-0.000 000 000 742 148 23 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 742 148 23(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
-0.000 000 000 742 148 23(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:

|-0.000 000 000 742 148 23| = 0.000 000 000 742 148 23


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: 0.000 000 000 742 148 23.

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) 0.000 000 000 742 148 23 × 2 = 0 + 0.000 000 001 484 296 46;
  • 2) 0.000 000 001 484 296 46 × 2 = 0 + 0.000 000 002 968 592 92;
  • 3) 0.000 000 002 968 592 92 × 2 = 0 + 0.000 000 005 937 185 84;
  • 4) 0.000 000 005 937 185 84 × 2 = 0 + 0.000 000 011 874 371 68;
  • 5) 0.000 000 011 874 371 68 × 2 = 0 + 0.000 000 023 748 743 36;
  • 6) 0.000 000 023 748 743 36 × 2 = 0 + 0.000 000 047 497 486 72;
  • 7) 0.000 000 047 497 486 72 × 2 = 0 + 0.000 000 094 994 973 44;
  • 8) 0.000 000 094 994 973 44 × 2 = 0 + 0.000 000 189 989 946 88;
  • 9) 0.000 000 189 989 946 88 × 2 = 0 + 0.000 000 379 979 893 76;
  • 10) 0.000 000 379 979 893 76 × 2 = 0 + 0.000 000 759 959 787 52;
  • 11) 0.000 000 759 959 787 52 × 2 = 0 + 0.000 001 519 919 575 04;
  • 12) 0.000 001 519 919 575 04 × 2 = 0 + 0.000 003 039 839 150 08;
  • 13) 0.000 003 039 839 150 08 × 2 = 0 + 0.000 006 079 678 300 16;
  • 14) 0.000 006 079 678 300 16 × 2 = 0 + 0.000 012 159 356 600 32;
  • 15) 0.000 012 159 356 600 32 × 2 = 0 + 0.000 024 318 713 200 64;
  • 16) 0.000 024 318 713 200 64 × 2 = 0 + 0.000 048 637 426 401 28;
  • 17) 0.000 048 637 426 401 28 × 2 = 0 + 0.000 097 274 852 802 56;
  • 18) 0.000 097 274 852 802 56 × 2 = 0 + 0.000 194 549 705 605 12;
  • 19) 0.000 194 549 705 605 12 × 2 = 0 + 0.000 389 099 411 210 24;
  • 20) 0.000 389 099 411 210 24 × 2 = 0 + 0.000 778 198 822 420 48;
  • 21) 0.000 778 198 822 420 48 × 2 = 0 + 0.001 556 397 644 840 96;
  • 22) 0.001 556 397 644 840 96 × 2 = 0 + 0.003 112 795 289 681 92;
  • 23) 0.003 112 795 289 681 92 × 2 = 0 + 0.006 225 590 579 363 84;
  • 24) 0.006 225 590 579 363 84 × 2 = 0 + 0.012 451 181 158 727 68;
  • 25) 0.012 451 181 158 727 68 × 2 = 0 + 0.024 902 362 317 455 36;
  • 26) 0.024 902 362 317 455 36 × 2 = 0 + 0.049 804 724 634 910 72;
  • 27) 0.049 804 724 634 910 72 × 2 = 0 + 0.099 609 449 269 821 44;
  • 28) 0.099 609 449 269 821 44 × 2 = 0 + 0.199 218 898 539 642 88;
  • 29) 0.199 218 898 539 642 88 × 2 = 0 + 0.398 437 797 079 285 76;
  • 30) 0.398 437 797 079 285 76 × 2 = 0 + 0.796 875 594 158 571 52;
  • 31) 0.796 875 594 158 571 52 × 2 = 1 + 0.593 751 188 317 143 04;
  • 32) 0.593 751 188 317 143 04 × 2 = 1 + 0.187 502 376 634 286 08;
  • 33) 0.187 502 376 634 286 08 × 2 = 0 + 0.375 004 753 268 572 16;
  • 34) 0.375 004 753 268 572 16 × 2 = 0 + 0.750 009 506 537 144 32;
  • 35) 0.750 009 506 537 144 32 × 2 = 1 + 0.500 019 013 074 288 64;
  • 36) 0.500 019 013 074 288 64 × 2 = 1 + 0.000 038 026 148 577 28;
  • 37) 0.000 038 026 148 577 28 × 2 = 0 + 0.000 076 052 297 154 56;
  • 38) 0.000 076 052 297 154 56 × 2 = 0 + 0.000 152 104 594 309 12;
  • 39) 0.000 152 104 594 309 12 × 2 = 0 + 0.000 304 209 188 618 24;
  • 40) 0.000 304 209 188 618 24 × 2 = 0 + 0.000 608 418 377 236 48;
  • 41) 0.000 608 418 377 236 48 × 2 = 0 + 0.001 216 836 754 472 96;
  • 42) 0.001 216 836 754 472 96 × 2 = 0 + 0.002 433 673 508 945 92;
  • 43) 0.002 433 673 508 945 92 × 2 = 0 + 0.004 867 347 017 891 84;
  • 44) 0.004 867 347 017 891 84 × 2 = 0 + 0.009 734 694 035 783 68;
  • 45) 0.009 734 694 035 783 68 × 2 = 0 + 0.019 469 388 071 567 36;
  • 46) 0.019 469 388 071 567 36 × 2 = 0 + 0.038 938 776 143 134 72;
  • 47) 0.038 938 776 143 134 72 × 2 = 0 + 0.077 877 552 286 269 44;
  • 48) 0.077 877 552 286 269 44 × 2 = 0 + 0.155 755 104 572 538 88;
  • 49) 0.155 755 104 572 538 88 × 2 = 0 + 0.311 510 209 145 077 76;
  • 50) 0.311 510 209 145 077 76 × 2 = 0 + 0.623 020 418 290 155 52;
  • 51) 0.623 020 418 290 155 52 × 2 = 1 + 0.246 040 836 580 311 04;
  • 52) 0.246 040 836 580 311 04 × 2 = 0 + 0.492 081 673 160 622 08;
  • 53) 0.492 081 673 160 622 08 × 2 = 0 + 0.984 163 346 321 244 16;
  • 54) 0.984 163 346 321 244 16 × 2 = 1 + 0.968 326 692 642 488 32;

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:


0.000 000 000 742 148 23(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0011 0000 0000 0000 0010 01(2)

6. Positive number before normalization:

0.000 000 000 742 148 23(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0011 0000 0000 0000 0010 01(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:


0.000 000 000 742 148 23(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0011 0000 0000 0000 0010 01(2) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0011 0000 0000 0000 0010 01(2) × 20 =

1.1001 1000 0000 0000 0001 001(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 1000 0000 0000 0001 001


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 1100 0000 0000 0000 1001 =

100 1100 0000 0000 0000 1001


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 1100 0000 0000 0000 1001


Decimal number -0.000 000 000 742 148 23 converted to 32 bit single precision IEEE 754 binary floating point representation:

1 - 0110 0000 - 100 1100 0000 0000 0000 1001

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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