 16) 0.000 027 492 352 × 2 = 0 + 0.000 054 984 704;
  • 17) 0.000 054 984 704 × 2 = 0 + 0.000 109 969 408;
  • 18) 0.000 109 969 408 × 2 = 0 + 0.000 219 938 816;
  • 19) 0.000 219 938 816 × 2 = 0 + 0.000 439 877 632;
  • 20) 0.000 439 877 632 × 2 = 0 + 0.000 879 755 264;
  • 21) 0.000 879 755 264 × 2 = 0 + 0.001 759 510 528;
  • 22) 0.001 759 510 528 × 2 = 0 + 0.003 519 021 056;
  • 23) 0.003 519 021 056 × 2 = 0 + 0.007 038 042 112;
  • 24) 0.007 038 042 112 × 2 = 0 + 0.014 076 084 224;
  • 25) 0.014 076 084 224 × 2 = 0 + 0.028 152 168 448;
  • 26) 0.028 152 168 448 × 2 = 0 + 0.056 304 336 896;
  • 27) 0.056 304 336 896 × 2 = 0 + 0.112 608 673 792;
  • 28) 0.112 608 673 792 × 2 = 0 + 0.225 217 347 584;
  • 29) 0.225 217 347 584 × 2 = 0 + 0.450 434 695 168;
  • 30) 0.450 434 695 168 × 2 = 0 + 0.900 869 390 336;
  • 31) 0.900 869 390 336 × 2 = 1 + 0.801 738 780 672;
  • 32) 0.801 738 780 672 × 2 = 1 + 0.603 477 561 344;
  • 33) 0.603 477 561 344 × 2 = 1 + 0.206 955 122 688;
  • 34) 0.206 955 122 688 × 2 = 0 + 0.413 910 245 376;
  • 35) 0.413 910 245 376 × 2 = 0 + 0.827 820 490 752;
  • 36) 0.827 820 490 752 × 2 = 1 + 0.655 640 981 504;
  • 37) 0.655 640 981 504 × 2 = 1 + 0.311 281 963 008;
  • 38) 0.311 281 963 008 × 2 = 0 + 0.622 563 926 016;
  • 39) 0.622 563 926 016 × 2 = 1 + 0.245 127 852 032;
  • 40) 0.245 127 852 032 × 2 = 0 + 0.490 255 704 064;
  • 41) 0.490 255 704 064 × 2 = 0 + 0.980 511 408 128;
  • 42) 0.980 511 408 128 × 2 = 1 + 0.961 022 816 256;
  • 43) 0.961 022 816 256 × 2 = 1 + 0.922 045 632 512;
  • 44) 0.922 045 632 512 × 2 = 1 + 0.844 091 265 024;
  • 45) 0.844 091 265 024 × 2 = 1 + 0.688 182 530 048;
  • 46) 0.688 182 530 048 × 2 = 1 + 0.376 365 060 096;
  • 47) 0.376 365 060 096 × 2 = 0 + 0.752 730 120 192;
  • 48) 0.752 730 120 192 × 2 = 1 + 0.505 460 240 384;
  • 49) 0.505 460 240 384 × 2 = 1 + 0.010 920 480 768;
  • 50) 0.010 920 480 768 × 2 = 0 + 0.021 840 961 536;
  • 51) 0.021 840 961 536 × 2 = 0 + 0.043 681 923 072;
  • 52) 0.043 681 923 072 × 2 = 0 + 0.087 363 846 144;
  • 53) 0.087 363 846 144 × 2 = 0 + 0.174 727 692 288;
  • 54) 0.174 727 692 288 × 2 = 0 + 0.349 455 384 576;

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:


0.000 000 000 839(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 1001 1010 0111 1101 1000 00(2)

6. Positive number before normalization:

0.000 000 000 839(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 1001 1010 0111 1101 1000 00(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:


0.000 000 000 839(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 1001 1010 0111 1101 1000 00(2) =

0.0000 0000 0000 0000 0000 0000 0000 0011 1001 1010 0111 1101 1000 00(2) × 20 =

1.1100 1101 0011 1110 1100 000(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):
1.1100 1101 0011 1110 1100 000


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. 110 0110 1001 1111 0110 0000 =

110 0110 1001 1111 0110 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) =
0110 0000

Mantissa (23 bits) =
110 0110 1001 1111 0110 0000


Decimal number -0.000 000 000 839 converted to 32 bit single precision IEEE 754 binary floating point representation:

1 - 0110 0000 - 110 0110 1001 1111 0110 0000

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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