-0.000 000 111 110 010 031 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 111 110 010 031(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
-0.000 000 111 110 010 031(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:

|-0.000 000 111 110 010 031| = 0.000 000 111 110 010 031


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: 0.000 000 111 110 010 031.

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) 0.000 000 111 110 010 031 × 2 = 0 + 0.000 000 222 220 020 062;
  • 2) 0.000 000 222 220 020 062 × 2 = 0 + 0.000 000 444 440 040 124;
  • 3) 0.000 000 444 440 040 124 × 2 = 0 + 0.000 000 888 880 080 248;
  • 4) 0.000 000 888 880 080 248 × 2 = 0 + 0.000 001 777 760 160 496;
  • 5) 0.000 001 777 760 160 496 × 2 = 0 + 0.000 003 555 520 320 992;
  • 6) 0.000 003 555 520 320 992 × 2 = 0 + 0.000 007 111 040 641 984;
  • 7) 0.000 007 111 040 641 984 × 2 = 0 + 0.000 014 222 081 283 968;
  • 8) 0.000 014 222 081 283 968 × 2 = 0 + 0.000 028 444 162 567 936;
  • 9) 0.000 028 444 162 567 936 × 2 = 0 + 0.000 056 888 325 135 872;
  • 10) 0.000 056 888 325 135 872 × 2 = 0 + 0.000 113 776 650 271 744;
  • 11) 0.000 113 776 650 271 744 × 2 = 0 + 0.000 227 553 300 543 488;
  • 12) 0.000 227 553 300 543 488 × 2 = 0 + 0.000 455 106 601 086 976;
  • 13) 0.000 455 106 601 086 976 × 2 = 0 + 0.000 910 213 202 173 952;
  • 14) 0.000 910 213 202 173 952 × 2 = 0 + 0.001 820 426 404 347 904;
  • 15) 0.001 820 426 404 347 904 × 2 = 0 + 0.003 640 852 808 695 808;
  • 16) 0.003 640 852 808 695 808 × 2 = 0 + 0.007 281 705 617 391 616;
  • 17) 0.007 281 705 617 391 616 × 2 = 0 + 0.014 563 411 234 783 232;
  • 18) 0.014 563 411 234 783 232 × 2 = 0 + 0.029 126 822 469 566 464;
  • 19) 0.029 126 822 469 566 464 × 2 = 0 + 0.058 253 644 939 132 928;
  • 20) 0.058 253 644 939 132 928 × 2 = 0 + 0.116 507 289 878 265 856;
  • 21) 0.116 507 289 878 265 856 × 2 = 0 + 0.233 014 579 756 531 712;
  • 22) 0.233 014 579 756 531 712 × 2 = 0 + 0.466 029 159 513 063 424;
  • 23) 0.466 029 159 513 063 424 × 2 = 0 + 0.932 058 319 026 126 848;
  • 24) 0.932 058 319 026 126 848 × 2 = 1 + 0.864 116 638 052 253 696;
  • 25) 0.864 116 638 052 253 696 × 2 = 1 + 0.728 233 276 104 507 392;
  • 26) 0.728 233 276 104 507 392 × 2 = 1 + 0.456 466 552 209 014 784;
  • 27) 0.456 466 552 209 014 784 × 2 = 0 + 0.912 933 104 418 029 568;
  • 28) 0.912 933 104 418 029 568 × 2 = 1 + 0.825 866 208 836 059 136;
  • 29) 0.825 866 208 836 059 136 × 2 = 1 + 0.651 732 417 672 118 272;
  • 30) 0.651 732 417 672 118 272 × 2 = 1 + 0.303 464 835 344 236 544;
  • 31) 0.303 464 835 344 236 544 × 2 = 0 + 0.606 929 670 688 473 088;
  • 32) 0.606 929 670 688 473 088 × 2 = 1 + 0.213 859 341 376 946 176;
  • 33) 0.213 859 341 376 946 176 × 2 = 0 + 0.427 718 682 753 892 352;
  • 34) 0.427 718 682 753 892 352 × 2 = 0 + 0.855 437 365 507 784 704;
  • 35) 0.855 437 365 507 784 704 × 2 = 1 + 0.710 874 731 015 569 408;
  • 36) 0.710 874 731 015 569 408 × 2 = 1 + 0.421 749 462 031 138 816;
  • 37) 0.421 749 462 031 138 816 × 2 = 0 + 0.843 498 924 062 277 632;
  • 38) 0.843 498 924 062 277 632 × 2 = 1 + 0.686 997 848 124 555 264;
  • 39) 0.686 997 848 124 555 264 × 2 = 1 + 0.373 995 696 249 110 528;
  • 40) 0.373 995 696 249 110 528 × 2 = 0 + 0.747 991 392 498 221 056;
  • 41) 0.747 991 392 498 221 056 × 2 = 1 + 0.495 982 784 996 442 112;
  • 42) 0.495 982 784 996 442 112 × 2 = 0 + 0.991 965 569 992 884 224;
  • 43) 0.991 965 569 992 884 224 × 2 = 1 + 0.983 931 139 985 768 448;
  • 44) 0.983 931 139 985 768 448 × 2 = 1 + 0.967 862 279 971 536 896;
  • 45) 0.967 862 279 971 536 896 × 2 = 1 + 0.935 724 559 943 073 792;
  • 46) 0.935 724 559 943 073 792 × 2 = 1 + 0.871 449 119 886 147 584;
  • 47) 0.871 449 119 886 147 584 × 2 = 1 + 0.742 898 239 772 295 168;

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:


0.000 000 111 110 010 031(10) =

0.0000 0000 0000 0000 0000 0001 1101 1101 0011 0110 1011 111(2)

6. Positive number before normalization:

0.000 000 111 110 010 031(10) =

0.0000 0000 0000 0000 0000 0001 1101 1101 0011 0110 1011 111(2)

7. Normalize the binary representation of the number.

Shift the decimal mark 24 positions to the right, so that only one non zero digit remains to the left of it:


0.000 000 111 110 010 031(10) =

0.0000 0000 0000 0000 0000 0001 1101 1101 0011 0110 1011 111(2) =

0.0000 0000 0000 0000 0000 0001 1101 1101 0011 0110 1011 111(2) × 20 =

1.1101 1101 0011 0110 1011 111(2) × 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 representation:

Sign 1 (a negative number)

Exponent (unadjusted): -24

Mantissa (not normalized):
1.1101 1101 0011 0110 1011 111


9. Adjust the exponent.

Use the 8 bit excess/bias notation:


Exponent (adjusted) =

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

-24 + 2(8-1) - 1 =

(-24 + 127)(10) =

103(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;
  • 103 ÷ 2 = 51 + 1;
  • 51 ÷ 2 = 25 + 1;
  • 25 ÷ 2 = 12 + 1;
  • 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) =

103(10) =

0110 0111(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 1110 1001 1011 0101 1111 =

110 1110 1001 1011 0101 1111


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 0111

Mantissa (23 bits) =
110 1110 1001 1011 0101 1111


Decimal number -0.000 000 111 110 010 031 converted to 32 bit single precision IEEE 754 binary floating point representation:

1 - 0110 0111 - 110 1110 1001 1011 0101 1111

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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