-0.000 000 000 742 147 673 25 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 742 147 673 25(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
-0.000 000 000 742 147 673 25(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:

|-0.000 000 000 742 147 673 25| = 0.000 000 000 742 147 673 25


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: 0.000 000 000 742 147 673 25.

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) 0.000 000 000 742 147 673 25 × 2 = 0 + 0.000 000 001 484 295 346 5;
  • 2) 0.000 000 001 484 295 346 5 × 2 = 0 + 0.000 000 002 968 590 693;
  • 3) 0.000 000 002 968 590 693 × 2 = 0 + 0.000 000 005 937 181 386;
  • 4) 0.000 000 005 937 181 386 × 2 = 0 + 0.000 000 011 874 362 772;
  • 5) 0.000 000 011 874 362 772 × 2 = 0 + 0.000 000 023 748 725 544;
  • 6) 0.000 000 023 748 725 544 × 2 = 0 + 0.000 000 047 497 451 088;
  • 7) 0.000 000 047 497 451 088 × 2 = 0 + 0.000 000 094 994 902 176;
  • 8) 0.000 000 094 994 902 176 × 2 = 0 + 0.000 000 189 989 804 352;
  • 9) 0.000 000 189 989 804 352 × 2 = 0 + 0.000 000 379 979 608 704;
  • 10) 0.000 000 379 979 608 704 × 2 = 0 + 0.000 000 759 959 217 408;
  • 11) 0.000 000 759 959 217 408 × 2 = 0 + 0.000 001 519 918 434 816;
  • 12) 0.000 001 519 918 434 816 × 2 = 0 + 0.000 003 039 836 869 632;
  • 13) 0.000 003 039 836 869 632 × 2 = 0 + 0.000 006 079 673 739 264;
  • 14) 0.000 006 079 673 739 264 × 2 = 0 + 0.000 012 159 347 478 528;
  • 15) 0.000 012 159 347 478 528 × 2 = 0 + 0.000 024 318 694 957 056;
  • 16) 0.000 024 318 694 957 056 × 2 = 0 + 0.000 048 637 389 914 112;
  • 17) 0.000 048 637 389 914 112 × 2 = 0 + 0.000 097 274 779 828 224;
  • 18) 0.000 097 274 779 828 224 × 2 = 0 + 0.000 194 549 559 656 448;
  • 19) 0.000 194 549 559 656 448 × 2 = 0 + 0.000 389 099 119 312 896;
  • 20) 0.000 389 099 119 312 896 × 2 = 0 + 0.000 778 198 238 625 792;
  • 21) 0.000 778 198 238 625 792 × 2 = 0 + 0.001 556 396 477 251 584;
  • 22) 0.001 556 396 477 251 584 × 2 = 0 + 0.003 112 792 954 503 168;
  • 23) 0.003 112 792 954 503 168 × 2 = 0 + 0.006 225 585 909 006 336;
  • 24) 0.006 225 585 909 006 336 × 2 = 0 + 0.012 451 171 818 012 672;
  • 25) 0.012 451 171 818 012 672 × 2 = 0 + 0.024 902 343 636 025 344;
  • 26) 0.024 902 343 636 025 344 × 2 = 0 + 0.049 804 687 272 050 688;
  • 27) 0.049 804 687 272 050 688 × 2 = 0 + 0.099 609 374 544 101 376;
  • 28) 0.099 609 374 544 101 376 × 2 = 0 + 0.199 218 749 088 202 752;
  • 29) 0.199 218 749 088 202 752 × 2 = 0 + 0.398 437 498 176 405 504;
  • 30) 0.398 437 498 176 405 504 × 2 = 0 + 0.796 874 996 352 811 008;
  • 31) 0.796 874 996 352 811 008 × 2 = 1 + 0.593 749 992 705 622 016;
  • 32) 0.593 749 992 705 622 016 × 2 = 1 + 0.187 499 985 411 244 032;
  • 33) 0.187 499 985 411 244 032 × 2 = 0 + 0.374 999 970 822 488 064;
  • 34) 0.374 999 970 822 488 064 × 2 = 0 + 0.749 999 941 644 976 128;
  • 35) 0.749 999 941 644 976 128 × 2 = 1 + 0.499 999 883 289 952 256;
  • 36) 0.499 999 883 289 952 256 × 2 = 0 + 0.999 999 766 579 904 512;
  • 37) 0.999 999 766 579 904 512 × 2 = 1 + 0.999 999 533 159 809 024;
  • 38) 0.999 999 533 159 809 024 × 2 = 1 + 0.999 999 066 319 618 048;
  • 39) 0.999 999 066 319 618 048 × 2 = 1 + 0.999 998 132 639 236 096;
  • 40) 0.999 998 132 639 236 096 × 2 = 1 + 0.999 996 265 278 472 192;
  • 41) 0.999 996 265 278 472 192 × 2 = 1 + 0.999 992 530 556 944 384;
  • 42) 0.999 992 530 556 944 384 × 2 = 1 + 0.999 985 061 113 888 768;
  • 43) 0.999 985 061 113 888 768 × 2 = 1 + 0.999 970 122 227 777 536;
  • 44) 0.999 970 122 227 777 536 × 2 = 1 + 0.999 940 244 455 555 072;
  • 45) 0.999 940 244 455 555 072 × 2 = 1 + 0.999 880 488 911 110 144;
  • 46) 0.999 880 488 911 110 144 × 2 = 1 + 0.999 760 977 822 220 288;
  • 47) 0.999 760 977 822 220 288 × 2 = 1 + 0.999 521 955 644 440 576;
  • 48) 0.999 521 955 644 440 576 × 2 = 1 + 0.999 043 911 288 881 152;
  • 49) 0.999 043 911 288 881 152 × 2 = 1 + 0.998 087 822 577 762 304;
  • 50) 0.998 087 822 577 762 304 × 2 = 1 + 0.996 175 645 155 524 608;
  • 51) 0.996 175 645 155 524 608 × 2 = 1 + 0.992 351 290 311 049 216;
  • 52) 0.992 351 290 311 049 216 × 2 = 1 + 0.984 702 580 622 098 432;
  • 53) 0.984 702 580 622 098 432 × 2 = 1 + 0.969 405 161 244 196 864;
  • 54) 0.969 405 161 244 196 864 × 2 = 1 + 0.938 810 322 488 393 728;

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:


0.000 000 000 742 147 673 25(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0010 1111 1111 1111 1111 11(2)

6. Positive number before normalization:

0.000 000 000 742 147 673 25(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0010 1111 1111 1111 1111 11(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:


0.000 000 000 742 147 673 25(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0010 1111 1111 1111 1111 11(2) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0010 1111 1111 1111 1111 11(2) × 20 =

1.1001 0111 1111 1111 1111 111(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 0111 1111 1111 1111 111


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 1011 1111 1111 1111 1111 =

100 1011 1111 1111 1111 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 0000

Mantissa (23 bits) =
100 1011 1111 1111 1111 1111


Decimal number -0.000 000 000 742 147 673 25 converted to 32 bit single precision IEEE 754 binary floating point representation:

1 - 0110 0000 - 100 1011 1111 1111 1111 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