-0.000 000 000 742 147 668 5 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 742 147 668 5(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
-0.000 000 000 742 147 668 5(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:

|-0.000 000 000 742 147 668 5| = 0.000 000 000 742 147 668 5


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: 0.000 000 000 742 147 668 5.

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) 0.000 000 000 742 147 668 5 × 2 = 0 + 0.000 000 001 484 295 337;
  • 2) 0.000 000 001 484 295 337 × 2 = 0 + 0.000 000 002 968 590 674;
  • 3) 0.000 000 002 968 590 674 × 2 = 0 + 0.000 000 005 937 181 348;
  • 4) 0.000 000 005 937 181 348 × 2 = 0 + 0.000 000 011 874 362 696;
  • 5) 0.000 000 011 874 362 696 × 2 = 0 + 0.000 000 023 748 725 392;
  • 6) 0.000 000 023 748 725 392 × 2 = 0 + 0.000 000 047 497 450 784;
  • 7) 0.000 000 047 497 450 784 × 2 = 0 + 0.000 000 094 994 901 568;
  • 8) 0.000 000 094 994 901 568 × 2 = 0 + 0.000 000 189 989 803 136;
  • 9) 0.000 000 189 989 803 136 × 2 = 0 + 0.000 000 379 979 606 272;
  • 10) 0.000 000 379 979 606 272 × 2 = 0 + 0.000 000 759 959 212 544;
  • 11) 0.000 000 759 959 212 544 × 2 = 0 + 0.000 001 519 918 425 088;
  • 12) 0.000 001 519 918 425 088 × 2 = 0 + 0.000 003 039 836 850 176;
  • 13) 0.000 003 039 836 850 176 × 2 = 0 + 0.000 006 079 673 700 352;
  • 14) 0.000 006 079 673 700 352 × 2 = 0 + 0.000 012 159 347 400 704;
  • 15) 0.000 012 159 347 400 704 × 2 = 0 + 0.000 024 318 694 801 408;
  • 16) 0.000 024 318 694 801 408 × 2 = 0 + 0.000 048 637 389 602 816;
  • 17) 0.000 048 637 389 602 816 × 2 = 0 + 0.000 097 274 779 205 632;
  • 18) 0.000 097 274 779 205 632 × 2 = 0 + 0.000 194 549 558 411 264;
  • 19) 0.000 194 549 558 411 264 × 2 = 0 + 0.000 389 099 116 822 528;
  • 20) 0.000 389 099 116 822 528 × 2 = 0 + 0.000 778 198 233 645 056;
  • 21) 0.000 778 198 233 645 056 × 2 = 0 + 0.001 556 396 467 290 112;
  • 22) 0.001 556 396 467 290 112 × 2 = 0 + 0.003 112 792 934 580 224;
  • 23) 0.003 112 792 934 580 224 × 2 = 0 + 0.006 225 585 869 160 448;
  • 24) 0.006 225 585 869 160 448 × 2 = 0 + 0.012 451 171 738 320 896;
  • 25) 0.012 451 171 738 320 896 × 2 = 0 + 0.024 902 343 476 641 792;
  • 26) 0.024 902 343 476 641 792 × 2 = 0 + 0.049 804 686 953 283 584;
  • 27) 0.049 804 686 953 283 584 × 2 = 0 + 0.099 609 373 906 567 168;
  • 28) 0.099 609 373 906 567 168 × 2 = 0 + 0.199 218 747 813 134 336;
  • 29) 0.199 218 747 813 134 336 × 2 = 0 + 0.398 437 495 626 268 672;
  • 30) 0.398 437 495 626 268 672 × 2 = 0 + 0.796 874 991 252 537 344;
  • 31) 0.796 874 991 252 537 344 × 2 = 1 + 0.593 749 982 505 074 688;
  • 32) 0.593 749 982 505 074 688 × 2 = 1 + 0.187 499 965 010 149 376;
  • 33) 0.187 499 965 010 149 376 × 2 = 0 + 0.374 999 930 020 298 752;
  • 34) 0.374 999 930 020 298 752 × 2 = 0 + 0.749 999 860 040 597 504;
  • 35) 0.749 999 860 040 597 504 × 2 = 1 + 0.499 999 720 081 195 008;
  • 36) 0.499 999 720 081 195 008 × 2 = 0 + 0.999 999 440 162 390 016;
  • 37) 0.999 999 440 162 390 016 × 2 = 1 + 0.999 998 880 324 780 032;
  • 38) 0.999 998 880 324 780 032 × 2 = 1 + 0.999 997 760 649 560 064;
  • 39) 0.999 997 760 649 560 064 × 2 = 1 + 0.999 995 521 299 120 128;
  • 40) 0.999 995 521 299 120 128 × 2 = 1 + 0.999 991 042 598 240 256;
  • 41) 0.999 991 042 598 240 256 × 2 = 1 + 0.999 982 085 196 480 512;
  • 42) 0.999 982 085 196 480 512 × 2 = 1 + 0.999 964 170 392 961 024;
  • 43) 0.999 964 170 392 961 024 × 2 = 1 + 0.999 928 340 785 922 048;
  • 44) 0.999 928 340 785 922 048 × 2 = 1 + 0.999 856 681 571 844 096;
  • 45) 0.999 856 681 571 844 096 × 2 = 1 + 0.999 713 363 143 688 192;
  • 46) 0.999 713 363 143 688 192 × 2 = 1 + 0.999 426 726 287 376 384;
  • 47) 0.999 426 726 287 376 384 × 2 = 1 + 0.998 853 452 574 752 768;
  • 48) 0.998 853 452 574 752 768 × 2 = 1 + 0.997 706 905 149 505 536;
  • 49) 0.997 706 905 149 505 536 × 2 = 1 + 0.995 413 810 299 011 072;
  • 50) 0.995 413 810 299 011 072 × 2 = 1 + 0.990 827 620 598 022 144;
  • 51) 0.990 827 620 598 022 144 × 2 = 1 + 0.981 655 241 196 044 288;
  • 52) 0.981 655 241 196 044 288 × 2 = 1 + 0.963 310 482 392 088 576;
  • 53) 0.963 310 482 392 088 576 × 2 = 1 + 0.926 620 964 784 177 152;
  • 54) 0.926 620 964 784 177 152 × 2 = 1 + 0.853 241 929 568 354 304;

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:


0.000 000 000 742 147 668 5(10) =

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

6. Positive number before normalization:

0.000 000 000 742 147 668 5(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:


0.000 000 000 742 147 668 5(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 -0.000 000 000 742 147 668 5 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