-0.000 000 000 272 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 272(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
-0.000 000 000 272(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:

|-0.000 000 000 272| = 0.000 000 000 272


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: 0.000 000 000 272.

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) 0.000 000 000 272 × 2 = 0 + 0.000 000 000 544;
  • 2) 0.000 000 000 544 × 2 = 0 + 0.000 000 001 088;
  • 3) 0.000 000 001 088 × 2 = 0 + 0.000 000 002 176;
  • 4) 0.000 000 002 176 × 2 = 0 + 0.000 000 004 352;
  • 5) 0.000 000 004 352 × 2 = 0 + 0.000 000 008 704;
  • 6) 0.000 000 008 704 × 2 = 0 + 0.000 000 017 408;
  • 7) 0.000 000 017 408 × 2 = 0 + 0.000 000 034 816;
  • 8) 0.000 000 034 816 × 2 = 0 + 0.000 000 069 632;
  • 9) 0.000 000 069 632 × 2 = 0 + 0.000 000 139 264;
  • 10) 0.000 000 139 264 × 2 = 0 + 0.000 000 278 528;
  • 11) 0.000 000 278 528 × 2 = 0 + 0.000 000 557 056;
  • 12) 0.000 000 557 056 × 2 = 0 + 0.000 001 114 112;
  • 13) 0.000 001 114 112 × 2 = 0 + 0.000 002 228 224;
  • 14) 0.000 002 228 224 × 2 = 0 + 0.000 004 456 448;
  • 15) 0.000 004 456 448 × 2 = 0 + 0.000 008 912 896;
  • 16) 0.000 008 912 896 × 2 = 0 + 0.000 017 825 792;
  • 17) 0.000 017 825 792 × 2 = 0 + 0.000 035 651 584;
  • 18) 0.000 035 651 584 × 2 = 0 + 0.000 071 303 168;
  • 19) 0.000 071 303 168 × 2 = 0 + 0.000 142 606 336;
  • 20) 0.000 142 606 336 × 2 = 0 + 0.000 285 212 672;
  • 21) 0.000 285 212 672 × 2 = 0 + 0.000 570 425 344;
  • 22) 0.000 570 425 344 × 2 = 0 + 0.001 140 850 688;
  • 23) 0.001 140 850 688 × 2 = 0 + 0.002 281 701 376;
  • 24) 0.002 281 701 376 × 2 = 0 + 0.004 563 402 752;
  • 25) 0.004 563 402 752 × 2 = 0 + 0.009 126 805 504;
  • 26) 0.009 126 805 504 × 2 = 0 + 0.018 253 611 008;
  • 27) 0.018 253 611 008 × 2 = 0 + 0.036 507 222 016;
  • 28) 0.036 507 222 016 × 2 = 0 + 0.073 014 444 032;
  • 29) 0.073 014 444 032 × 2 = 0 + 0.146 028 888 064;
  • 30) 0.146 028 888 064 × 2 = 0 + 0.292 057 776 128;
  • 31) 0.292 057 776 128 × 2 = 0 + 0.584 115 552 256;
  • 32) 0.584 115 552 256 × 2 = 1 + 0.168 231 104 512;
  • 33) 0.168 231 104 512 × 2 = 0 + 0.336 462 209 024;
  • 34) 0.336 462 209 024 × 2 = 0 + 0.672 924 418 048;
  • 35) 0.672 924 418 048 × 2 = 1 + 0.345 848 836 096;
  • 36) 0.345 848 836 096 × 2 = 0 + 0.691 697 672 192;
  • 37) 0.691 697 672 192 × 2 = 1 + 0.383 395 344 384;
  • 38) 0.383 395 344 384 × 2 = 0 + 0.766 790 688 768;
  • 39) 0.766 790 688 768 × 2 = 1 + 0.533 581 377 536;
  • 40) 0.533 581 377 536 × 2 = 1 + 0.067 162 755 072;
  • 41) 0.067 162 755 072 × 2 = 0 + 0.134 325 510 144;
  • 42) 0.134 325 510 144 × 2 = 0 + 0.268 651 020 288;
  • 43) 0.268 651 020 288 × 2 = 0 + 0.537 302 040 576;
  • 44) 0.537 302 040 576 × 2 = 1 + 0.074 604 081 152;
  • 45) 0.074 604 081 152 × 2 = 0 + 0.149 208 162 304;
  • 46) 0.149 208 162 304 × 2 = 0 + 0.298 416 324 608;
  • 47) 0.298 416 324 608 × 2 = 0 + 0.596 832 649 216;
  • 48) 0.596 832 649 216 × 2 = 1 + 0.193 665 298 432;
  • 49) 0.193 665 298 432 × 2 = 0 + 0.387 330 596 864;
  • 50) 0.387 330 596 864 × 2 = 0 + 0.774 661 193 728;
  • 51) 0.774 661 193 728 × 2 = 1 + 0.549 322 387 456;
  • 52) 0.549 322 387 456 × 2 = 1 + 0.098 644 774 912;
  • 53) 0.098 644 774 912 × 2 = 0 + 0.197 289 549 824;
  • 54) 0.197 289 549 824 × 2 = 0 + 0.394 579 099 648;
  • 55) 0.394 579 099 648 × 2 = 0 + 0.789 158 199 296;

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:


0.000 000 000 272(10) =

0.0000 0000 0000 0000 0000 0000 0000 0001 0010 1011 0001 0001 0011 000(2)

6. Positive number before normalization:

0.000 000 000 272(10) =

0.0000 0000 0000 0000 0000 0000 0000 0001 0010 1011 0001 0001 0011 000(2)

7. Normalize the binary representation of the number.

Shift the decimal mark 32 positions to the right, so that only one non zero digit remains to the left of it:


0.000 000 000 272(10) =

0.0000 0000 0000 0000 0000 0000 0000 0001 0010 1011 0001 0001 0011 000(2) =

0.0000 0000 0000 0000 0000 0000 0000 0001 0010 1011 0001 0001 0011 000(2) × 20 =

1.0010 1011 0001 0001 0011 000(2) × 2-32


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): -32

Mantissa (not normalized):
1.0010 1011 0001 0001 0011 000


9. Adjust the exponent.

Use the 8 bit excess/bias notation:


Exponent (adjusted) =

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

-32 + 2(8-1) - 1 =

(-32 + 127)(10) =

95(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;
  • 95 ÷ 2 = 47 + 1;
  • 47 ÷ 2 = 23 + 1;
  • 23 ÷ 2 = 11 + 1;
  • 11 ÷ 2 = 5 + 1;
  • 5 ÷ 2 = 2 + 1;
  • 2 ÷ 2 = 1 + 0;
  • 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) =

95(10) =

0101 1111(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. 001 0101 1000 1000 1001 1000 =

001 0101 1000 1000 1001 1000


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) =
0101 1111

Mantissa (23 bits) =
001 0101 1000 1000 1001 1000


Decimal number -0.000 000 000 272 converted to 32 bit single precision IEEE 754 binary floating point representation:

1 - 0101 1111 - 001 0101 1000 1000 1001 1000

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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