-0.000 000 000 742 148 24 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 742 148 24(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 24(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 24| = 0.000 000 000 742 148 24


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 24.

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 24 × 2 = 0 + 0.000 000 001 484 296 48;
  • 2) 0.000 000 001 484 296 48 × 2 = 0 + 0.000 000 002 968 592 96;
  • 3) 0.000 000 002 968 592 96 × 2 = 0 + 0.000 000 005 937 185 92;
  • 4) 0.000 000 005 937 185 92 × 2 = 0 + 0.000 000 011 874 371 84;
  • 5) 0.000 000 011 874 371 84 × 2 = 0 + 0.000 000 023 748 743 68;
  • 6) 0.000 000 023 748 743 68 × 2 = 0 + 0.000 000 047 497 487 36;
  • 7) 0.000 000 047 497 487 36 × 2 = 0 + 0.000 000 094 994 974 72;
  • 8) 0.000 000 094 994 974 72 × 2 = 0 + 0.000 000 189 989 949 44;
  • 9) 0.000 000 189 989 949 44 × 2 = 0 + 0.000 000 379 979 898 88;
  • 10) 0.000 000 379 979 898 88 × 2 = 0 + 0.000 000 759 959 797 76;
  • 11) 0.000 000 759 959 797 76 × 2 = 0 + 0.000 001 519 919 595 52;
  • 12) 0.000 001 519 919 595 52 × 2 = 0 + 0.000 003 039 839 191 04;
  • 13) 0.000 003 039 839 191 04 × 2 = 0 + 0.000 006 079 678 382 08;
  • 14) 0.000 006 079 678 382 08 × 2 = 0 + 0.000 012 159 356 764 16;
  • 15) 0.000 012 159 356 764 16 × 2 = 0 + 0.000 024 318 713 528 32;
  • 16) 0.000 024 318 713 528 32 × 2 = 0 + 0.000 048 637 427 056 64;
  • 17) 0.000 048 637 427 056 64 × 2 = 0 + 0.000 097 274 854 113 28;
  • 18) 0.000 097 274 854 113 28 × 2 = 0 + 0.000 194 549 708 226 56;
  • 19) 0.000 194 549 708 226 56 × 2 = 0 + 0.000 389 099 416 453 12;
  • 20) 0.000 389 099 416 453 12 × 2 = 0 + 0.000 778 198 832 906 24;
  • 21) 0.000 778 198 832 906 24 × 2 = 0 + 0.001 556 397 665 812 48;
  • 22) 0.001 556 397 665 812 48 × 2 = 0 + 0.003 112 795 331 624 96;
  • 23) 0.003 112 795 331 624 96 × 2 = 0 + 0.006 225 590 663 249 92;
  • 24) 0.006 225 590 663 249 92 × 2 = 0 + 0.012 451 181 326 499 84;
  • 25) 0.012 451 181 326 499 84 × 2 = 0 + 0.024 902 362 652 999 68;
  • 26) 0.024 902 362 652 999 68 × 2 = 0 + 0.049 804 725 305 999 36;
  • 27) 0.049 804 725 305 999 36 × 2 = 0 + 0.099 609 450 611 998 72;
  • 28) 0.099 609 450 611 998 72 × 2 = 0 + 0.199 218 901 223 997 44;
  • 29) 0.199 218 901 223 997 44 × 2 = 0 + 0.398 437 802 447 994 88;
  • 30) 0.398 437 802 447 994 88 × 2 = 0 + 0.796 875 604 895 989 76;
  • 31) 0.796 875 604 895 989 76 × 2 = 1 + 0.593 751 209 791 979 52;
  • 32) 0.593 751 209 791 979 52 × 2 = 1 + 0.187 502 419 583 959 04;
  • 33) 0.187 502 419 583 959 04 × 2 = 0 + 0.375 004 839 167 918 08;
  • 34) 0.375 004 839 167 918 08 × 2 = 0 + 0.750 009 678 335 836 16;
  • 35) 0.750 009 678 335 836 16 × 2 = 1 + 0.500 019 356 671 672 32;
  • 36) 0.500 019 356 671 672 32 × 2 = 1 + 0.000 038 713 343 344 64;
  • 37) 0.000 038 713 343 344 64 × 2 = 0 + 0.000 077 426 686 689 28;
  • 38) 0.000 077 426 686 689 28 × 2 = 0 + 0.000 154 853 373 378 56;
  • 39) 0.000 154 853 373 378 56 × 2 = 0 + 0.000 309 706 746 757 12;
  • 40) 0.000 309 706 746 757 12 × 2 = 0 + 0.000 619 413 493 514 24;
  • 41) 0.000 619 413 493 514 24 × 2 = 0 + 0.001 238 826 987 028 48;
  • 42) 0.001 238 826 987 028 48 × 2 = 0 + 0.002 477 653 974 056 96;
  • 43) 0.002 477 653 974 056 96 × 2 = 0 + 0.004 955 307 948 113 92;
  • 44) 0.004 955 307 948 113 92 × 2 = 0 + 0.009 910 615 896 227 84;
  • 45) 0.009 910 615 896 227 84 × 2 = 0 + 0.019 821 231 792 455 68;
  • 46) 0.019 821 231 792 455 68 × 2 = 0 + 0.039 642 463 584 911 36;
  • 47) 0.039 642 463 584 911 36 × 2 = 0 + 0.079 284 927 169 822 72;
  • 48) 0.079 284 927 169 822 72 × 2 = 0 + 0.158 569 854 339 645 44;
  • 49) 0.158 569 854 339 645 44 × 2 = 0 + 0.317 139 708 679 290 88;
  • 50) 0.317 139 708 679 290 88 × 2 = 0 + 0.634 279 417 358 581 76;
  • 51) 0.634 279 417 358 581 76 × 2 = 1 + 0.268 558 834 717 163 52;
  • 52) 0.268 558 834 717 163 52 × 2 = 0 + 0.537 117 669 434 327 04;
  • 53) 0.537 117 669 434 327 04 × 2 = 1 + 0.074 235 338 868 654 08;
  • 54) 0.074 235 338 868 654 08 × 2 = 0 + 0.148 470 677 737 308 16;

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 24(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0011 0000 0000 0000 0010 10(2)

6. Positive number before normalization:

0.000 000 000 742 148 24(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0011 0000 0000 0000 0010 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:


0.000 000 000 742 148 24(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0011 0000 0000 0000 0010 10(2) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0011 0000 0000 0000 0010 10(2) × 20 =

1.1001 1000 0000 0000 0001 010(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 010


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 1010 =

100 1100 0000 0000 0000 1010


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 1010


Decimal number -0.000 000 000 742 148 24 converted to 32 bit single precision IEEE 754 binary floating point representation:

1 - 0110 0000 - 100 1100 0000 0000 0000 1010

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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